The Schnapsen Log
All the Aces (appendix 2)
Martin Tompa
We want to count the total number of possible hands Hans could hold. There are 7 cards you haven’t seen (since you have seen ♣K, the 8th card, in Hans’s hand) of which any 4 could be in Hans’s hand to accompany his ♣K. There are 7 cards that could be at the top of the stock, 6 that could be next in the stock, and 5 that could be next. This means that the number of arrangements of the face-down cards in the stock is 7⋅6⋅5. Once the stock cards are chosen, Hans’s hand is completely determined to be the remaining 4 cards plus ♣K. But since we only care what cards are in Hans’s hand, it doesn’t matter in what order the stock’s face-down cards occur. The number of possible reorderings of those 3 cards is 3⋅2⋅1, since any of 3 cards can be moved to the top, any of the other 2 cards can be moved to the next position, and the remaining 1 card must be moved to the last position. Thus, there are (7⋅6⋅5)/(3⋅2⋅1) = 35 possible hands Hans could hold.
© 2012 Martin Tompa. All rights reserved.
