Thread: Hold'em Probability Puzzler View Single Post
#3
 PseudoPserious Senior Member Join Date: Oct 2002 Posts: 151 Possible Solution

I'll give it a shot...

Let's call N the number of cards you've already seen. Basically, we're looking for the average number of hands it takes to go from N=0 to N=52.

Call A(N) the average number of hands it takes to see a new card, given that you've already seen N cards.

Call P(N) the probability of having seen exactly N cards at some point during the game.

A(N)*P(N) is the expected number of hands you'll be dealt for each value of N.

If you sum A(N)*P(N) over N=0 to 51, you'll get the average number of hands it takes to see the entire deck.

Tackling A(N) first:

Given that you've seen N cards, the probability of being dealt two cards you've already seen is (N/52)*((N-1)/51). Thus, the probability on any one deal of seeing a new card is:

M(N) = 1 - (N/52)*((N-1)/51)).

We're looking for the average number of independent trials (hands) it takes for us to see our first success (our first hand with a new card). This is the geometric distribution. The mean of the geometric distribution with probability of success p is 1/p.

So, A(N) = 1/M(N)

Now for P(N):

You will NOT see exaclty N cards at some time during the game if and only if (a) you have seen exactly N-1 cards and (b) the next time you are dealt an unseen card, you are dealt two unseen cards instead of just one.

We've already defined the probability of (a) as P(N-1).

The probability of (b) is the fraction of times that you receive two unseen cards when you are dealt at least one unseen card, given that you've seen N-1 cards. We'll call this T(N-1).

Given that you've seen (x) cards, on any one deal the probability of being dealt two unseen cards is P2(x) = ((52-x)*(51-x))/(52*51). The probability of being dealt exactly one unseen card is P1(x) = 2x(52-x)/(52*51).

So, T(N-1) = P2(N-1) / (P1(N-1) + P2(N-1))

Thus, we can find the chance of skipping a value of N by P(N-1)*T(N-1). This gives us an iterative formula for the chance of seeing exactly N cards:

P(N) = 1 - P(N-1)*T(N-1)

Along with P(0) = 1 (since we have seen no cards before the first deal) and P(1) = 0 (since the first deal will always give us two unseen cards), this allows us to calculate P(N) for all values of N.

Using the above definitions, if you sum A(N)*P(N) from N=0 to 51, you get 117.09.

So, I'd have to say just over 117 hands or so...

Cya,
PP

P.S. A quickie computer sim of 20,000 games gave an average of 117.65 hands.