The Mind Flayer Paradox

[wriggz]

Hello all,

I have no strategy article today so I put together something to get everyone thinking about the New Mind Flayer

The Mind Flayer Paradox

You have placed all your OM's on the Mind Flayer this round hoping to remove all your opponents Order makers. You look at your Opponents order markers A, B, C, and D all on the 4th Mass.

Suppose You have just successfully Psionic Blasted a 4th mass figure, and get to remove a OM from a choice of Four Order makers. Of course one is the X OM; and the others are OM 1, 2, and 3. Your opponent is going to give you a boost by showing you one of the order marker that is not OM 1, but you must choose which one you will remove first.

You pick an OM, say "A", hoping it is OM 1 (which would rob your opponent of their first turn), and your opponat shows you another OM, say D, which is OM 3. He then says to you, "Do you want to pick OM B or C instead?"

Is it to your advantage to switch your choice?

[That Guy0715]

yes, it is.
When theres three order markers left for you to choose, there is a 33% chance it is in any one of them. Then, when you choose OM A, theres a 33% chance its OM 1, and a 66% chance that OM 1 is in one of the other two OM's. Your opponent is going to show you an order marker that is not OM1, and not the one you've chosen. When he/she gives you the oppertunity to switch, there is still a 33% chance its the one youve already chosen, and a 66% its in one of the other two, which has now been eliminated down to just one.

[TheLorax]

http://en.wikipedia.org/wiki/Monty_Hall_problem

[Aldin]

Is it to your advantage to switch your choice?

No it is not. Though your odds of hitting OM1 are higher, your odds of hitting OMX are higher as well. On your first pick you had a 75% chance to hit a valid OM. The switch drops that to 67%. Reducing your chance of removing a viable OM in order to increase your chance of getting a "better" OM just doesn't make any sense.

~Aldin, running the numbers

[Lord Pyre]

Is it to your advantage to switch your choice?

No it is not. Though your odds of hitting OM1 are higher, your odds of hitting OMX are higher as well. On your first pick you had a 75% chance to hit a valid OM. The switch drops that to 67%. Reducing your chance of removing a viable OM in order to increase your chance of getting a "better" OM just doesn't make any sense.

~Aldin, running the numbers

Actually, Thatguy is right.

yes, it is.
When theres three order markers left for you to choose, there is a 33% chance it is in any one of them. Then, when you choose OM A, theres a 33% chance its OM 1, and a 66% chance that OM 1 is in one of the other two OM's. Your opponent is going to show you an order marker that is not OM1, and not the one you've chosen. When he/she gives you the oppertunity to switch, there is still a 33% chance its the one youve already chosen, and a 66% its in one of the other two, which has now been eliminated down to just one.

This is a classic question that has been around game shows for years! Very clever, wriggs!

[dok]

Aldin is 100% correct. Those of you saying he is wrong are assuming this is identical to the classic problem, in stead of thinking critically about what he wrote and actually doing out the probability.

If your goal is to get the "1", you should switch. That's the classic Monty Hall problem.

If your goal is to NOT get the "X", you should not switch. And since that should be your goal, don't switch. It's a "reverse Monty Hall" problem, if you will.

[Knight of Scape]

Is it to your advantage to switch your choice?

No it is not. Though your odds of hitting OM1 are higher, your odds of hitting OMX are higher as well. On your first pick you had a 75% chance to hit a valid OM. The switch drops that to 67%. Reducing your chance of removing a viable OM in order to increase your chance of getting a "better" OM just doesn't make any sense.

~Aldin, running the numbers
You want OM 1. There's a 25% chance it's the first one you picked, and a 75% chance it's one of the other three. He reveals one of the other three that is not OM 1. That means that the remaining two each have a 37.5% chance of being OM1, while your original choice still has only a 25% chance. BUT the odds of any of them being OM X is only 33.3%. Why you ask? It's because the OM he revealed was not randomly chosen. If he randomly chose the OM, and it wasn't OM 1, there'd be no advantage to switching. But he specifically chose one that wasn't OM1. That isn't true for OM X. I probably did a terrible job of explaining that, and wouldn't be surprised if someone ninja'd me with a better explanation, but I'm pretty sure I'm correct.

[Robotech Master]

Isn't that more of a 'conundrum?'

I always thought Paradox was something that somehow trapped itself in a loop of nonsense-ery.

Like:

The Sentence Below is True.
The Sentence Above is False.

Maybe I'm thinking of something else.

[dok]

Is it to your advantage to switch your choice?

No it is not. Though your odds of hitting OM1 are higher, your odds of hitting OMX are higher as well. On your first pick you had a 75% chance to hit a valid OM. The switch drops that to 67%. Reducing your chance of removing a viable OM in order to increase your chance of getting a "better" OM just doesn't make any sense.

~Aldin, running the numbers
You want OM 1. There's a 25% chance it's the first one you picked, and a 75% chance it's one of the other three. He reveals one of the other three that is not OM 1. That means that the remaining two each have a 37.5% chance of being OM1, while your original choice still has only a 25% chance. BUT the odds of any of them being OM X is only 33.3%. Why you ask? It's because the OM he revealed was not randomly chosen. If he randomly chose the OM, and it wasn't OM 1, there'd be no advantage to switching. But he specifically chose one that wasn't OM1. That isn't true for OM X. I probably did a terrible job of explaining that, and wouldn't be surprised if someone ninja'd me with a better explanation, but I'm pretty sure I'm correct.It's true that if your opponent is willing to reveal the "X", that there's an advantage to switching (assuming you prefer the "1"). However, it seems ridiculous to think your opponent would reveal the "X".

[TheLorax]

This situation is a bit different from the Monty Hall problem, because there is a gradient of "wins," not just a win or lose. Getting the 1 order marker is valued higher than the 2 or 3. However, as Aldin said, gettin a 3 OM is much better than an X, so switching doesn't make sense.

On a side note, I don't know why you would put 3 OMs on a Mind Flayer, unless you have some huge initiative advantage.

[Aldin]

Thanks dok,

Think of it as Monty Hall with three cars and a goat. One car is nicer than the others, but you really, really don't want the goat. Having a car revealed increases your chances of getting the goat if you switch. Stay with the pick you made and you are more likely to have a car - even if it's the BMW instead of the Porsche.

~Aldin, drivingly

[Lord Pyre]

Ah, I get it!
I wasn't thinking that way! :oops:

[Dredd Stev]

...what if you want the goat?

[jschild]

...what if you want the goat?

Then you deserve to lose. :P

[Grungebob]

...what if you want the goat?

Then you deserve to lose. :PDon't listen to him he's just trying to get your goat.

[killercactus]

I've heard this problem (The Monty Hall problem) hashed out a bunch of times, and I've never agreed with its reasoning. I probably never will, no matter how or how often it is explained to me.

I pick #1 out of 3, and you show me #3 is wrong, and ask if I want to switch to #2. I have two options now - #1 and #2. I can only pick #1 or #2. At this point, it's a 50/50 shot because the known facts have changed. There is exactly as much chance for the car to be behind #1 as there is to be behind #2, and my choice is to pick one or the other. In my mind, it makes no difference that switching to #2 is like me picking #2 and #3 from the get-go - #3 no longer matters.

I look at it just like texas hold'em on TV. If I have 4-4 against an opponent's 9-10 suited, I have (very close to) a 50% chance to win the hand. But, those same 2 hands against each other after a 9 comes out on the flop gives me around a 10% chance (assuming I don't pick up a straight or flush draw and no one folded / burned any relevant cards). The percentages change as the facts do.

I'm sure this is a classic example of someone trying to unsuccessfully refute this theory, but there it is.

[TheLorax]

Killer, I've had trouble with this in the past. The key issue, that is often over looked, is what the "host" or revealer knows. If he doesn't know any more than the chooser, then it doesn't matter, switching is no better than not switching.

