Cards, decks and hands

Transcription

1 Cards, decks and hands Blackjack The cards (in the mathematical sense) are the cards, aces have weight, cards from 2 to 9 have their point value as weight, 0s, jacks, queens and kings have weight 0, so d d 2 =... = d 9 =, d 0 = 4, d r = 0 for r> 0. (In this model we do not distinguish between suits.) Therefore the 2-variable hand-enumerator is H 9 x r y r x 0 y x y x 2 y x 3 y x 4 y x 5 y x 6 y x 7 y x 8 y x 9 y x 0 y 4 4 To find the number of hands of value 8 consisting of 4 cards, we need to extract the coefficient of x 8 y 4. SeriesCoefficient H, x, 0, 8, y, 0, 4 5 The money-changing problem Question: In how many ways can one pay a certain amount using the given coins? E.g., in how many way can one pay 37p using p, 2p, 5p, 0p coins using exactly 2 coins or using an arbitrary number of coins? (Each coin can be used arbitrarily many times.) Now the cards are the coins, and the weight of each coin is its value. So in this example, d d 2 d 5 = d 0 =, and d r 0 for other values of r. Therefore the 2-variable

2 2 Cards.nb hand-enumerator is H x y x 2 y x 5 y x 0 y x y x 2 y x 5 y x 0 y To find the number of ways one can pay 37p using exactly 2 coins, we need to extract the coefficient of x 37 y 2. SeriesCoefficient H, x, 0, 37, y, 0, 2 7 If we do not care about the number of coins used, we calculate the -variable hand-enumerator by substituting y= into H, and then extract the coefficient of x 37. SeriesCoefficient H. y, x, 0, BT payphones The payphones accept 5p, 0p, 20p and 50p coins, and to make a call you need to insert at least 60p using at most 4 coins. The cards are the coins and the weight of each coin is its value, just as in the previous example. Therefore d 5 d 0 d 20 = d 50 =, and d r 0 for other values of r. The 2- variable hand-enumerator is H x 5 y x 0 y x 20 y x 50 y x 5 y x 0 y x 20 y x 50 y The number of possible combinations of coins is the sum of the coefficients x n y k for n 60 and k 4. First we discard all the terms of degree 5 in y, then the terms of degree <60 in x.

3 Cards.nb 3 J Normal Series H, y, 0, 4 x 5 x 0 x 20 x 50 y x 0 x 5 x 20 x 25 x 30 x 40 x 55 x 60 x 70 x 00 y 2 x 5 x 20 x 25 2 x 30 x 35 x 40 x 45 x 50 2 x 60 x 65 x 70 x 75 x 80 x 90 x 05 x 0 x 20 x 50 y 3 x 20 x 25 x 30 2 x 35 2 x 40 x 45 2 x 50 x 55 x 60 2 x 65 2 x 70 x 75 3 x 80 x 85 x 90 x 95 x 00 2 x 0 x 5 x 20 x 25 x 30 x 40 x 55 x 60 x 70 x 200 y 4 K J Normal Series J, x, 0, 59 Expand x 60 y 2 x 70 y 2 x 00 y 2 2 x 60 y 3 x 65 y 3 x 70 y 3 x 75 y 3 x 80 y 3 x 90 y 3 x 05 y 3 x 0 y 3 x 20 y 3 x 50 y 3 x 60 y 4 2 x 65 y 4 2 x 70 y 4 x 75 y 4 3 x 80 y 4 x 85 y 4 x 90 y 4 x 95 y 4 x 00 y 4 2 x 0 y 4 x 5 y 4 x 20 y 4 x 25 y 4 x 30 y 4 x 40 y 4 x 55 y 4 x 60 y 4 x 70 y 4 x 200 y 4 The answer is the sum of the coefficients of the above polynomial, which we can obtain easily by substituting x=y= into it. K. x, y 38 Partitions Let p(n) be the number of ways in which a non-negative integer can be written as the sum of positive integers, without regard to the order of the terms. E.g., 4=3+=2+2=2++=+++, so p(4)=5. This can be considered as a special case of the money changing problem when we have one coin for each positive integer, so d r for each positive integer r. As we have not specified the number of terms, we write down the -variable handenumerator. H r x r QPochhammer x, x To find the number of partitions of any particular number

4 4 Cards.nb n, we need to extract the coefficient of x n from the Maclaurin series of this function. For example, if n=200, SeriesCoefficient H, x, 0,

5 Cards.nb 5 Restricted problems Blackjack The previous model assumed that each card may be used arbitrarily many times which corresponds to playing with infinitely many decks. The right model for calculating the optimal strategy for a single deck of card is to distinguish the suits and to allow each card to be used at most once, so d d 2 =... = d 9 = 4, d 0 = 6, d r = 0 for r> 0, and the restriction set is W={0,}. Therefore the 2-variable hand - enumerator is H 9 x r y 4 x 0 y 6 r x y 4 x 2 y 4 x 3 y 4 x 4 y 4 x 5 y 4 x 6 y 4 x 7 y 4 x 8 y 4 x 9 y 4 x 0 y 6 To find the number of hands worth 8 points consisting of 4 cards, we again need to extract the coefficient of x 8 y 4. SeriesCoefficient H, x, 0, 8, y, 0, The money-changing problem Let us consider the problem of paying 37p using p, 2p, 5p, 0p coins using exactly 2 coins or using an arbitrary number of coins with the additional restriction that at most 5 of each coin may be used. d d 2 d 5 = d 0 =, and d r 0 for other values of r as

6 6 Cards.nb before and W={0,,2,3,4,5}. Therefore the 2-variable hand-enumerator is H x 6 y 6 x 2 y 6 x 30 y 6 x 60 y 6 x y x 2 y x 5 y x 0 y x 6 y 6 x 2 y 6 x 30 y 6 x 60 y 6 x y x 2 y x 5 y x 0 y To find the number of ways one can pay 37p using exactly 2 coins with the restriction, we need to extract the coefficient of x 37 y 2. SeriesCoefficient H, x, 0, 37, y, 0, 2 2 If we do not care about the number of coins used, we calculate the -variable hand-enumerator by substituting y= into H and then extract the coefficient of x 37. SeriesCoefficient H. y, x, 0, 37 2