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View Full Version : Short handed poker, Ed Miller, Bunching Effect, The Price is Right

08-17-2005, 07:21 AM
Ed Miller wrote that the bunching effect is miniscule between acting first in a four- handed game and acting one of the button in a ten-handed game after six folds. (see footnote),

But is there some other effect that comes into play?

The Price is Right puzzle from the Ask Marilyn column:

One prize is behind door A, B or C, you are asked to choose and you guess door A. The host tells you truthfully that the prize is not behind door B and now gives you the option to change your choice. Should you change your choice?

The answer is yes you should. If you change to door C you’ll have a 66% chance of receiving the prize, if you stick with door A you’ll have a 33% chance of receiving the prize.

I hope we don’t discuss this, trust me this is correct. Math professors challenged this answer saying it was wrong. Marilyn then proved the answer to their satisfaction.

Can we apply this same logic to poker?

An extreme example: You’re in the BB with KK. Your opponents only play AA except the SB who’ll steal with every hand if he’s 1st in. It has been folded to the SB who raises. What’s the chance that the SB has AA?

Let’s forget the bunching effect here. My question is: Is there a “Price is Right” effect? I realize it won’t be as dramatic, but does it exist. And if it does, does it follow that the hand qualities, not considering bunching, are better in a ten-handed game after six folds than in a four-handed game?

Footnote: By the way, Ed Miller did explain why there are differences between the four-handed game and the ten-handed after six folds. In this post I’m looking at something different.

08-17-2005, 08:34 AM
I'm guessing not

LetYouDown
08-17-2005, 09:04 AM
Thank God. It's been far too long since someone included the 3 doors problem in a post. One of the longer stretches that I can recall.

IGMorton
08-17-2005, 09:07 AM
hehe, funny you bring this up. the ol' Let's make a deal statistical obscurity. funny how many absurd arguments this brings up amongst intelligent folks over cocktail partys about causality, the difference between Bayesians and Frequentists, etc.

In let's make a deal... you have only 3 possible outcomes to look at. You count the number of times a swithch could be beneficial and the nubmer of times it couldn't.

the problem with poker is that calculating the probability any of 9 opponents (or even 5) in front of you might have had AA is a huge multi tiered decision tree in of combinatorics. i suppose it might be workable for soley AA within a few hours. but... how does that help your game?

in reallity we are much more likely to worry about say holding somehting like JJ and raising 3 preflop callers who between them might hold 2-3 overcards (a combinatorial problem of absurd magnitude with assumptions about the likelyhood of certains players to call with A6o, etc) and then evaluate the massive possible post flop scenarios.

ugh /images/graemlins/smile.gif

LetYouDown
08-17-2005, 09:25 AM
Figuring out whether one of your opponents has A-A does not qualify as a "huge multi tiered decision tree in of combinatorics"...even though that phrase makes no sense anyway.

This thread is like a bad joke involving the most commonly asked questions in the history of this forum.

IGMorton
08-17-2005, 10:07 AM
well excuse me for being sophomoric in my first post in this form. jees, other forums at 2+2 welcome a first time poster. have you forgotten the first time you heard the "let's make a deal problem"? it may seem banal to you now, but it still fascinates the masses. why criticize someone for their curiosity?

but, to evaluate the likelyhood of 9 guys having AA, don't you do a decision tree...

ie... assume player 1 has KK.

probability Player 2 might have AA:

=4/50*3/49

probability player 3 might have AA:

if player 1 held no aces

=4/48*3/47

if player 1 held 1 ace

=3/48*2/47

probability player 4 has AA gets much more complicated:

if neither 1 or 2 held an ace

=4/46*4/45

if 1 held an ace

=3/46*2/45

if both held an ace

=2/46*1/45

as you go up the chain of players, you have more conditions to examine... the complex probability grows immensely as you add players.

LetYouDown
08-17-2005, 10:40 AM
If you hold K-K at a 10 person table, the probability that one of your opponents has A-A is 9 * 6/C(50,2) - C(9,2)/C(50,4) or about 4.39%. Not exactly complex.

That question, along with the 3 doors problem, was asked basically every other day for a year. It has been analyzed to death, brought back from death, and analyzed to death again, more times than you can imagine. Any other question is met with open arms. The only ones that are problematic are the questions that have 100 threads pertaining to them, and are easily found with the search function.

IGMorton
08-17-2005, 10:58 AM
ok fair enough. i understand the regulars might get annoyed at a new poster who fails to search the archives. i apologize, but i was drawn here from another forum. btw, having taken 1 senior level combinatorics class at university i never encountered this inclusion/exclusion principle you linked me to... THANKS!

08-17-2005, 11:04 AM
Sorry letyoudown, I thought this was a unique question. And I didn't know the protocal was to do a search before posting a question.

LetYouDown
08-17-2005, 11:09 AM
Not a problem. There are questions that appear every month or so which don't really bother anyone...it's just that specific problem. It is brought up by someone way too frequently. Your actual question was valid.

Sorry for being pissy, I guess losing the over/under on when the 3 doors problem would surface again upset me.

\$ock
08-18-2005, 06:49 PM
Fascinating puzzle. U gain information + equity by not opening door B.

Reminds of the kind of situation in NLHE where u have a medium/high pocket pair and there are 2 all-ins in front of u. U don't like your chances, until someone/something comes along and tells u they're both holding AK. Now u gladly decide to call.