Jon Perry asked:
split a pack of cards into 2*26.
In each pack, count the number of numbers where all 4 of the suits are present, e.g. in pack A, increase the count if say 3C, 3D, 3H and 3S are all present.
What is the expected final count when both packs have been fully examined?
First calculate the probability that, say, the first four cards in the deck are all of the same rank. That is, the second matches the first, the third matches the first two, and the fourth matches the first three; probability is 3/51 * 2/50 * 1/49. Now you care about four same-rank cards in lots of different places, not just the top four places in the deck. In fact, the number of sets of four locations which you want to count is just 2 * binomial(26,4) --- any four slots all in the first half or all in the second half of the deck. Expected values add, even if things aren't independent (yay). So we just multiply the probability times the count, and get E(# foursomes in same half deck) = 3/51 * 2/50 * 1/49 * 2 * 14950 = 1196 / 833, which is around 1.4. --Michael Kleber kleber@brandeis.edu