If the host does know, then your initial chance to correctly pick the car is 1/3. Once he reveals the "goat", again assuming he knows where the "car" is, the odds now shift to 1/2. However, if you do not switch, your odds remain at the original 1/3. So if you change your answer, your odds are now 1/2.

It is worth noting that switching does not guarantee a "win," it just increases your odds.

[Aldin]

I've heard this problem (The Monty Hall problem) hashed out a bunch of times, and I've never agreed with its reasoning. I probably never will, no matter how or how often it is explained to me.

I pick #1 out of 3, and you show me #3 is wrong, and ask if I want to switch to #2. I have two options now - #1 and #2. I can only pick #1 or #2. At this point, it's a 50/50 shot because the known facts have changed. There is exactly as much chance for the car to be behind #1 as there is to be behind #2, and my choice is to pick one or the other. In my mind, it makes no difference that switching to #2 is like me picking #2 and #3 from the get-go - #3 no longer matters.

I look at it just like texas hold'em on TV. If I have 4-4 against an opponent's 9-10 suited, I have (very close to) a 50% chance to win the hand. But, those same 2 hands against each other after a 9 comes out on the flop gives me around a 10% chance (assuming I don't pick up a straight or flush draw and no one folded / burned any relevant cards). The percentages change as the facts do.

I'm sure this is a classic example of someone trying to unsuccessfully refute this theory, but there it is.

Here's the easy proof:

There are three possible door arrangements (T=Goat, C=Car)

A) 1T 2T 3C
B) 1T 2C 3T
C) 1C 2T 3T

Let's say you always pick door 1.

In example A door 2 gets revealed and you switch to door 3 and win.
In example B door 3 gets revelaed and you switch to door 2 and win.
Only in example C do you switch from the car to the goat.

Thus, two thirds of the time switching is the better option. It works because the revelation is not random and cannot consider whatever you originally chose. By introducing a measure that eliminates the only bad option 2/3rds of the time, switching becomes the best choice.

~Aldin, who hashed this out with his brother in law a couple of years ago

[killercactus]

Thanks, Aldin. I get all that.

But what I contend is that the original choice makes absolutely no difference IF one of the wrong doors is ALWAYS going to be revealed. I didn't make that clear before.

If the game is such that one of the wrong results will always be revealed to you, then your odds are 1/2 no matter which door you pick, and no matter whether or not you switch, because this is really a game of picking between 2 choices, not 3.

If one of the wrong answers will not always be revealed but one of them is revealed, then I agree that your odds go from 1/3 to 1/2 by switching.

[Rich10]

I've heard this problem (The Monty Hall problem) hashed out a bunch of times, and I've never agreed with its reasoning. I probably never will, no matter how or how often it is explained to me.

I pick #1 out of 3, and you show me #3 is wrong, and ask if I want to switch to #2. I have two options now - #1 and #2. I can only pick #1 or #2. At this point, it's a 50/50 shot because the known facts have changed. There is exactly as much chance for the car to be behind #1 as there is to be behind #2, and my choice is to pick one or the other. In my mind, it makes no difference that switching to #2 is like me picking #2 and #3 from the get-go - #3 no longer matters.

I look at it just like texas hold'em on TV. If I have 4-4 against an opponent's 9-10 suited, I have (very close to) a 50% chance to win the hand. But, those same 2 hands against each other after a 9 comes out on the flop gives me around a 10% chance (assuming I don't pick up a straight or flush draw and no one folded / burned any relevant cards). The percentages change as the facts do.

I'm sure this is a classic example of someone trying to unsuccessfully refute this theory, but there it is.I don't know if you watched "Lets make a deal" when it was on TV, but there is a trick here. The trick is that the host would only show a bad prize to the contestant.

Think of it this way, I put down 3 cards, 2 red and 1 black. You want to pick the black card and you pick the first one. If I showed you the third card (recognizing that I would always show you a red card), do you benefit from picking the second card? In other words, if the third card were black, I wouldn't show you that card. As a result, there is no benefit to switching cards.

This is different from your example in which a random card is shown which does move the odds.

[dok]

killercactus, Aldin's explanation proves the correctness of switching as clearly as anything can.

However, if you want a qualitative explanation why that works, here it is: the host will never reveal that the door you already picked is the one with the goat.

Think of it this way: let's say you pick door one. Then the host says, "would you prefer door one, or would you prefer the better outcome between door two and door three"? In that case, obviously you would pick the latter. Well, that's exactly what he's doing when he reveals a door. If he shows you the goat behind door two, he's telling you that door three is the better outcome between doors two and three.

You're not picking door two. You're picking doors two and three.

Similar classic problem: let's say I pick a random family with two children. If one of the children is a boy, I will tell you that. If I tell you one child is a boy, what is the probability that the other child is a boy?

Answer:
The probability is one in three. 25% of two-child populations are boy-boy, and 50% are girl-boy. Bayes' law FTW.

[killercactus]

Think of it this way: let's say you pick door one. Then the host says, "would you prefer door one, or would you prefer the better outcome between door two and door three"? In that case, obviously you would pick the latter. Well, that's exactly what he's doing when he reveals a door. If he shows you the goat behind door two, he's telling you that door three is the better outcome between doors two and three.

See, I don't understand how that situation isn't a coin flip either.

You are offering me a choice between #1 and #2 + #3. But, because I know that either #2 or #3 is worth nothing, it's a 50/50 shot that I pick the correct one, regardless of which choice I make.


Similar classic problem: let's say I pick a random family with two children. If one of the children is a boy, I will tell you that. If I tell you one child is a boy, what is the probability that the other child is a boy?

50% (even though I knew the "answer" to this problem is 1/3 - I can't wrap my head around it because knowledge changes the facts).

[wriggz]

Lets say I have 100 OM instead of 4. Now there is a 1 in 100 that you will get the dreaded X. If I let you Pick 1 and then reveal 98 that are not X leaving with the one you picked and the one that's left that maybe the X.

Now the Odds you picked the X is 1 in 100, the oddes I left the X is 50/50 since I eliminated all the others. It would be foolish to switch since there as a good chance the X was in the 99 you did not pick, and now i have revealed all but one.

[dok]

Think of it this way: let's say you pick door one. Then the host says, "would you prefer door one, or would you prefer the better outcome between door two and door three"? In that case, obviously you would pick the latter. Well, that's exactly what he's doing when he reveals a door. If he shows you the goat behind door two, he's telling you that door three is the better outcome between doors two and three.

See, I don't understand how that situation isn't a coin flip either.

You are offering me a choice between #1 and #2 + #3. But, because I know that either #2 or #3 is worth nothing, it's a 50/50 shot that I pick the correct one, regardless of which choice I make.OK, then that's the disconnect, right there.

Let's say that there's not three doors, but one thousand. Behind one of them is a car, and behind 999 of them is nothing.

If you had a choice between getting what's behind door number one, or getting your pick of what's behind doors 2-1000, you wouldn't pick the latter?

ETA: or make it poker. Say it's my set of Kings against your set of deuces with one card to come. You have a choice between taking the next card off the top, or setting that card aside and picking any other card left in the deck. You wouldn't pick the latter?

IMO, boy-girl and girl-boy are the same outcome, which means there are only 2 outcomes to choose from - boy-boy and boy-girl Then how do you explain how there are twice as many boy/girl families in the general population as there are boy/boy families?

[TheLorax]

See, I don't understand how that situation isn't a coin flip either.

You are offering me a choice between #1 and #2 + #3. But, because I know that either #2 or #3 is worth nothing, it's a 50/50 shot that I pick the correct one, regardless of which choice I make.



You don't know which of the 3 is worth nothing until the reveal. The initial choice is a 1 in 3 shot. You only know which of the remaining three choices is worth nothing until after the reveal. It's not #1 and #2 + #3, its #1, #2, #3. After the reveal, it becomes a coin toss, but only if you switch your choice, otherwise you are still at the 1 in 3 odds, because your initial choice was based on old information.

