Sunday, January 19, 2014
At first glance, this appears to be two pretty ordinary fouled hands in a game of open-face Chinese poker.
But look again. There's something peculiar about the hand on the right. Go ahead--take a closer look. I'll wait.
Do you see it? She got 13 cards with no pair--a 13-card straight!
If you dealt out 13 cards from a well-shuffled deck, what is the probability that you'd get that result? I don't know, but let's figure it out.
The first card can obviously be anything. To state the trivially obvious, the probability of getting a non-pairing first card is 1.00, or 100%. The second card can be anything except a pair to the first. So with 51 cards left in the deck, 3 kill our hopes for the Big Straight, and 48 keep it going. The probability of a card that doesn't pair the first one, then, is 48/51, or 0.941, or 94.1%.
The third card must not pair either of the first two. There are 50 cards left in the deck, of which 6 will create a pair and 44 will not. So the probability of getting a non-pairing card is 44/50, or 0.880, or 88.0%.
We continue this pattern. When we're ready to deal the 13th card, there are 40 cards left in the deck, of which just 4 will complete our lovely 13-card straight, while 36 will pair one of the first 12 cards now on the board. The probability of hitting one of our "outs" is therefore 4/40, or 0.100, or 10.0%.
Because every one of the 13 events we've described must occur in order to end up with the 13-card-no-pair hand, we multiply all of the individual probabilities together: 1.00 x 0.941 x 0.880 ... x 0.100. The final product, if I didn't screw up the calculation, is 0.000106, or 0.0106%. To put it in the conceptually easiest terms, this will occur about 1 in 9460 hands. If I've ever seen this happen before in an OFC game, I didn't notice it.
You may be tempted to think that knowing 13 other cards along the way (i.e., the ones dealt to me in the picture above) changes this probability, but it doesn't. It would change our ongoing estimate, card by card, as to whether this hand will play out as a 13-card straight. But the question I've really asked is how often the deck will have been arranged by a random shuffle so the 13 cards destined to go to player X will contain no pairs. When asked that way, it makes no difference what happens to the other 39 cards, whether they are seen or unseen. The answer does change if instead we ask the probability of a shuffled deck yielding a 13-card straight to somebody when two (or three, or four) players are dealt in, but it makes my head hurt to think about how to calculate that, so I'll leave it to somebody else.