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EasilyFound
03-17-2005, 09:34 AM
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.
03-17-2005, 01:31 PM
[ 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
03-17-2005, 04:49 PM
You are correct. I read the chart (http://www.doylesroom.com/education_odds4.html) wrong.

What is "C" in your answer?

Dave H.
03-17-2005, 06:14 PM
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 /images/graemlins/spade.gif7 /images/graemlins/diamond.gif is the same as 7 /images/graemlins/diamond.gif8 /images/graemlins/spade.gif

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
03-17-2005, 06:24 PM
Is C(48,10) / C(52,10) the same?

EasilyFound
03-17-2005, 07:01 PM
[ QUOTE ]
Make sense?

[/ QUOTE ]

I'm a lawyer, so no. /images/graemlins/confused.gif 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
03-17-2005, 08:20 PM
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
03-17-2005, 08:43 PM
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.
03-17-2005, 10:23 PM
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
03-17-2005, 11:13 PM
When attempting to determine the probability of someone else being dealt a certain hand, how do you take into account the number of players?

EasilyFound
03-17-2005, 11:40 PM
I'll just have to learn to play by the gut.

Dave H.
03-18-2005, 12:57 PM
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
03-18-2005, 07:46 PM
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.