[dok]

See, I don't understand how that situation isn't a coin flip either.

You are offering me a choice between #1 and #2 + #3. But, because I know that either #2 or #3 is worth nothing, it's a 50/50 shot that I pick the correct one, regardless of which choice I make.



You don't know which of the 3 is worth nothing until the reveal. The initial choice is a 1 in 3 shot. You only know which of the remaining three choices is worth nothing until after the reveal. It's not #1 and #2 + #3, its #1, #2, #3. After the reveal, it becomes a coin toss, but only if you switch your choice, otherwise you are still at the 1 in 3 odds, because your initial choice was based on old information.
That's wrong, Lorax. It is #2 + #3. If you switch it is not 1/2, it is 2/3. Again, just look at what Aldin wrote. The idea that switching could make it 1/2 where not switching would leave it at 1/3 is completely illogical - the prizes are not re-scrambled after you switch.

[wriggz]

Yet I wonder does this translate to game play.

1. You are trying to avoid the X.
2. Each round your opponent will reveal something other than the X.

So

I. During your first round you pick D to be the X. You have a 1 in 4 of hitting the X, thus you remove B which is the second ordermarker.

II. Your opponet reveals OM1 in the A position.

III. Durring the Second Round you only have the 3 and the X left. You Know that OM1 and OM2 have be removed.

Do you continue to believe that OMX is in the "D" position or do you switch base on the fact that the "C" OM has a 50% instead of the original 25%?

[dok]

Yet I wonder does this translate to game play.

1. You are trying to avoid the X.
2. Each round your opponent will reveal something other than the X.

So

I. During your first round you pick D to be the X. You have a 1 in 4 of hitting the X, thus you remove B which is the second ordermarker.

II. Durring the Second Round you only have the 3 and the X left. Knowing that OM1 and OM2 have be removed. Do you continue to believe that OMX is in the "D" position or do you switch base on the fact that the "C" OM has a 50% instead of the original 25%.
Nope, it doesn't translate at all. The reason is that if 1 or 2 were in position "D", your opponent would have revealed it. So unlike the Monty Hall problem, no information is given with each reveal.

The analogy to the Monty Hall problem would be if Monty randomly chose any door without the car to reveal - including the one you already picked. In that case, you should switch doors if he reveals the door you picked (obviously); otherwise it doesn't matter if you switch or not.

[killercactus]

Think of it this way: let's say you pick door one. Then the host says, "would you prefer door one, or would you prefer the better outcome between door two and door three"? In that case, obviously you would pick the latter. Well, that's exactly what he's doing when he reveals a door. If he shows you the goat behind door two, he's telling you that door three is the better outcome between doors two and three.

See, I don't understand how that situation isn't a coin flip either.

You are offering me a choice between #1 and #2 + #3. But, because I know that either #2 or #3 is worth nothing, it's a 50/50 shot that I pick the correct one, regardless of which choice I make.OK, then that's the disconnect, right there.

Let's say that there's not three doors, but one thousand. Behind one of them is a car, and behind 999 of them is nothing.

If you had a choice between getting what's behind door number one, or getting your pick of what's behind doors 2-1000, you wouldn't pick the latter?

ETA: or make it poker. Say it's my set of Kings against your set of deuces with one card to come. You have a choice between taking the next card off the top, or setting that card aside and picking any other card left in the deck. You wouldn't pick the latter?

I don't care if there are a million doors and you offer me a choice between #1 and the other 999,999. There is a qualitative factor to consider in this context, because I know that only 1 out of the 999,999 are good, and I have a choice between 2 alternatives, not 1 million. And, in the Monty Hall problem, if the host always is going to show me 999,998 of the wrong ones before I make my final decision, then it was really only ever a choice of 1 or 2.

What if you offered me a choice between either the 1 or the 999,999 and I pick the 999,999, and then you show me that 999,998 of the ones I picked are wrong and ask if I want to switch - what are my odds then? And how is that different from the first case?

I feel like I'm playing Deal or No Deal (which is a classic example of this, by the way).

IMO, boy-girl and girl-boy are the same outcome, which means there are only 2 outcomes to choose from - boy-boy and boy-girl Then how do you explain how there are twice as many boy/girl families in the general population as there are boy/boy families?

Because I know that girl/girl exists, and it makes up the difference.

That's a different question. In the first situation, I already knew there was a boy. That eliminates girl-girl from the equation, and means all that's left are boy-boy and boy-girl.



See, I don't understand how that situation isn't a coin flip either.

You are offering me a choice between #1 and #2 + #3. But, because I know that either #2 or #3 is worth nothing, it's a 50/50 shot that I pick the correct one, regardless of which choice I make.



You don't know which of the 3 is worth nothing until the reveal. The initial choice is a 1 in 3 shot. You only know which of the remaining three choices is worth nothing until after the reveal. It's not #1 and #2 + #3, its #1, #2, #3. After the reveal, it becomes a coin toss, but only if you switch your choice, otherwise you are still at the 1 in 3 odds, because your initial choice was based on old information.

dok was specifically giving me a choice between #1 and #2 + #3, which was (sort of, but not really) different than the original problem.

[TheLorax]

I was trying to explain it in a more qualitative way. I understand that the prizes aren't getting scrambled. I was trying to focus on the way new information changes the odds, but only if the host is required to reveal a wrong answer.

The situation doesn't change after a reveal (what is behind which door), but what you know about it does (that the car is not behind a door you didn't pick).

[Aldin]

Yet I wonder does this translate to game play.

1. You are trying to avoid the X.
2. Each round your opponent will reveal something other than the X.

So

I. During your first round you pick D to be the X. You have a 1 in 4 of hitting the X, thus you remove B which is the second ordermarker.

II. Your opponet reveals OM1 in the A position.

III. Durring the Second Round you only have the 3 and the X left. You Know that OM1 and OM2 have be removed.

Do you continue to believe that OMX is in the "D" position or do you switch base on the fact that the "C" OM has a 50% instead of the original 25%.

Doesn't work because "D" is just as subject to being revealed as "A" or "C". Also because no bad options are being removed, only good ones.

~Aldin, odds shifitingly

[Aldin]

Hmmm, superninja'd but want to go back to kc with real world terms.

I have five envelopes and tell you that one of them has a $20 in it and the other four are empty. I tell you that you can take anywhere from one to four envelopes leaving me with the rest. How many envelopes would you take?

~Aldin, hashingly

[killercactus]

Hmmm, superninja'd but want to go back to kc with real world terms.

I have five envelopes and tell you that one of them has a $20 in it and the other four are empty. I tell you that you can take anywhere from one to four envelopes leaving me with the rest. How many envelopes would you take?

~Aldin, hashingly

That depends - are you going to show me 3 empty ones and offer me another choice after I originally choose?

[dok]

I don't care if there are a million doors and you offer me a choice between #1 and the other 999,999. There is a qualitative factor to consider in this context, because I know that only 1 out of the 999,999 are good, and I have a choice between 2 alternatives, not 1 million.At this point, I think you are letting your fixation on the problem cloud your basic intuition about how things work. Aldin's example is the same as this.

How about this example: say there is a raffle. We know one number will be drawn at the end of the day. Say you have a choice between one ticket, or the rest of the roll. You think the probability is even? After all, only one is going to win...

Say there's one black jellybean in a jar with 500 jellybeans. You get to reach in blindfolded and grab one, and I get the rest of the jar. You think we have an even chance of getting the black jellybean? After all, there's only one black jellybean...


What if you offered me a choice between either the 1 or the 999,999 and I pick the 999,999, and then you show me that 999,998 of the ones I picked are wrong and ask if I want to switch - what are my odds then? And how is that different from the first case?It's not different. You have a 99.9999% chance of having the prize if you don't switch.

