I still don't think odds for a flush draw is correct

[PokerEveryDay]

I'm thinking about this again. You have 2 suited cards and the flop comes giving you a flush draw. So you think you have 9 outs but do you really? Let's say in a long handed game of 10 people you and one other are still in to see the flop and you have a flush draw. Not including the guy still in there are 16 cards in the muck. On avg. 4.5 of your suits will be in the muck and your opponent leaving you 4.5 outs of 29 un dealt cards for 15.5%, not 20% for the turn. In HU play 9 outs seems to make more sense.

[Jovial]

Play 2 million hands and proof your thoughts.

[pzhon]

Suppose you are trying to figure out the probability of dealing a spade from a freshly shuffled deck. It's 13/52 = .25.

Ok, now deal a card face down. What is the probability that the next card is a spade? 1/4 of the time, the first card was a spade, and there are 12 left in the 51 cards. 3/4 of the time, there are 13 spades left in the 51 cards. The probability that the second card is a spade is (1/4)(12/51)+(3/4)(13/51) = .25. The unseen card does not affect the probability.

Back to the flush draw: It's possible that there are fewer cards of your suit left in the deck. However, there are also fewer cards left in the deck. When you have no additional information about the suits of the folded cards, these effects precisely cancel, giving you a probability of hitting of 9/(# unseen cards), whether the unseen cards are in your opponents' hands, in the muck, or in the remainder of the deck.

[Mike Haven]

[ QUOTE ]
I'm thinking about this again. You have 2 suited cards and the flop comes giving you a flush draw. So you think you have 9 outs but do you really? Let's say in a long handed game of 10 people you and one other are still in to see the flop and you have a flush draw. Not including the guy still in there are 16 cards in the muck. On avg. 4.5 of your suits will be in the muck and your opponent leaving you 4.5 outs of 29 un dealt cards for 15.5%, not 20% for the turn. In HU play 9 outs seems to make more sense.

[/ QUOTE ]

You made a couple of minor mistakes in your synopsis that threw you off a little in your conclusion.

Because you know 9 of your suits are in the 47 unknown (to you) cards then 1 of your suit is in every 5.22 unknown cards, (47 / 9).

There are 16 unknown cards in the muck, so, on average, 3.06 of these are your suits, (16 / 5.22), leaving 5.94, (9 - 3.06), of your suits in the 31 cards, (47 - 16), remaining in the deck. (For the purpose of clarity I have ignored the two cards in your single opponent's hand, but you could treat these in a similar way as above, and reach the same conclusion as below.)

You can now see that the odds of you catching one of these 5.94 cards on the turn are [(31 - 5.94) / 5.94] = 4.22 to 1, the same as your original [(47 - 9) / 9] = 4.22 to 1 chance.

So, your method was "correct", but the simple arithmetic errors gave you the wrong final answer.

[PokerEveryDay]

Thanks for the clarification, I guess there is no need to play out 2 million hands.

[binions]

With your poker brain, you don't have enough money to play 2 million hands.

[PokerEveryDay]

[ QUOTE ]
With your poker brain, you don't have enough money to play 2 million hands.

[/ QUOTE ]

Jerk

[Questje]

You could play 2 million hands for play money, for fun at home or let a computer play them for you /images/graemlins/smile.gif
Easy as that /images/graemlins/smile.gif

[vypremik]

[ QUOTE ]
Suppose you are trying to figure out the probability of dealing a spade from a freshly shuffled deck. It's 13/52 = .25.

Ok, now deal a card face down. What is the probability that the next card is a spade? 1/4 of the time, the first card was a spade, and there are 12 left in the 51 cards. 3/4 of the time, there are 13 spades left in the 51 cards. The probability that the second card is a spade is (1/4)(12/51)+(3/4)(13/51) = .25. The unseen card does not affect the probability.

Back to the flush draw: It's possible that there are fewer cards of your suit left in the deck. However, there are also fewer cards left in the deck. When you have no additional information about the suits of the folded cards, these effects precisely cancel, giving you a probability of hitting of 9/(# unseen cards), whether the unseen cards are in your opponents' hands, in the muck, or in the remainder of the deck.

[/ QUOTE ]

As a Stats teacher, I can assure you that pzhon is correct. As a matter of fact, you can always count the unknown cards as part of the same pool. It doesn't matter if they are mucked, discarded, in the deck, or in your opponent's hand - they are unknown. The results will always be the same whether you calculate it simply or take each pile differently.

[weaklyatomic]

This idea might have more merit say in a small stakes game where you are looking to pair an overcard ace as an out.

Given that a high percentage of these players will play any ace, if they have mucked to you on the button and you open with AK, then perhaps you can really consider the remainder of the deck to be "ace rich" since many opponents would probably have limped in with any ace.

Otherwise as has been pointed out, your arithmetic simply led you to draw an erroneous conclusion.