Ben
United States Ann Arbor Michigan

Calling all mathematicians! (Or other generally smart folk.)
I'm trying to think through a game, potentially for a review, and it would be very helpful for me to understand card draw probabilities.
The Setup: 108 card deck 16 each of 1s 2s and 3s 12 each of 4s 5s and 6s 8 each of 7s 8s and 9s
What I Need To Know: In an average five card hand, how many pairs are likely? What about a sixcard hand?
I'm hoping this isn't too hard to people who know what they're doing. for any help.




Howdy! An interesting challenge, but I need to ask you a couple questions to get a mathematically precise description of the problem.
How do you wish to count 3+ of a kind? Three of a kind is often considered three pairs and four of a kind six pairs (as in cribbage). Those are the number of possible pairs you can draw from three and four cards, respectively. By extension, five of a kind would be 15 pairs. If you wish to count 3, 4, 5 and 6 of a kind all separately, the problem gets complicated.
Also, when you say "how many pairs are likely" do you mean what is the expected number (i.e., mean) number of pairs in a hand? Would you like the entire probability distribution, i.e., a list of the chances of 0 pairs, 1 pair, etc.?




One more thing: Do you want an analytical (exact) answer, or is an approximation close enough? For example, it would take me quite a while to calculate all the probabilities mathematically, but I could write and run a simulation to estimate the chances to, say, three significant digits pretty easily.


Ben
United States Ann Arbor Michigan

robigo wrote: Howdy! An interesting challenge, but we first need a mathematically precise description of the problem.
How do you wish to count 3+ of a kind? Three of a kind is traditionally considered three pairs and four of a kind six pairs (as in cribbage). Those are the number of possible pairs you can draw from three and four cards, respectively. By extension, five of a kind would be 15 pairs.
Also, when you say "how many pairs are likely" do you mean what is the expected number (i.e., mean) number of pairs in a hand? Would you like the entire probability distribution, i.e., a list of the chances of 0 pairs, 1 pair, etc.? Wow, so this is potentially more complicated than I thought.
First off, approximations are fine. My goal is to get a sense of the average hand (e.g. "nearly 60% of the time, you will start with one pair and three other distinct cards"). To that end, I would like to treat 3 of a kind as its own thing as with 4 of a kind (let me know if that just complicates things).
So, with a five card hand, I guess I'm asking for approximate odds of the following hands:
5 distinct cards. One pair and three other distinct cards. Two distinct pairs and one other card. Three of a kind and two other distinct cards. Four of a kind and one other card.
Does that make sense?


K Septyn
United States Unspecified Michigan
SEKRIT MESSAGE SSSHHHHHHHHH

robigo wrote: Three of a kind is traditionally considered three pairs and four of a kind six pairs (as in cribbage). Those are the number of possible pairs you can draw from three and four cards, respectively. By extension, five of a kind would be 15 pairs.
I agree with everything except "traditionally considered". But I do suspect the OP is looking for "one pair" hands as are usually found in poker.




Septyn wrote: robigo wrote: Three of a kind is traditionally considered three pairs and four of a kind six pairs (as in cribbage). Those are the number of possible pairs you can draw from three and four cards, respectively. By extension, five of a kind would be 15 pairs. I agree with everything except "traditionally considered". But I do suspect the OP is looking for "one pair" hands as are usually found in poker.
Yeah—changed that to "often considered".




chally wrote: Wow, so this is potentially more complicated than I thought.
First off, approximations are fine. My goal is to get a sense of the average hand (e.g. "nearly 60% of the time, you will start with one pair and three other distinct cards"). To that end, I would like to treat 3 of a kind as its own thing as with 4 of a kind (let me know if that just complicates things).
So, with a five card hand, I guess I'm asking for approximate odds of the following hands:
5 distinct cards. One pair and three other distinct cards. Two distinct pairs and one other card. Three of a kind and two other distinct cards. Four of a kind and one other card.
Does that make sense?
Yep—five of a kind will be possible, too. That's not as many possibilities as I'd thought at first. The sixcard case would include (where each number means "n of a kind of a unique rank"):
111111 21111 2211 222 3111 321 33 411 42 51 6




Here's the results from a quick pilot study of 100,000 draws of 5 cards each. The numbers of each possible hand were:
Five different: 25466 One pair: 51277 Two pair: 12813 Three of a kind: 8608 Full house: 1201 Four of a kind: 618 Five of a kind: 17
(where "full house" is 3 + 2)
The estimated percentage of each kind of hand is, of course, just (the number drawn) / 100,000. The standard error of each estimate is (the square root of the number drawn) / 100,000. For example, the probability for three of a kind is 0.08608 ± 0.00093 (estimate ± SE). That's an error of a couple percent of the estimate, which is fairly decent.
The probability of 5 of a kind isn't very well approximated here, since there were only N instances. However, the exact probability is easy to calculate for 5 of a kind. The number of ways to draw five 1s, 2s or 3s is 16!/(5!11!) = 4368 each, of 4s, 5s or 6s = 12!/(5!7!) = 792, and of 7s, 8s or 9s = 8!/(5!3!) = 56 each. And the total number of ways to draw any hand (including identicallooking ones) is 108!/(5!103!) = 111,469,176. So, the probability of drawing 5 of a kind is
(3 * 4368 + 3 * 792 + 3 * 56) / 111469176 = 15648/111469176 ~ 0.00014
That's within error of the estimate above, 0.00017 ± 0.00004.




jmzero, I bow before your superior probability calculatin' skillz. How did you do that?
Edit: Never mind—I think I've figured it out.
Edit 2: Ah, that's sneaky. Somewhat different, and more elegant, than my solution.


K Septyn
United States Unspecified Michigan
SEKRIT MESSAGE SSSHHHHHHHHH

jmzero wrote: For 5 cards, the odds are:
2111  0.512380427033938
Excellent work, sir. Your number for the onepair case matches what I found. My slow method was to categorize 13, 46, 79 into different classes, then count the number of ways to make a onepair hand for each of 27 different class combinations. I stopped back in to see if anyone beat me to the punch, and thank goodness someone did.


Ben
United States Ann Arbor Michigan

Thanks, all! This is tremendously helpful.




Thank you, and good luck!


Luis Fernandez
Venezuela Caracas Miranda

two word for you
Card Weaving!
Nah that´s trap, but if you place the cards in convenient places in the deck and then shuffle well, they would tend to have very variable positions.