I feel like I'm playing Deal or No Deal (which is a classic example of this, by the way).No, no, no, no, no it's not. The "do you want to swich cases" bull-hooey at the end changes nothing. At no point is any information being revealed to the player.

IMO, boy-girl and girl-boy are the same outcome, which means there are only 2 outcomes to choose from - boy-boy and boy-girl Then how do you explain how there are twice as many boy/girl families in the general population as there are boy/boy families?

Because I know that girl/girl exists, and it makes up the difference.

That's a different question. In the first situation, I already knew there was a boy. That eliminates girl-girl from the equation, and means all that's left are boy-boy and boy-girl.Right, and you know that boy/girl is twice as common as boy/boy in the general population. They're not equally common. So picking the more common one is more likely to be right.

[dok]

I was trying to explain it in a more qualitative way. I understand that the prizes aren't getting scrambled. I was trying to focus on the way new information changes the odds, but only if the host is required to reveal a wrong answer.

The situation doesn't change after a reveal (what is behind which door), but what you know about it does (that the car is not behind a door you didn't pick).But you actually got the probabilities wrong. It's 2/3 if you switch, and 1/3 if you do not. It's not 1/2 if you switch and 1/3 if you do not. That's impossible - for starters, it doesn't add to 1, as a probability distribution must.

[dok]

Hmmm, superninja'd but want to go back to kc with real world terms.

I have five envelopes and tell you that one of them has a $20 in it and the other four are empty. I tell you that you can take anywhere from one to four envelopes leaving me with the rest. How many envelopes would you take?

~Aldin, hashingly

That depends - are you going to show me 3 empty ones and offer me another choice after I originally choose?No, he's just giving you a choice of taking four envelopes or one.

...and the two versions are the same, but let's just establish a baseline here.

[padlock]

Thanks, Aldin. I get all that.

But what I contend is that the original choice makes absolutely no difference IF one of the wrong doors is ALWAYS going to be revealed. I didn't make that clear before.

If the game is such that one of the wrong results will always be revealed to you, then your odds are 1/2 no matter which door you pick, and no matter whether or not you switch, because this is really a game of picking between 2 choices, not 3.

If one of the wrong answers will not always be revealed but one of them is revealed, then I agree that your odds go from 1/3 to 1/2 by switching.

The bolded part is incorrect. The fact that one of the wrong answers is revealed does not suddenly change the odds of your initial guess, you've still only got a 1/3 chance of getting the car.

That's because the person revealing the answer knows that they're always revealing an incorrect answre. If a completely random answer is revealed (which might actually actually have been the car) but is in fact a goat, then you're absolutely correct that switching will not matter. You would now have a 1/2 chance of being correct either way.

[TheLorax]

I was trying to explain it in a more qualitative way. I understand that the prizes aren't getting scrambled. I was trying to focus on the way new information changes the odds, but only if the host is required to reveal a wrong answer.

The situation doesn't change after a reveal (what is behind which door), but what you know about it does (that the car is not behind a door you didn't pick).But you actually got the probabilities wrong. It's 2/3 if you switch, and 1/3 if you do not. It's not 1/2 if you switch and 1/3 if you do not. That's impossible - for starters, it doesn't add to 1, as a probability distribution must.


Again, I was trying to break it down into two different scenarios. The initial game is 1/3, and then 2/3s after a reveal, if a goat is revealed.

I'm trying to break it down into two different games to better explain it. The first phase of the problem is a choice of 1 in 3. That is game one. The second part of the game can be expressed as a choice between two doors, because one of the choices has been removed as a possibility (it has been revealed with a goat).

So if you stay with the choice you made in the first game, your chance remains at 1 in 3, because you are still playing the first game. If you change your choice, after the reveal, you are essentially switching the game to a 1 in 2 chance. I would prefer to play a game with a 50% chance at winning rather than a 33.3%

Again, this goes back to what I was saying about knowing the host has to reveal a wrong answer. If his reveal choice is just as random as your original one, then the game never changes, and your first choice is just as likely to win as the second (this also means the host might reveal a car).

The 1/3, 2/3, 2/3 scenario best expresses the game as a whole, but some people have trouble understanding it that way.

[lefton4ya]

Aldin is 100% correct. Those of you saying he is wrong are assuming this is identical to the classic problem, in stead of thinking critically about what he wrote and actually doing out the probability.

If your goal is to get the "1", you should switch. That's the classic Monty Hall problem.

If your goal is to NOT get the "X", you should not switch. And since that should be your goal, don't switch. It's a "reverse Monty Hall" problem, if you will.

Every post on this thread agrees with the above conclusion. Except not getting the X should not always be the goal, but you must add Heroscape situations to debate the value of a chance of X over a chance of 1 IN A GAME. Each game may be different, so value of trying to get a 1 over the chance of getting an X may be different and therefore not solvable in one simple equation or statement without taking Heroscape mechanics into play. Most of the time I would not switch unless if if someone got to use their "1" turn they would assuredly win a game or kill a larger point hero or character that I have a lot of order marks on this round, so I would risk going after the "1" since a "2", "3" or "x" would be moot anyway.

Interesting discussion still - thanks Wriggs and everyone who contributed.

[dok]

So if you stay with the choice you made in the first game, your chance remains at 1 in 3, because you are still playing the first game. If you change your choice, after the reveal, you are essentially switching the game to a 1 in 2 chance. I would prefer to play a game with a 50% chance at winning rather than a 33.3%See, this is simply wrong, which is why I keep telling you that you are wrong. You are not switching to a 1 in 2 chance. You are switching to a 2 in 3 chance.

[TheLorax]

So if you stay with the choice you made in the first game, your chance remains at 1 in 3, because you are still playing the first game. If you change your choice, after the reveal, you are essentially switching the game to a 1 in 2 chance. I would prefer to play a game with a 50% chance at winning rather than a 33.3%See, this is simply wrong, which is why I keep telling you that you are wrong. You are not switching to a 1 in 2 chance. You are switching to a 2 in 3 chance.


No, its not.

I have two doors. One has a car behind it, and one doesn't. What is the likelihood that someone will pick a door with a car behind it?

[killercactus]

How about this example: say there is a raffle. We know one number will be drawn at the end of the day. Say you have a choice between one ticket, or the rest of the roll. You think the probability is even? After all, only one is going to win...

Say there's one black jellybean in a jar with 500 jellybeans. You get to reach in blindfolded and grab one, and I get the rest of the jar. You think we have an even chance of getting the black jellybean? After all, there's only one black jellybean...

These two kind of hammer it home for me, but I still contend that, in the end, it's a choice between 2 tickets, or 2 jelly beans, but the educated decision would be to switch since the person doing the revealing knew they were revealing incorrect ones.


I feel like I'm playing Deal or No Deal (which is a classic example of this, by the way).No, no, no, no, no it's not. The "do you want to swich cases" bull-hooey at the end changes nothing. At no point is any information being revealed to the player.

You sure? If I pick a case in Deal or No Deal and, by the end of the game all that's left is the $1,000,000 and $.01, you're saying I shouldn't switch? Did I not have a 1/26 chance to pick the $1,000,000 at the beginning, and a 25/26 chance not to pick it? Why don't I now have a 25/26 chance of winning the $1,000,000 by switching? Because the cases revealed weren't knowingly not the $1,000,000?

IMO, boy-girl and girl-boy are the same outcome, which means there are only 2 outcomes to choose from - boy-boy and boy-girl Then how do you explain how there are twice as many boy/girl families in the general population as there are boy/boy families?

Because I know that girl/girl exists, and it makes up the difference.

That's a different question. In the first situation, I already knew there was a boy. That eliminates girl-girl from the equation, and means all that's left are boy-boy and boy-girl.Right, and you know that boy/girl is twice as common as boy/boy in the general population. They're not equally common. So picking the more common one is more likely to be right.

But we aren't dealing with a general population - we're dealing with a specific instance of which I already know half of the answer. Given boy, the probability of boy-boy is 50%, not 25% or 33 1/3%.

