Ten players participate in the first round of a tennis tournament: 2 females and

Question

Ten players participate in the first round of a tennis tournament: 2 females
and 8 males. Five single matches are fixed at random by successively drawing,
without replacement, the names of all 10 players from an urn: the player drawn
first plays against the one whose name comes up second, the third against the
fourth, etc.
a. What is the probability that there will not be a single match involving
two female players? Is this probability smaller, equal to, or larger than the
corresponding probability with 20 females and 80 males?
b. Try to answer the general case in which there are 2n players, of whom
2 = k = n are female. What is the probability p(k, n) that among the n
matches there will not be a single one involving two female players?

Explanation / Answer

Let us say a given player is at rank i if his or her name is the ith one drawn from the urn. The total number of different ways to distribute the k female players among the 2n ranks is 2n k . Clearly, not all of them are favorable; for example, if a female player is drawn first (rank 1) and another female player is drawn second (rank 2), then the condition mentioned in the problem is not satisfied. How many different favorable ways are there to distribute the k females players among the 2n ranks? Imagine we put together successive pairs of ranks, namely (1, 2), (3, 4), . . . , (2n - 1, 2n). Of these n pairs we may select k to associate exactly one female player with each of them; the number of ways to select these k pairs from a total of n pairs is n k. However, within each of the k pairs chosen, the respective female may be placed either at the odd or at the even rank of that pair, yielding a total of 2k · n k different ways to distribute k female players among 2n ranks such that each of the n pairs has at most one female. p(k, n) then is the ratio of the favorable over all possible cases, i.e., p(k, n) = 2k · n k 2n k = 2n-k n 2n n · 2k Note that the formula also covers the trivial cases of k = 0, 1. For a constant fraction of female players p(k, n) declines rapidly as n increases. For example, with k = 2 females among 2n = 10 players, chances are 8/9 = 88.9% that there will be no match involving only females (i.e., that those two females will not have to play against each other). However, with k = 20 females among 2n = 100 players (thus, again a fraction of 20% females), the corresponding probability drops to about 9.2%. That is, with a probability of more than 90% there will among all 50 matches be at least one match with two female players.

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