Cross-Post : Probability Problem

xPeru · #1 02-09-2007, 04:07 PM

Posted this in Probability forum - no answers, I know I can rely on STTers to get the right answer before lock.

Classic Monty Hall problem: 3 doors, one prize. Contestant selects a door. Monty opens one of the other doors and shows no prize. Contestant is asked if he wants to switch to the remaining door or stick with original choice. Probability of winning is 1/3 if he sticks, 2/3 if he switches.

Are player folds equivalent to Monty opening a door and showing no prize? eg 9 players NLHE. Probability of a AA being dealt to one player is 9/221. 7 folds. What is the probability of one of the two remaining players having As? 2/221 or 9/221?

Slim Pickens · #2 02-09-2007, 04:27 PM

People play hands other than AA. You've gained a little bit of information with those 7 folds since a lot of playable hands contain an ace and there are no playable hands within those 7 folds. The probability is about that of dealing AA in one of two hands out of a 38-card deck with 4 aces in it, assuming ZOMG A2o UTG limp.

Kibby · #3 02-09-2007, 04:35 PM

How you estimate how likely or not it is based on the folding ranges of the players that have already acted. If you know that the folding players all play any ace then you know that there are 14 cards removed that can't be aces. Typically there is a .4525% chance a single person gets aces. With 14 cards removed there is a .8535% chance a single person gets aces or ~1.7% that either person has aces. Say you're the SB and don't have an ace in your hand. That would be 16 cards removed and the BB would have a .9524% chance of having rockets.

Obviously you're rarely going to be at a table where you can count on each person always playing any ace. So you have the weight the chances of them folding hands that did have aces. I'm not going to attempt the math for that one right now.

jafeather · #4 02-09-2007, 04:54 PM

[ QUOTE ]
Probability of winning is 1/3 if he sticks, 2/3 if he switches.

[/ QUOTE ]

(buzzer sounds) Sorry....wrong answer. Probability is 50/50 no matter what he does now....the other third you refer to no longer exists.

citanul · #5 02-09-2007, 04:58 PM

ja,

the probability of winning if you do not switch remains 1/3.

Kibby · #6 02-09-2007, 05:01 PM

Sorry, OP is correct. Don't you see?

[ QUOTE ]
[ QUOTE ]
Probability of winning is 1/3 if he sticks, 2/3 if he switches.

[/ QUOTE ]

(buzzer sounds) Sorry....wrong answer. Probability is 50/50 no matter what he does now....the other third you refer to no longer exists.

[/ QUOTE ]

alanbrown · #7 02-09-2007, 05:04 PM

No they aren't. The reason is that we don't know that there is an AA out there. If we knew there was then everyone who folded would increase the odds of each of the other people having AA. But in this case the odds for each person is 1/221. If there are 9 people that is 9/221, but if one of 'em folds then the odds of being against an AA is now 8/221

Slim Pickens · #8 02-09-2007, 05:09 PM

[ QUOTE ]
[ QUOTE ]
Probability of winning is 1/3 if he sticks, 2/3 if he switches.

[/ QUOTE ]

(buzzer sounds) Sorry....wrong answer. Probability is 50/50 no matter what he does now....the other third you refer to no longer exists.

[/ QUOTE ]

You got information about the one remaining door. Monty could have opened it but didn't. Either he chose not to (because you picked the correct door to start and it didn't matter which of the other two he opened) or he couldn't (because it has the big prize behind it). That problem is the classic Bayesian mind-[censored].

alanbrown · #9 02-09-2007, 05:09 PM

This is actually a very famous problem first presented by Marilyn vos Savant. You're not alone in presuming it's 50/50 but you're also not correct. Go look it up on wikipedia. It's a great story as well as a great puzzle. OP is right. Odds are 2/3 that you'll get the car if you change your pick. However, the Monty Hall problem has nothing to do with his AA scenario. The 'trick' in the Monty Hall problem was that Monty knew there was a car and he knew where it was so he could always show a door with a goat behind it. That's not the case in this situation as we don't know there's AA out there for sure.

citanul · #10 02-09-2007, 05:13 PM

if you could know that they fold means that they never have any ace in their hand at all, you could do this very easilly, and yes, if you take the more abstract view of the problem:

if it is 7 folds to you on the button, chances are better than normal that the blinds hold stronger than average hands.