My wife and I have a son. Are you telling me that if my wife gets pregnant again (hopefully not to be due at GenCon time), that she has a 2/3 chance of having a girl, just because that's what's generally happened? She'd be very excited to hear that....

[wriggz]

Aldin is 100% correct. Those of you saying he is wrong are assuming this is identical to the classic problem, in stead of thinking critically about what he wrote and actually doing out the probability.

If your goal is to get the "1", you should switch. That's the classic Monty Hall problem.

If your goal is to NOT get the "X", you should not switch. And since that should be your goal, don't switch. It's a "reverse Monty Hall" problem, if you will.

Every post on this thread agrees with the above conclusion. Except not getting the X should not always be the goal, but you must add Heroscape situations to debate the value of a chance of X over a chance of 1 IN A GAME. Each game may be different, so value of trying to get a 1 over the chance of getting an X may be different and therefore not solvable in one simple equation or statement without taking Heroscape mechanics into play. Most of the time I would not switch unless if someone got their "1" turn they would assuredly win a game or kill a larger point hero or character that I have a lot of order marks on this round.

Interesting discussion still - thanks Wriggs and everyone who contributed.


I like this point. The Mind Flayer could take more OM's off, so going for the 1 is the best plan, since you may have a chance to removed 2 and 3 (thus totally negating your opponents turn). A chance at OM 1 might be better then the sure OM 3.

So I think this means we need a filler unit that Shows OM's instead of removing them to help reveal plans and make this entire thread valuable.

[killercactus]

So if you stay with the choice you made in the first game, your chance remains at 1 in 3, because you are still playing the first game. If you change your choice, after the reveal, you are essentially switching the game to a 1 in 2 chance. I would prefer to play a game with a 50% chance at winning rather than a 33.3%See, this is simply wrong, which is why I keep telling you that you are wrong. You are not switching to a 1 in 2 chance. You are switching to a 2 in 3 chance.


No, its not.

I have two doors. One has a car behind it, and one doesn't. What is the likelihood that someone will pick a door with a car behind it?

Actually, I'm falling in line with dok on this one.

Let's take the choice out of it. Say I have 7 cups and 1 ball. I hide the ball under one of the cups and don't show you which one it is. Assuming I don't take the ball away:

1) What is the probability that there is a ball under one of the 7 cups?

Next, I pick up 6 of the cups - none of which have a ball under them.

2) What is the probability that there is a ball under the last cup?

The answer to both of these is 100%, and what I'm trying to illustrate is that the probability doesn't change when someone knowingly removes a wrong outcome. I think it's the same principle as this thread....

OK - you guys got me. I was wrong.

[padlock]

So if you stay with the choice you made in the first game, your chance remains at 1 in 3, because you are still playing the first game. If you change your choice, after the reveal, you are essentially switching the game to a 1 in 2 chance. I would prefer to play a game with a 50% chance at winning rather than a 33.3%See, this is simply wrong, which is why I keep telling you that you are wrong. You are not switching to a 1 in 2 chance. You are switching to a 2 in 3 chance.

edit: never mind. I recant my previous statement :)

[dok]

How about this example: say there is a raffle. We know one number will be drawn at the end of the day. Say you have a choice between one ticket, or the rest of the roll. You think the probability is even? After all, only one is going to win...

Say there's one black jellybean in a jar with 500 jellybeans. You get to reach in blindfolded and grab one, and I get the rest of the jar. You think we have an even chance of getting the black jellybean? After all, there's only one black jellybean...

These two kind of hammer it home for me, but I still contend that, in the end, it's a choice between 2 tickets, or 2 jelly beans, but the educated decision would be to switch since the person doing the revealing knew they were revealing incorrect ones.:excited: YES, THAT IS RIGHT.

I feel like I'm playing Deal or No Deal (which is a classic example of this, by the way).No, no, no, no, no it's not. The "do you want to swich cases" bull-hooey at the end changes nothing. At no point is any information being revealed to the player.

You sure? If I pick a case in Deal or No Deal and, by the end of the game all that's left is the $1,000,000 and $.01, you're saying I shouldn't switch? Did I not have a 1/26 chance to pick the $1,000,000 at the beginning, and a 25/26 chance not to pick it? Why don't I now have a 25/26 chance of winning the $1,000,000 by switching? Because the cases revealed weren't knowingly not the $1,000,000?:excited:

YES, THAT IS PRECISELY THE REASON.

IMO, boy-girl and girl-boy are the same outcome, which means there are only 2 outcomes to choose from - boy-boy and boy-girl Then how do you explain how there are twice as many boy/girl families in the general population as there are boy/boy families?

Because I know that girl/girl exists, and it makes up the difference.

That's a different question. In the first situation, I already knew there was a boy. That eliminates girl-girl from the equation, and means all that's left are boy-boy and boy-girl.Right, and you know that boy/girl is twice as common as boy/boy in the general population. They're not equally common. So picking the more common one is more likely to be right.

But we aren't dealing with a general population - we're dealing with a specific instance of which I already know half of the answer. Given boy, the probability of boy-boy is 50%, not 25% or 33 1/3%.

My wife and I have a son. Are you telling me that if my wife gets pregnant again (hopefully not to be due at GenCon time), that she has a 2/3 chance of having a girl, just because that's what's generally happened? She'd be very excited to hear that....No, because if we know the order, then we're just looking at one child - the second. You're really just asking the chance of a given child being a boy, and that's obviously 50%.

As I presented the problem, when I tell you "one child is a boy", I'm not looking at the firstborn child and seeing if that child is a boy. I'm looking at both children and seeing if either child is a boy. It's much like Monty Hall looking behind both doors and seeing if either has the goat.

[Aldin]

So if you stay with the choice you made in the first game, your chance remains at 1 in 3, because you are still playing the first game. If you change your choice, after the reveal, you are essentially switching the game to a 1 in 2 chance. I would prefer to play a game with a 50% chance at winning rather than a 33.3%See, this is simply wrong, which is why I keep telling you that you are wrong. You are not switching to a 1 in 2 chance. You are switching to a 2 in 3 chance.

No, TheLorax is correct. Switching after the first reveal gives you a 1/2 chance of winning the car. I'm not sure why you would think otherwise.

Because you are getting two out of the first three doors. When you pick your door you pick it knowing that your odds are one in three. You know that the odds that the winning door is one of the other two is two in three. The opportunity to switch essentially gives you both those doors - even though a "bad one" has been removed, it doesn't even out the odds and you're still at 66.7% to win with the switch just like you would have been taking BOTH doors in the first place.

~Aldin, summing up

[padlock]

So if you stay with the choice you made in the first game, your chance remains at 1 in 3, because you are still playing the first game. If you change your choice, after the reveal, you are essentially switching the game to a 1 in 2 chance. I would prefer to play a game with a 50% chance at winning rather than a 33.3%See, this is simply wrong, which is why I keep telling you that you are wrong. You are not switching to a 1 in 2 chance. You are switching to a 2 in 3 chance.

No, TheLorax is correct. Switching after the first reveal gives you a 1/2 chance of winning the car. I'm not sure why you would think otherwise.

Because you are getting two out of the first three doors. When you pick your door you pick it knowing that your odds are one in three. You know that the odds that the winning door is one of the other two is two in three. The opportunity to switch essentially gives you both those doors - even though a "bad one" has been removed, it doesn't even out the odds and you're still at 66.7% to win with the switch just like you would have been taking BOTH doors in the first place.

~Aldin, summing up

After thinking about it some more I see that dok, and you, are absolutely right. I was incorrect.

[TheLorax]

You guys aren't understanding what I'm saying.

I'm not even addressing the probability of the initial game or choices. I'm saying the game itself changes, but only if two things are true:

1) The host must reveal a goat

2) You change your answer after the reveal

The likelihood of winning if you change your choice, relative to your first choice, if both those are true are 2/3, just as everyone has said.

