The Mind Flayer Paradox

[padlock]

Originally Posted by 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.

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]

Originally Posted by dok

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]

Originally Posted by 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.

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]

Originally Posted by 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.

[TheLorax]

Originally Posted by dok

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]

Originally Posted by 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.

Originally Posted by dok

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?

Originally Posted by dok

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]

Originally Posted by lefton4ya

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]

Originally Posted by TheLorax

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?

Spoiler Alert!

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]

Originally Posted by dok

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]

Originally Posted by killercactus

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.

YES, THAT IS RIGHT.

Originally Posted by killercactus

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?

YES, THAT IS PRECISELY THE REASON.

Originally Posted by killercactus

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]

Originally Posted by padlock

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]

Originally Posted by Aldin

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.