Where are the flaws?

[Nick Royale]

Where are the flaws in these calculations? Any comments appreciated.

Hero hold QQ preflop. 5 more players to the flop. Flop comes A x x. First player bet, 3 callers, one fold, Hero...
(I count the 2 folded cards as seen non ace cards)
What are the probability one player holds an ace?

There are 45 unseen cards in the deck.
Among those there are 3 aces.
In other words every 15th card is an ace on average (45/3=15).
There are 8 cards in the hands of our opponents that might be an ace.

To make this calculation accurate you have to find out how likely you are to NOT hit an ace when drawing 8 cards from these 15 containing 1 Ace. And that's a quite easy calculation:

If you draw only 1 card the you'll NOT hit an ace 93% of the times. 14/15=0.93
If you draw only 2 cards you'll NOT hit an ace 87% of the times. (14/15)*(13/14)=0.87
If you draw 8 cards you'll NOT hit an ace 47% of the times. ((14!)/(6!))/((15!)/7!)) EDIT: = 7/15
EDIT: If you wan't to know the probability they did hit (yes, you wanna know!) you could simply take 8/15= 53% . Much more simple

In other words you're opponents will hit an ace 53% of the times. Our poor Hero will be behind the majority of the times.

I suspect something goes wrong when I divide 45 with 3 (haven't had the time to think that part through entirely), but feel free to tell me my whole calcultion stinks /images/graemlins/tongue.gif

[BruceZ]

[ QUOTE ]
Where are the flaws in these calculations? Any comments appreciated.

Hero hold QQ preflop. 5 more players to the flop. Flop comes A x x. First player bet, 3 callers, one fold, Hero...
(I count the 2 folded cards as seen non ace cards)
What are the probability one player holds an ace?

There are 45 unseen cards in the deck.
Among those there are 3 aces.
In other words every 15th card is an ace on average (45/3=15).
There are 8 cards in the hands of our opponents that might be an ace.

To make this calculation accurate you have to find out how likely you are to NOT hit an ace when drawing 8 cards from these 15 containing 1 Ace. And that's a quite easy calculation:

If you draw only 1 card the you'll NOT hit an ace 93% of the times. 14/15=0.93
If you draw only 2 cards you'll NOT hit an ace 87% of the times. (14/15)*(13/14)=0.87
If you draw 8 cards you'll NOT hit an ace 47% of the times. ((14!)/(6!))/((15!)/7!)) EDIT: = 7/15
EDIT: If you wan't to know the probability they did hit (yes, you wanna know!) you could simply take 8/15= 53% . Much more simple

In other words you're opponents will hit an ace 53% of the times. Our poor Hero will be behind the majority of the times.

I suspect something goes wrong when I divide 45 with 3 (haven't had the time to think that part through entirely), but feel free to tell me my whole calcultion stinks /images/graemlins/tongue.gif

[/ QUOTE ]

You have computed the probability of drawing 8 non-aces out of 15 cards with 1 ace, but this is not the same as drawing 8 non-aces out of 45 cards with 3 aces. You want [42!/(42-8)!] / [45!/(45-8)!] = 54.8%. Note that this is the same as C(42,8)/C(45,8). So the probability that an ace is out is 45.2%.

Also note that this is not exactly the same as 8*3/45 = 53.3%. That only works if just one person can have the ace. Otherwise this estimate is too high since it overcounts the times that more than one player has the ace.

[Nick Royale]

Thanks alot Bruce! That explains everything very clear and will ceratainly kill this thread. Guess you're used to much more complicated problems than that, so I appreciate that you took your time.

/Nick

[BruceZ]

Nick,

I had some typos in my original response which are now corrected, so you might want to check it again now.

-Bruce