Phil Pilgrim (PhiPi)

02-13-2012, 12:40 AM

I play poker with a group of friends on a fairly regular basis. We've been playing together for more than thirty years. In that time, we've seen some odd runs of the cards, but nothing like what transpired in our most recent get-together. One of the games we play involves dealing out cards two-at-a-time, face up. This time, however, the dealer dealt out six pairs in a row! The obvious question was, "What are the odds of that ever happening?" So, being the computer nerd of the group, I was elected to compute those odds.

In order to simplify the problem somewhat, I've restated it as, "What is the probability that, starting from a shuffled deck of 52 cards, and dealing two cards at a time from the top, that the first six duets will be consist of six different pairs?" We can start by computing the probability for the first pair. Since each rank (ace through king) has four suits, there are six distinct and equally-probable pairs that can be formed from them. This is given by the formula for combinations of n things, taken m at a time:

C(n, m) = n! / [m! (n - m)!)]

So the total number of first pairs from all 52 cards is 13 * 6, or 78. The total number of distinct two-card combinations, pairs and non-pairs included, is

C(52, 2) = 52! / (2! 50!) = 52 * 51 / (2 * 1) = 1326

So the probability that the first two cards dealt out will be a pair is

78 / 1326 = 0.0588235 or 5.88%

For the second duet, assuming the first was a pair, the of ranks available to form a different pair now number 12, and the number of cards remaining is 50. So the probability that it will be a pair is:

12 * 6 / [50! / (2! 48!)] = 72 / 1225 = .05877551

The probability that the first two duets will be different pairs is the product of the two probabilities computed thus far, or .00345738.

We can continue in this fashion for the next four duets dealt and arrive at the final answer:

(78 / 1326) * (72 / 1225) * (66 / 1128) * (60 * 1036) * (54 / 946) * (48 / 861) = 3.72833E-8

That's one chance in 26.8 million. I don't think we'll ever see that happen again.

-Phil

In order to simplify the problem somewhat, I've restated it as, "What is the probability that, starting from a shuffled deck of 52 cards, and dealing two cards at a time from the top, that the first six duets will be consist of six different pairs?" We can start by computing the probability for the first pair. Since each rank (ace through king) has four suits, there are six distinct and equally-probable pairs that can be formed from them. This is given by the formula for combinations of n things, taken m at a time:

C(n, m) = n! / [m! (n - m)!)]

So the total number of first pairs from all 52 cards is 13 * 6, or 78. The total number of distinct two-card combinations, pairs and non-pairs included, is

C(52, 2) = 52! / (2! 50!) = 52 * 51 / (2 * 1) = 1326

So the probability that the first two cards dealt out will be a pair is

78 / 1326 = 0.0588235 or 5.88%

For the second duet, assuming the first was a pair, the of ranks available to form a different pair now number 12, and the number of cards remaining is 50. So the probability that it will be a pair is:

12 * 6 / [50! / (2! 48!)] = 72 / 1225 = .05877551

The probability that the first two duets will be different pairs is the product of the two probabilities computed thus far, or .00345738.

We can continue in this fashion for the next four duets dealt and arrive at the final answer:

(78 / 1326) * (72 / 1225) * (66 / 1128) * (60 * 1036) * (54 / 946) * (48 / 861) = 3.72833E-8

That's one chance in 26.8 million. I don't think we'll ever see that happen again.

-Phil