Sticky Thirteens is a card game which is played at two of my local pubs.

You are given 13 random playing cards. The dealer then takes his own complete, shuffled deck of 52 and turns them over one at a time, calling out the values. When all of your 13 cards have appeared, you win.

Obviously there will be many other players, each with their own random 13 taken from different decks (each 13-pack is marked and distributed by the pub so nobody can surreptitiously substitute their own fixed deck). So, like Bingo, it is a game of chance, you just wait and see who wins. There is no skill element.

Ordinarily the dealer just keeps going until somebody calls. In this case you are assured of an eventual winner, though there may be a tie. However, at one of the pubs I realised that a different game was being played. The dealer turned over all but 25 cards, and if nobody won, the game was over. The jackpot prize was held until the following week, when the dealer would turn over all but 24 cards, and so on, until, naturally, the number got low enough that somebody won.

I studied the game for a while, wondering what the expected winnings were if I were to enter. I tried to figure out some probabilities but didn't have the space to calculate, but alarm bells were ringing.

Eventually I got home and, the lazy bum that I am, punched the figures into an Excel spreadsheet and hit "Fill" to figure out the actual probabilities. Guess what? Nobody is going to win for a long time.

Formula

Let's say that you start with 13 cards in your hand, and the deck has 52 cards in it.

As the game progresses, the deck will be turned over one card at a time. Let's call the number of cards remaining in the deck, A. A is 52 to begin with.

Every time a card is turned over, you may find yourself able to discard a card from your hand. Let's call the number of cards remaining in your hand B. B is 13 to begin with.

Initial conditions are (A,B) = (52,13). You are trying to get to (A,0) for some value of A.

Let's call P(A,B) the probability of arriving at situation (A,B). Since we know we start at (52,13) we know that P(52,13)=1.

Let's say for the sake of argument that the dealer keeps turning cards until all 52 have been revealed, regardless of what happens. By the end of the deck, you know that all the cards in your hand will have been discarded. So, we know that P(0,0)=1 too.

I won't bother you with the details of the tedious calculation which led to this formula, but I found that in general:

                39! 13! (52-A)! A!
P(A,B) = ---------------------------------
          52! (A-B)! B! (13-B)! (39-A+B)!

Table

This gives the following table of probabilities:

These figures are to five decimal places. Several probabilities are so low that they register as 0 on this scale, however, only the blank cells represent actually impossible events.

Simplified formula

The important column is, as I've said, P(A,0), which is the far right side.

Setting B=0 allows us to simplify the formula to:

          39! (52-A)!
P(A,0) = -------------
          52! (39-A)!

Simplified table

The right two columns of this assume 15 players each week.

In other words, at A=25, the odds of a given person winning are around 32,000 to 1! It looks like we're going to be here until we get down to around 14 cards at the earliest, or another 10 weeks. Still, by that time, maybe the jackpot will be large enough to make the investment worthwhile, hmm? I wonder if there's a limit on how many times you can enter each week.

Generalised formula

For a deck of size X and a hand of size Y, the formula is:

                 (X-Y)! Y! (X-A)! A!
P(A,B) = ------------------------------------
          X! (A-B)! B! (Y-B)! ((X-Y)-(A-B))!