What I'm saying is that it can be broken down into two games, a game where you have a 1/3 chance of winning if you don't change your answer, or a game with a 1/2 chance of winning if you do change your answer. These probabilities are isolated, not relative to each other. You don't have a 50% chance of winning relative to your first choice, because there is no first choice. Again, it is asking someone if they want to play a game with a 1/3 chance of winning, or a 1/2 chance of winning.

[killercactus]

OK - I'll submit to all of that other stuff, but now the boy/girl thing has me hung up.


IMO, boy-girl and girl-boy are the same outcome, which means there are only 2 outcomes to choose from - boy-boy and boy-girl Then how do you explain how there are twice as many boy/girl families in the general population as there are boy/boy families?

Because I know that girl/girl exists, and it makes up the difference.

That's a different question. In the first situation, I already knew there was a boy. That eliminates girl-girl from the equation, and means all that's left are boy-boy and boy-girl.Right, and you know that boy/girl is twice as common as boy/boy in the general population. They're not equally common. So picking the more common one is more likely to be right.

But we aren't dealing with a general population - we're dealing with a specific instance of which I already know half of the answer. Given boy, the probability of boy-boy is 50%, not 25% or 33 1/3%.

My wife and I have a son. Are you telling me that if my wife gets pregnant again (hopefully not to be due at GenCon time), that she has a 2/3 chance of having a girl, just because that's what's generally happened? She'd be very excited to hear that....No, because if we know the order, then we're just looking at one child - the second. You're really just asking the chance of a given child being a boy, and that's obviously 50%.

As I presented the problem, when I tell you "one child is a boy", I'm not looking at the firstborn child and seeing if that child is a boy. I'm looking at both children and seeing if either child is a boy. It's much like Monty Hall looking behind both doors and seeing if either has the goat.

I think this situation is different than the Monty Hall problem.

I don't care about the order of the children in this problem, nor do I care about the general population. Telling me that one child is a boy and asking me to predict the second is not the same as asking me how many combinations of two children are there that include at least one boy.

I underlined that statement above - if we know that one child is a boy (regardless of which one it is), aren't we just looking at the child that we don't know the sex of?

[dok]

You guys aren't understanding what I'm saying.

I'm not even addressing the probability of the initial game or choices. I'm saying the game itself changes, but only if two things are true:

1) The host must reveal a goat

2) You change your answer after the reveal

The likelihood of winning if you change your choice, relative to your first choice, if both those are true are 2/3, just as everyone has said.

What I'm saying is that it can be broken down into two games, a game where you have a 1/3 chance of winning if you don't change your answer, or a game with a 1/2 chance of winning if you do change your answer. These probabilities are isolated, not relative to each other. You don't have a 50% chance of winning relative to your first choice, because there is no first choice. Again, it is asking someone if they want to play a game with a 1/3 chance of winning, or a 1/2 chance of winning.
You don't have a 1/2 chance of winning if you change. You have a 2/3 chance of winning if you change. Again, Aldin broke it down (http://www.heroscapers.com/community/showpost.php?p=1124536&postcount=18). It's not 2/3 "relative to your first choice". It's just a 2/3 chance, full stop.

[killercactus]

What I'm saying is that it can be broken down into two games, a game where you have a 1/3 chance of winning if you don't change your answer, or a game with a 1/2 chance of winning if you do change your answer. These probabilities are isolated, not relative to each other. You don't have a 50% chance of winning relative to your first choice, because there is no first choice. Again, it is asking someone if they want to play a game with a 1/3 chance of winning, or a 1/2 chance of winning.

That's how I was thinking earlier. But it isn't really broken down into 2 games. The fact that the host knows he will show you the worst of the 2 doors you didn't pick doesn't change the fact that there were still 2/3 of the doors that you didn't pick.

[Aldin]

You guys aren't understanding what I'm saying.

I'm not even addressing the probability of the initial game or choices. I'm saying the game itself changes, but only if two things are true:

1) The host must reveal a goat

2) You change your answer after the reveal

The likelihood of winning if you change your choice, relative to your first choice, if both those are true are 2/3, just as everyone has said.

What I'm saying is that it can be broken down into two games, a game where you have a 1/3 chance of winning if you don't change your answer, or a game with a 1/2 chance of winning if you do change your answer. These probabilities are isolated, not relative to each other. You don't have a 50% chance of winning relative to your first choice, because there is no first choice. Again, it is asking someone if they want to play a game with a 1/3 chance of winning, or a 1/2 chance of winning.

If I understand you correctly you are saying "let's take the existing scenario, but now the chooser is FORCED to switch which makes it 50/50". That is also not correct because they will be forced to switch to what is effectively the other TWO choices, doubling their chance of winning.

The other thing I think you might have meant is "let's take the existing scenario, but the chooser doesn't get to choose until after one of the choices is eliminated". Then it would be 50/50 because there never was an option to pick one of three things - only the illusion of that option. It is the same effect as the host saying here are three doors but you can only pick door one or door two - 50/50.

~Aldin, dividedly

[padlock]

There are 4 equally likely possiblities if you have 2 kids

Boy, Boy
Girl, Girl
Boy, Girl
Girl, Boy

If I tell that I have 2 children and 1 child is a boy, that removes the girl, girl possibility. That means that there's a 33% chance it's once of the other combinations.

Given that 2 of the 3 remanining possibilities include a girl, and I have revealed that one of my children is a boy, you can deduce that there's a 66% chance that my other child is a girl.

[dok]

OK - I'll submit to all of that other stuff, but now the boy/girl thing has me hung up.


IMO, boy-girl and girl-boy are the same outcome, which means there are only 2 outcomes to choose from - boy-boy and boy-girl Then how do you explain how there are twice as many boy/girl families in the general population as there are boy/boy families?

Because I know that girl/girl exists, and it makes up the difference.

That's a different question. In the first situation, I already knew there was a boy. That eliminates girl-girl from the equation, and means all that's left are boy-boy and boy-girl.Right, and you know that boy/girl is twice as common as boy/boy in the general population. They're not equally common. So picking the more common one is more likely to be right.

But we aren't dealing with a general population - we're dealing with a specific instance of which I already know half of the answer. Given boy, the probability of boy-boy is 50%, not 25% or 33 1/3%.

My wife and I have a son. Are you telling me that if my wife gets pregnant again (hopefully not to be due at GenCon time), that she has a 2/3 chance of having a girl, just because that's what's generally happened? She'd be very excited to hear that....No, because if we know the order, then we're just looking at one child - the second. You're really just asking the chance of a given child being a boy, and that's obviously 50%.

As I presented the problem, when I tell you "one child is a boy", I'm not looking at the firstborn child and seeing if that child is a boy. I'm looking at both children and seeing if either child is a boy. It's much like Monty Hall looking behind both doors and seeing if either has the goat.

I think this situation is different than the Monty Hall problem.

I don't care about the order of the children in this problem, nor do I care about the general population. Telling me that one child is a boy and asking me to predict the second is not the same as asking me how many combinations of two children are there that include at least one boy.

I underlined that statement above - if we know that one child is a boy (regardless of which one it is), aren't we just looking at the child that isn't a boy?No, because the baseline probability matters. We're not talking about an average child. We're talking about an average child who has a brother.

Say it's a 2-child family in the USA. I tell you the first child is Jewish. Are you saying the odds the second child is Jewish is between 2% and 3%, or something higher?

The importance of the baseline probability is obvious in the second case, but just as present in the original problem. The average sibling of a Jew is a Jew, overwhelmingly. Similarly, two thirds of children (in two child families) who have brothers are girls.

[ElvenEnvy]

I tend to agree with Killercactus on this problem. You start with a 1/3 chance and when an incorrect choice is revealed your chances move to 1/2.

I don't think that the large scale example using 1,000,000 doors applies to the basic problem. In the basic problem you are revealing just one choice and all but one other choice. In the large scale you are only revealing all but one other choice.

