Specific Hand Probability

[vqchuang]

When attempting to determine the probability of someone else being dealt a certain hand, how do you take into account the number of players?

[Dave H.]

Sorry, but that's too vague a question. Do you mean something like:

What's the probability, if it's heads up, that I will have exactly one ace and you won't have any ace in the pocket?

I would determine the probability that I have an ace as

(4*48)/C(52,2)

Then I would calculate the probability that you do NOT have an ace as C(47,2)/C(50,2)

Now you would just multiply those two probabilities. I don't know if this is what you wanted, but you should probably start a new post and get some of the real probability gurus to look at it.

[EasilyFound]

I meant this: the probability that there will be no ace in any of the 5 pockets.

The probability that you will not have an Ace in your pocket is 85%. And the probability that none of five players will have and Ace is roughly 41%.

I wanted to know whether there was a quick formula to follow to get from the first figure to the second.

But after reading the replies here, I now understand that this stat thing is way too complicated for my head at my age.

I just need someone to break it down into a chart, like the one Brunson has in his book for the probability on the absence of Aces.

Thanks to all who took the time to reply. Will try to digest it all and see if I can grasp it.