# MATHEMATICS 154, SPRING 2010 PROBABILITY THEORY Outline #3 (Combinatorics, bridge, poker)

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1 Last modified: February, 00 References: MATHEMATICS 5, SPRING 00 PROBABILITY THEORY Outline # (Combinatorics, bridge, poker) PRP(Probability and Random Processes, by Grimmett and Stirzaker), Section Ex(One Thousand Exercises in Probability, by Grimmett and Stirzaker), page 5. Web pages on poker and bridge from Principles of counting In the following items, use these principles of counting, taken for granted in PRP. Principle : Multiply for sequential counting. If we are forming a set of n-tuples where there are k choices for the first element x, k for x, and so on, the number of n-tuples x x...x n in the set is k k...k n Principle : Divide to correct systematic overcounting. If M different n-tuples all correspond to the same element in our sample space, count the n-tuples and then divide by the number of times M that each was overcounted. Principle : Divide and Conquer. To count the number of elements in a union of disjoint subsets, count each subset and sum the results. Principle : Subtract off special cases. To count the number of elements in a difference A \ B where B A, count each set and take the difference. Classic example of principles and : counting the number of k-element subsets of an n-element set. First count the k-tuples whose elements are all different: n(n )(n )...(n k + ). Then divide by k! to correct for overcounting. The result is the familiar combinations formula ( ) n n! = k k!(n k)!

2 Using binomial coefficients whenever possible is the safest way to avoid accidental overcounting. For the following topics, the easiest way to get numerical answers is to use Maple or Mathematica. Any Harvard undergraduate can download a copy for free. For details, go to and follow instructions.. Counting poker hands Count the number of ways to get each of the following types of 5-card poker hands, using a deck of 5 cards with cards of each of ranks, and use this to calculate the probability of being dealt such a hand. of a kind(four cards of one rank, the fifth of a different rank) a full house(three cards of one rank, two of another) of a kind(three cards of one rank, two others of different ranks) Solutions: How many distinct 5-card poker hands can be dealt from a 5-card deck? Answer: Select 5 cards sequentially: the number of ways to do this is But this generates each distinct hand 5! = 0 times. So the number of distinct hands is ( 5 5 ) = 5! 7!5! =, 598, 960 How many distinct ways are there to get of a kind, and what is the probability? choices for the rank of the, 8 choices for the other card. P = 8, 58, 960 = A full house ( of one rank, of another) choices for the first rank (with cards), for the suit that is missing. choices for the second rank (with cards), ( ) = 6 for the pair of suits that have the second rank. 6 P =, 58, 960 = Three of a kind ( of one rank, the other two are of different ranks) ( choices for the first rank (with cards), for the suit that is missing. ) = 66 choices for the pair of ranks of the other two cards; 6 choices for the suit of each. So the number is 66 6 = 5, 9 and P = 5, 9, 58, 960 = 0.085

3 . Bridge problems A bridge hand consists of cards from a 5-card deck. Count the number of bridge hands with 6 spades, hearts, diamonds, and club. Count the number of bridge hands with 6--- suit distribution (6 cards in the longest suit, in the second-longest, in the third-longest) Count the number of bridge hands with --- or --- suit distribution, and show that the former has a higher probability. Solution: How many ways are there to select 6 cards from the spades? ( ) =! 6 7! 6! How many distinct hands have 6 spades, hearts, diamonds, club? Apply the same analysis to each suit in turn: ( )( )( )( ) =!!!! 6 7! 6! 9!!!!!! How many distinct hands have a 6--- distribution? Multiply the preceding number by the number of ways to choose the 6-card suit, the -card suit, etc, which is!=. How many distinct hands have a --- distribution? There are ways to select the -card suit followed by ways to select the -card suit, so we will multiply by. ( N = )( )( )( ) =!!!! 9!! 9!! 0!!!! How many distinct hands have a --- distribution? Now there are ways to select the -card suit, so we will multiply only by. ( N = The ratio is )( )( )( ) =!!!! 9!! 0!! 0!! 0!! P = N 0!! 0!! = 5 P N 9!!!!

4 . Dividing up a set with two type of objects Suppose we have a set S with n elements, m of one type and n m of the other. As a concrete example, suppose that Lisa has n = 6 Dunkin Munchkins, of which m = are chocolate and the remaining n m = are plain. She chooses n = of them at random and puts them in a bag for her son Thomas. The remaining three go into a bag for her daughter Catherine. In general we divide up S at random into set S and set S, each with n elements. The number of different ways of doing this is equal to the number of ways of selecting the n elements of S, namely N = ( ) n = (n)! n n!n! We are interested in the probability that k of the special objects (chocolate Munchkins) end up in set S. Assume that k m, k n, and k m n. The number of ways of selecting k special objects and n k non-special objects for set is ( )( ) m n m m! (n m)! M = = k n k k!(m k)! (n k)!(n + k m)! If we assume that each way of selecting the elements of S is equally likely, the ratio M gives the probability that set S N contains precisely m special objects. As a practical matter, it is much easier to repeat this reasoning in special cases than to memorize these formulas!