Opening 999,998 wrong doors and being left with the one you chose and the one left by the host applies. Chosing one door in 1 million, the host opening 1 wrong door, and then the host allowing you to chose to keep the door you chose or any one other 999,998 doors applies too. So you have a 1/1,000,000 or a 1/999,9998 choice. Rounded you essentially have the same choice whether you allow the host to influence you or not. Using this logic on the basic problem you have a 1/2 chance vs a 1/2 chance.

Since neither large scale model can replicate the way in which one choice is eliminated and all but one other choice is eliminated, its difficult to really decide.

Because the all but one door example leans heavily towards changing and the 1/999,998 for changing slightly edges the 1/million for not changing in the one wrong choice eliminated example; I think that changing is the smart choice. However I'm a firm believer in gut feelings and I'd rather chance it on a 1 in 3 shot. 1 in a million and I'll take all of the hosts help I can get.

Edit: I think that Killercactus is a person that sees the argument I brought to the table that 1/million is nearly the same as 1/999,988 and doesn't think that the the host revealing all but one door in a million is an accurate example.

[TheLorax]

I'm re-reading what I said, and I honestly can't express what I am trying to say. All I was trying to do was explain it in a way that has helped people in the past. I never said that you could have a 50% (or didn't mean to)chance to win the overall game, because that doesn't make sense in a game with 3 doors.

What I was trying to say is that it could be expressed as a choice between two separate games.

Obviously, I'm familiar with the overall problem, or I wouldn't have posted the link :p

Anyhow, my initial hang up with the problem when it was first presented to me, is that the person didn't include the stipulation that the host had to reveal a "goat" door.

[Aldin]

Just to be clear, I think this is likely the most counterintuitive theory in probability. It is also hard to accept from an investment perspective. You choose the door you feel has the highest liklihood of having the car behing it (hi, ElvenEnvy!) and since you have convinced yourself it is more likely to have the car than either of the other two doors it is actually quite easy to convince yourself it is more likely than both combined. In order to make the switch, you have to abandon your first choice - which feels like admitting it was wrong - and you have to do it knowing that being wrong in switching will feel more painful than being wrong in staying. There's something about making a choice and sticking to it that feels honorable whereas changing up can feel dishonorable or even flighty. Just the way we're wired.

~Aldin, short-circuitingly

[killercactus]

Say it's a 2-child family in the USA. I tell you the first child is Jewish. Are you saying the odds the second child is Jewish is between 2% and 3%, or something higher?

The importance of the baseline probability is obvious in the second case, but just as present in the original problem. The average sibling of a Jew is a Jew, overwhelmingly. Similarly, two thirds of children (in two child families) who have brothers are girls.

But, to answer the boy-girl question, do I not simply have to deduce the sex of the other child?

I have only two options to choose from - Boy or Girl, even though I know that boy/girl is twice as likely as boy/boy. But because this is a specific instance and I wasn't given the general population to choose from, it could be either. Doesn't the fact that Girl/Boy and Boy/Girl are the same make this completely about determining the sex of the unknown child?

[airdroppers]

wriggz
...look what you started

[ElvenEnvy]

I think your missing what I'm saying, people view this differently based on their grand example.

You are a solid veiwer from the perspective you have a million doors, you choose one, the host opens all of the doors except the one you chose and one left by the host. you had a 1/million shot and the host knows all, ofcourse you'll switch. You understand this example so I won't explain it fully.

The other view favored by people like KC is that you choose from a million doors and the host opens one wrong door. You can stick with what you chose out of a 1,000,000 or choose one of the other 999,998 doors. Essentially you have the same shot if you stay or choose again. That's where people get the idea of it being 50/50 because in the basic problem you have only 2 options remaining instead of 999,998.

Neither grand example replicates the basic problem perfectly. In the basic problem revealing one is the same as revealing all but one, that can't be replicated on a grand scale. Because of that you will decide to stay or switch based on your idea of which grand example is more accurate.

Grand example one says switch, grand example two says it doesn't matter. Gut-feeling people use example two to justify and stay. Non-gut-feeling people who see example two as correct feel it doesn't matter if outside forces sway them, because its the same either way.

Logically speaking you should switch because it's favors heavily in example one and ever so very slightly in example two.

[dok]

Just to be clear, I think this is likely the most counterintuitive theory in probability.I agree completely. Conditional probability is strongly counterintuitive for most people. Once you internalize this, though, all sort of other odd societal behaviors and common hang-ups start to make sense.


Say it's a 2-child family in the USA. I tell you the first child is Jewish. Are you saying the odds the second child is Jewish is between 2% and 3%, or something higher?

The importance of the baseline probability is obvious in the second case, but just as present in the original problem. The average sibling of a Jew is a Jew, overwhelmingly. Similarly, two thirds of children (in two child families) who have brothers are girls.

But, to answer the boy-girl question, do I not simply have to deduce the sex of the other child?

I have only two options to choose from - Boy or Girl, even though I know that boy/girl is twice as likely as boy/boy. But because this is a specific instance and I wasn't given the general population to choose from, it could be either. Doesn't the fact that Girl/Boy and Boy/Girl are the same make this completely about determining the sex of the unknown child?Of course it's about determining the sex of the unknown child, but the baseline probabilities still matter. As I said at the outset, the family was chosen at random, so general population statistics are significant.

Half of the children with older brothers are girls. Two thirds of children (in two-child families) with brothers are girls.

[rednax]

I started getting a headache around page 3. :razz:

[Grungebob]

I started getting a headache around page 3. :razz:Page 1 was the best. I even made a goat joke that went over the heads of all the nerds swarming this thread.

[Dysole]

Page 1 was the best. I even made a goat joke that went over the heads of all the nerds swarming this thread.
I chuckled.

I started getting a headache around page 3. :razz:

Well I'd've popped in earlier but dok and Aldin seemed to be covering everything I was going to say.

~Dysole, who loves probability problems

[ElvenEnvy]

Lorax's link to wikipedia did me in. Reading about the Monty Hall problem was fine, but then that linked to all the paradoxes. That stuff can really give you a headache.

[bumper15]

Say it's a 2-child family in the USA. I tell you the first child is Jewish. Are you saying the odds the second child is Jewish is between 2% and 3%, or something higher?

The importance of the baseline probability is obvious in the second case, but just as present in the original problem. The average sibling of a Jew is a Jew, overwhelmingly. Similarly, two thirds of children (in two child families) who have brothers are girls.

But, to answer the boy-girl question, do I not simply have to deduce the sex of the other child?

I have only two options to choose from - Boy or Girl, even though I know that boy/girl is twice as likely as boy/boy. But because this is a specific instance and I wasn't given the general population to choose from, it could be either. Doesn't the fact that Girl/Boy and Boy/Girl are the same make this completely about determining the sex of the unknown child?Of course it's about determining the sex of the unknown child, but the baseline probabilities still matter. As I said at the outset, the family was chosen at random, so general population statistics are significant.

Half of the children with older brothers are girls. Two thirds of children (in two-child families) with brothers are girls.

Kc is right on this, it is like flipping a coin, 1/4 for HH, 1/2 for HT, and 1/4 for TT. That is the base statistic, agreed? Now when flipping dice, the first flip in no way influences the second flip, it is still 50/50 either way. So if the first flip is heads, then the second flip has an equal chance of being heads or tails.

So in the Boy/Girl example, the birth of the first child does not influence the second so if the first is a boy, than the second has an equal chance of being a boy or a girl, so it is 50% boy-girl and 50% boy-boy. The reason that at the start the odds for boy-girl are 1/2 is because you can have that outcome regardless of what the first child was.

EDIT: This might explain it better than all that, you start with the normal odds. After the first flip, or birth, the odds go from out of four (boy-boy, boy-girl, girl-boy, and girl-girl (girl-boy and boy-girl are the considered the same, but are seperated for explanation.)) to two either boy-boy and boy-girl. The odds move from 1/4, 1/2, 1/4 to 1/2, 1/2.

