Re: Poker probability math is incorrect
 

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DiceyPlay, i am by no means an expert but Im fairly certain i disagree with your logic. There is no point in switching, as it has not changed the otehr to door to 2/3. Rather, that percentage is distributed over both remaining doors, changing them both to 1/2. I cant possibly believe that the other door is now more likely to have the vette, just doesnt work like that.

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wow are u serious?

Re: Poker probability math is incorrect
 

WCG, it's easier to realize Dicey is correct when you add doors to the problem.

Suppose there are 1000 doors. One corvette 999 donkeys. You pick door #537. Host shows you 998 donkeys, leaving door #537 and #84. If he offers you to switch doors from #537 to #84, would you do it?

It's probably even better explained in the wikipedia link so just read that one instead.

Edit: Wow there's a lot written on that page.

Re: Poker probability math is incorrect
 

Im usually so good with math, i feel like a retard right now. Mets messaged me explaining, and i still dont even remotely understand, i wishi could figure this out.

Re: Poker probability math is incorrect
 

Hi Aaron,

It's still not clear to me. In the Monty Hall problem we assign a probability to the door originally chosen and that probability doesn't change when new information arrives. When the new information is available we revise the probability of the other door. That makes sense.

In our hold'em game we deal the cards 2 to a player and 5 unseen cards on the board. We assign the probability of any unseen card being an ace as 4/50 (our hand has no ace). Now 3 of the 5 board cards are revealed and one of them is an ace. Do we revise the probability of all unseen cards or do we hold any of the originally assigned probabilities constant as we did in the Monty Hall game? Intuition tells me we revise them all to 3/47. But as in the Money Hall game we know that intuition can be misleading. How does the hold'em scenario differ from the Monty Hall game?

Re: Poker probability math is incorrect
 

I think Fabians example is the clearest way to see this phenomena for the first time. Read it again and ask yourself as if you're playing the game what your choice would be and why you are making that choice.

Re: Poker probability math is incorrect
 

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Im usually so good with math, i feel like a retard right now. Mets messaged me explaining, and i still dont even remotely understand, i wishi could figure this out.

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i sent you another pm hope that helps

Re: Poker probability math is incorrect
 

I think i got it now, although i will be frank, part of me has my reserves about it. But mets and fabian make a lot of sense, and im sure after sitting on it for a while it will seem to be a lot clearer.

Thanks.

Re: Poker probability math is incorrect
 

I found this the best way to get past the counter-intuitiveness of the Monty Hall problem. Every time you pick a wrong door (i.e., 2/3 of the time), the host will reveal the other wrong door, and switching will always reveal the prize. It's only when you've picked the prize door (1/3 of the time) that switching will be wrong.