Calculating Probability of a Specific Hand in Poker

[DyslexicHobo]

To ask my question, I will explain a scenario and ask for the number in which I'm trying to obtain. I think it might be called the statistical confidence, but I'm not sure.

A 20-sided die is rolled 50 times. How can I calculate the percent chance that a certain number, n, or a certain set of numbers, N, will be rolled within that 50 rolls?

The exact scenario that inspired the inquiry had to do with texas holdem. I'm trying to figure out, if you have x players that have all played n games, what percentage of them will have had a specific hand at least z times.

Would this be considered a confidence calculation?

 

[LCKurtz]

To ask my question, I will explain a scenario and ask for the number in which I'm trying to obtain. I think it might be called the statistical confidence, but I'm not sure.

A 20-sided die is rolled 50 times. How can I calculate the percent chance that a certain number, n, or a certain set of numbers, N, will be rolled within that 50 rolls?

Let p be the probability that your number or one of your set of numbers happens on any trial. If X represents the number of trials up to and including the first success, then, assuming the trials are independent, X has a negative binomial distribution:

P(X=x) = p(1-p)^{x-1}

The probability of at least one success in n trials would be

P(X\le n) = \sum_{x=1}^n p(1-p)^{x-1}= 1 - (1-p)^n

For example, for your dodecahedron die (one of the few platonic solid fair dice, you might want to use a spinner for other sizes ), say your successful numbers were {3,8,13,15,19} so p = 1/4. Then the probability of at least one success in 6 trials would be:

P(X\le 6) = 1 - (\frac 3 4 )^6 = .822

The exact scenario that inspired the inquiry had to do with texas holdem. I'm trying to figure out, if you have x players that have all played n games, what percentage of them will have had a specific hand at least z times.

Would this be considered a confidence calculation?

In your scenario, a difficulty arises because the trials aren't independent. Player A getting 4 aces and player B getting 4 aces aren't independent events because they are mutually exclusive. It all makes my head hurt.

 

[DyslexicHobo]

You don't need to worry about the fact that poker hands aren't independent events. The only cards I'm worried about in this case are my own two cards. In holdem you are dealt only two cards. The percentage you're dealt a specific hand can easily be calculated.

For example, the probability you're dealt two aces is p=(4/52)*(3/52)=~.0044

So that probability can be plugged into the equations you showed. However, one important part of my question was left unanswered. You explained very well the case of being dealt at least 1 hand of two aces within a certain number of hands.

Now I may be terribly wrong, so please point out my mistake if I'm incorrect. Instead of using poker as an example I'll use a coin. If I roll a two-sided die (flip a coin) three times, I can easily calculate the chance of getting heads at least once. Another way to calculate the chance is to list all combinations and count those that satisfy what I'm looking for (at least one heads) and divide by the total number of combinations.

If this is done for flipping a coin three times, these are the possible combinations:

1	H	H	H
2	H	H	T
3	H	T	H
4	T	H	H
5	T	T	H
6	T	H	T
7	H	T	T
8	T	T	T

So we take the number of combinations which heads appears at least once and divide by the total number of combinations. 7/8 = .875

Now instead of looking for at least one heads in 3 flips, we'll look for 2 heads in six flips. We can find all 64 combinations by combining the 8 strings of combinations in each possible way. This is done in the following matrix:

	1    	2    	3	4	5	6	7	8
1	11	21	31	41	51	61	71	81
2	12	22	32	42	52	62	72	82
3	13	23	33	43	53	63	73	83
4	14	24	34	44	54	64	74	84
5	15	25	35	45	55	65	75	85
6	16	26	36	46	56	66	76	86
7	17	27	37	47	57	67	77	87
8	18	28	38	48	58	68	78	88

For example, 11 would be the first combination of 3 flips repeated twice: HHH HHH. 34 would be the third combination followed by the fourth combination: HTH THH. Out of these 64 combinations, any combination without the number 8 must have two heads. Any combination with the numbers 1-4 must have two heads. Therefore the only numbers that do not have at least two heads are 58, 68, 78, 88, 87, 86, 85.

(64-7)/64 = .89 != .875

I believe I can conclude from this that the chance of, for example, getting dealt pocket aces 20 times within 10000 hands does not have the equivalent probability of being dealt two aces 1 time within 500, but slightly higher. If so, how can I calculate this probability?

PS: I have not taken any probability courses so please stick with symbolics that I'd have seen in calc I - calc IV. Thanks :D