[dok]

Say it's a 2-child family in the USA. I tell you the first child is Jewish. Are you saying the odds the second child is Jewish is between 2% and 3%, or something higher?

The importance of the baseline probability is obvious in the second case, but just as present in the original problem. The average sibling of a Jew is a Jew, overwhelmingly. Similarly, two thirds of children (in two child families) who have brothers are girls.

But, to answer the boy-girl question, do I not simply have to deduce the sex of the other child?

I have only two options to choose from - Boy or Girl, even though I know that boy/girl is twice as likely as boy/boy. But because this is a specific instance and I wasn't given the general population to choose from, it could be either. Doesn't the fact that Girl/Boy and Boy/Girl are the same make this completely about determining the sex of the unknown child?Of course it's about determining the sex of the unknown child, but the baseline probabilities still matter. As I said at the outset, the family was chosen at random, so general population statistics are significant.

Half of the children with older brothers are girls. Two thirds of children (in two-child families) with brothers are girls.

Kc is right on this, it is like flipping a coin, 1/4 for HH, 1/2 for HT, and 1/4 for TT. That is the base statistic, agreed? Now when flipping dice, the first flip in no way influences the second flip, it is still 50/50 either way. So if the first flip is heads, then the second flip has an equal chance of being heads or tails.

So in the Boy/Girl example, the birth of the first child does not influence the second so if the first is a boy, than the second has an equal chance of being a boy or a girl, so it is 50% boy-girl and 50% boy-boy. The reason that at the start the odds for boy-girl are 1/2 is because you can have that outcome regardless of what the first child was.Quite right, it is like flipping a coin.

Question 1: "If I flip a coin twice, and the first result is tails, what is the probability that the other result is tails?"

The answer is 1/2.

Question 2: "If I flip a coin twice, and one of the results is tails, what is the probability that the other result is tails?"

The correct answer is 1/3, not 1/2.

[bumper15]

OK, you are right on that fact, then the odds of the other result being heads is 2/3. I got it.

[BurnyFlame]

Question 2: "If I flip a coin twice, and one of the results is tails, what is the probability that the other result is tails?"

The correct answer is 1/3, not 1/2.
I read nothing in this thread except this line, and I disagree. The probability is 0, you said "one of the results" is tails, so the other result has to be heads. Perhaps you meant "at least one result" is tails?

[dok]

OK, you are right on that fact, then the odds of the other result being heads is 2/3. I got it.Right, and that's the exact situation I'm describing in my "one child is male, what is the chance the other is male" question. Ergo, 2/3.

Question 2: "If I flip a coin twice, and one of the results is tails, what is the probability that the other result is tails?"

The correct answer is 1/3, not 1/2.
I read nothing in this thread except this line, and I disagree. The probability is 0, you said "one of the results" is tails, so the other result has to be heads. Perhaps you meant "at least one result" is tails?Yes, you pedant, I mean "at least one". Don't you have more questions to ask about Warforged resolve? :roll:

[RobertDD]

This was a fun read.

It always all comes down from what point off view you are viewing a situation, and realizing you are not always seeing the whole picture. When you are only looking at 2 children families that have one child be a girl, you are excluding the 25% of the families that have 2 boys, which really askews the observation.

My professor used the following example once: for the average observer the beams on a railroad crossing are closed only maybe 5% of the time, but if you observe the world from a train they are closed 100% of the time.

[Fencerjared]

OK, you are right on that fact, then the odds of the other result being heads is 2/3. I got it.Right, and that's the exact situation I'm describing in my "one child is male, what is the chance the other is male" question. Ergo, 2/3.

Question 2: "If I flip a coin twice, and one of the results is tails, what is the probability that the other result is tails?"

The correct answer is 1/3, not 1/2.
I read nothing in this thread except this line, and I disagree. The probability is 0, you said "one of the results" is tails, so the other result has to be heads. Perhaps you meant "at least one result" is tails?Yes, you pedant, I mean "at least one". Don't you have more questions to ask about Warforged resolve? :roll:

The system won't let me rep you again. The system, man.

[Sarpedon]

This was a fun read.

:screwy:

Dok and Aldin's explanations are very good and, of course, mathematically correct. However, here's some ghetto street smarts: If your opponent should offer to let you choose again, then you would KNOW you had NOT picked the "X" order marker and you should NOT switch. Keep it real.

[Grungebob]

This was a fun read.

:screwy:

Dok and Aldin's explanations are very good and, of course, mathematically correct. However, here's some ghetto street smarts: If your opponent should offer to let you choose again, then you would KNOW you had NOT picked the "X" order marker and you should NOT switch. Keep it real.
Haha! Sarp for the win!!

[Dysole]

This was a fun read.

:screwy:

Dok and Aldin's explanations are very good and, of course, mathematically correct. However, here's some ghetto street smarts: If your opponent should offer to let you choose again, then you would KNOW you had NOT picked the "X" order marker and you should NOT switch. Keep it real.
Haha! Sarp for the win!!

While in fact this may often be the case with many people, my opponents would probably know me well enough to know that I might just do this because of the puzzle aspect of it.

~Dysole, who tends to mathatize just about anything

[Takanuva]

Question 2: "If I flip a coin twice, and one of the results is tails, what is the probability that the other result is tails?"



The Government wants tells us to belive the answer is 1/3. The rest have been fooled...Only I hold the answer...Only I can see through the lies...The answer is not 1/3! Coins flips are not an exact system! They lie! They cheat! They are Biased!!! The truth! The truth is that a coin shows up heads 2/3's of the time! The Answer is thus 1/3!...Oh wait...

[bumper15]

There is one major problem that people have forgotten. What if the coin lands on its side, or what if there is no X marker to pick from?

[Aldin]

There is one major problem that people have forgotten. What if the coin lands on its side, or what if there is no X marker to pick from?

"It is best to keep your mouth shut and let people think you're an idiot, then to open it and have them remove all doubt"- Mark Twain

~Aldin, :p

[bumper15]

That quote doesn't apply to me, everyone already knew I was an idiot. :D

[padlock]

[
Half of the children with older brothers are girls. Two thirds of children (in two-child families) with brothers are girls.

The bolded statement is incorrect.

I do however completely agree with the statement that in a two child family, if one child is a boy, then there's a 2 in 3 chance the other is a girl.

That's because in the 1 in 3 chance where the other child is a boy, you have 2 children who have a brother, whereas in the 2 in 3 chance the other child is a girl, there's only 1 child with a brother.

To demonstrate let's go back to the 4 equally likely possibilities in a 2 child family:

Boy, Boy
Girl, Girl
Boy, Girl,
Girl, Boy

To make it easier, let's give them all names:

Peter, Paul
Mary, Jane
Michael, Sarah
Anne, Tom

Now we can see that 4 of the children above have brother's
Peter, Paul, Sarah, and Anne. 2 Boys and 2 Girls.

So to summarize, the following 2 seemingly contradictory (but not) statements are true:

-In a 2 child family 50% of the children with brothers are boys
-In a 2 child family, if 1 child is a boy, there's a 2/3 chance the other is a girl

Unintuitive indeed.

[Tornado]

I guess I do not understand the point of this thread. The power states that an OM is removed at random which to me means you really do not get to choose. Am I wrong here? You can not say the OM to your far left because that would not be random. We always use a dice roll to determine what OM gets knocked down. I guess we all got a lesson in probability and it sounds like you all had fun with it... peace

[bmaczero]

Probability always makes me want to bang my head against a wall. I could never reconcile that stupid Monty Hall problem, and I was convinced that the book I'd read it in was talking utter bull. Eventually I wrote a program that played the game with itself. I ran it 1000 times and had it switch every time. Sure enough, it won 67% of the games. That drove me crazy :p.