5 5. Waring s Theorem This useful result is proved cleverly in chapter 5 of PRP. Here is a straightforward proof inspired by the solution to problem on page of PRP (solutions on page of 000Ex). It avoids the use of the result from problem. We have n events A i, i = {,,...n}. Examples: Toss coins; A i is the ith coin comes up heads. Place 5 cups of different colors randomly on 5 colored saucers; A i is the ith cup (blue) lands on the blue saucer. Buy 6 boxes of cereal, each containing a bust of a recent president; A i is at least i of your 6 busts is a Barack Obama. Let S be a subset of {,,...n} like the first and fourth coin tosses the red, blue, and yellow cups Clinton and Bush Define events: A S = all A i for i S occur, and maybe others too. B S = no event A i for i outside S occurs. So A S B S = all A i for i S occur, and no others. This is one specific way that precisely k events can occur. For different choices of S, such events are disjoint. Sum over S to get the probability of their union. N k = precisely k events occur. In many cases, P(A S ) is easy to calculate, but P(N k ) is a pain to determine. Waring s theorem relates the two: P(N k ) = ( ) k + P(A S ) k Proof: S =k T =(k+) ( ) k + P(A T )+ k T =(k+) P(A T ) Start, as usual, with a disjoint union: P(A S ) = P(A S B S ) + P(A S B c S) 5

6 P(A S B S ) = P(A S ) P(A S j / S A j ) P(A S B S ) = P(A S ) P( j / S(A S A j )). Now apply generalized inclusion-exclusion to the union in the right term: P(A S B S ) = P(A S ) j / S P(A S A j ) + j<k / S P(A S A j A k ) Finally, sum over all subsets of {,,...n}, for sets of size k, k+, k+,, n to get the theorem: P(N k ) = P(A S B S ) = S =k To avoid confusion, I used the generic symbol T for sets with more than k elements. Each set T with k + m elements will arise in the sum ( ) k+m k times, once from each of its k-element subsets that can serve as S. For the problem of placing R, B, Y, and G cups on matching saucers, the probability of getting exactly one cup on the saucer of the same color is, according to Waring s theorem, ( ) ( ) ( ) P(N ) = P(A ) 6 P(A A )+ P(A A A ) P(A A A A ) P(N ) = + = + 6 = 6

7 6. Waring s theorem applied to coin tossing What is the probability of getting exactly one head when you toss three fair coins? Fundamental events: A i is the event that coin i comes up heads. Clearly P(A i ) =. Since the tosses are independent events, P(A i A j ) = for i j, and P(A A A ) =. 8 Since the probability of all heads is the same for any set of coins of a given size, we need only consider probabilities like P(A ) and P(A A ). The summation required by Waring s theorem can be accomplished by multiplying by the number of relevant events. N is the event that exactly one coin comes up heads. Waring says ( ) P(N ) = P(A ) P(A A ) + P(N ) = = 8. ( ) P(A A A ). What is the probability of getting exactly two heads when you toss three fair coins? N is the event that exactly two coins comes up heads. Waring says P(N ) = P(A A ) 7. Waring s theorem applied to cards P(N ) = 8 = 8. ( ) P(A A A ). What is the probability of getting exactly one spade when you select two cards from a shuffled deck? A i is the event that card i is a spade. In this case P(A i ) = for each card. However, P(A A ) = 5 5 = 7. N is the event that exactly one card is a spade. Waring says P(N ) = P(A ) ( ) P(A A ). P(N ) = 7 =. 7

8 8. Waring s theorem applied to dice What is the probability of getting exactly one 6 when you roll three fair dice? A i is the event that die i comes up 6. Clearly P(A i ) =. Since the tosses are independent events, 6 P(A i A j ) = for i j, and P(A 6 A A ) =. 6 Since P(A ) = P(A ) = P(A ) and P(A A ) = P(A A ) = P(A A ), we need only consider probabilities like P(A ) and P(A A ). The summation required by Waring s theorem can be accomplished by multiplying by the number of relevant events. N is the event that exactly one die comes up 6. Waring says ( ) P(N ) = P(A ) P(A A ) + P(N ) = = ( ) P(A A A ). N is the event that exactly two dice come up 6. Waring says P(N ) = P(A A ) ( ) P(A A A ). P(N ) = 6 6 = 5 6. N 0 is the event that no die comes up 6. Waring says ( ) ( ) P(N 0 ) = P(A ) P(A A ) P(N 0 ) = = 5 6. ( ) P(A A A ). 0 8

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