Specific Hand Probability

[EasilyFound]

Hi. The probability of being dealt at least one Ace preflop is 14.93%. And I've seen the chart re: probability absence of Aces before prelop. Five handed, probability nobody will have an Ace is 48.6 %.

My question is what equation do you use to get that result or make that calculation--the probability that nobody will have at least one Ace preflop.

[Dave H.]

[ QUOTE ]
Five handed, probability nobody will have an Ace is 48.6 %.


[/ QUOTE ]

I believe you are misstating the above assuming you mean the probability that there will be no ace in any of the 5 pockets. The probability that no one will have an ace in 5 pockets is:

C(48,2)/C(52,2) * C(46,2)/C(50,2) * C(44,2)/C(48,2) * C(42,2)/C(46,2) * C(40,2)/C(44,2)

= (1128 * 1035 * 946 * 861 * 780) / (1326 * 1225 * 1128 * 1035 * 946)

= 41.34%

[EasilyFound]

You are correct. I read the chart wrong.

What is "C" in your answer?

[Dave H.]

C(52,2) stands for the number of ways 2 cards can be chosen from 52 cards (where order doesn't matter, i.e. 8 [img]/images/graemlins/spade.gif[/img]7 [img]/images/graemlins/diamond.gif[/img] is the same as 7 [img]/images/graemlins/diamond.gif[/img]8 [img]/images/graemlins/spade.gif[/img]

The general formula for C(n,r) is:

n!/(r!*(n-r)!)

So, C(52,2) = 52!/(2!*(52-2)! =

(52*51*50*...*1)/(2*1)*(50*49*48*...*1)

You can see that most of the terms in the numerator cancel with those in the denominator and you're left with:

(52*51)/2 = 26*51 = 1326

Make sense?

[DiceyPlay]

Is C(48,10) / C(52,10) the same?

[EasilyFound]

[ QUOTE ]
Make sense?

[/ QUOTE ]

I'm a lawyer, so no. [img]/images/graemlins/confused.gif[/img] It will take some time for me to digest it.

I know it should be obvious, but again, I'm a lawyer, so it ain't.

What does "n" and "r" represent. And what does the "!" represent?

[DiceyPlay]

If you were asked how many ways can you choose 2 items out of a collection of 5 distinct items a short hand way of asking this is to write C(5,2).

The formula to evaluate C(5,2) is 5!/(2! * (5-2)!) and ! is a factorial which is evaluated as follows:

To evaluate n! it is equal to n * (n-1) * (n-2) * (n-3) * ... * 3 * 2 * 1

so C(5,2) = 5!/(2! * (5-2)!) = 5 * 4 * 3! / (2 * 3!) = 5 * 4 / 2 = 10.

Therefore there are 10 ways to choose 2 items from a collection of 5 distinct items.

Notice the order in which you choose the items from the collection does not make a difference. This is a combination. If it did make a difference, it would be a permutation. The formula to evaluate a permutation is P(n,r) = n! / (n-r)!

Hope that makes sense.

[EasilyFound]

Wow. Stat Man am I not. I'll just stick to charts.

If it is 20.36 % that I am dealt a hand with a Pair or an Ace, what are the probabilities that, in a five-handed game, at least one person will be dealt a pair or an Ace?

Is there a chart for that?

[Dave H.]

n can be anything and so can r...

The formula n!/(r!*(n-r)!) represents the number of ways you can take n things r at a time where order doesn't matter. For example, if you have a deck of 52 cards and want to know how many two card pockets you could have, you are trying to decide how many ways you can take 52 cards (so n = 52) 2 (so r = 2) at a time. So n = 52 and r = 2 in this example.

The answer is 52!/(2!*(52 - 2)!)

The ! is called "factorial". Don't be scared by the word. It simply means this:

4! = 4*3*2*1
17! = 17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1
0! is the only weird one...it is 1 by definition.

So in the above example,
52!/(2!(52-2)!) =
52*51*50*49*...*1/(2*1)*(50*49*48*47*46*...*1)

If you write that out, you see that the 50 in the numerator cancels with the 50 in the denominator. The same goes for 49, 48, 47, etc.

So the only thing you're left with in the numerator that doesn't cancel with something in the denominator is:

52*51

And the denominator still has a 2 in it, but everything else cancelled.

So you're left with 52*51/2 = 26*51 = 1326

Now what if you had 52 cards and wanted to know how many THREE card pockets you could have?

In this case, n = 52 and r = 3
So the formula would be:
52!/(3!*(52-3)!) =
52!/(3!*49!) =

52*51*50*49*...*1/(3*2*1)*(49*48*47*46*...*1)

Notice how 49 in the numerator cancels with 49 in the denominator. The same for 47,46,45,44...all the way down to 1.

So you're left with 52*51*50/3*2*1 = 132600/6 = 22,100

Make more sense now?

[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?