Homework 4 Solutions - Statistics Department



Homework 4 Solutions

1.

Let [pic] be the probability that Andre wins a set against his father, and let [pic] be the probability that Andre wins a set against the club champion. Clearly, because the club champion is a better player than Andre’s father, we have [pic].

For sequence (a) father-champion-father, the probability of Andre winning at least 2 tennis sets in a row is just the probability of him winning the first 2 sets plus the probability of him losing the 1st set and winning the 2nd and 3rd sets:

[pic]

For sequence (b) champion-father-champion, the probability of Andre winning at least 2 tennis sets in a row is instead [pic].

Comparing the probabilities for (a) and (b), note that because [pic], it follows that [pic]; this means that Andre would have a higher probability of winning at least 2 sets in a row if he chose to play the champion-father-champion sequence.

2. Ross 3.80 p. 122

a) Let [pic] be the event that A plays in exactly i contests, where i can be any integer value between 1 nd n, inclusive. In order for A to play in exactly i contests, A must lose the i-th contests for values of i less than n ([pic]). Why does this not hold when [pic]? Because once A reaches the n-th contest, there are none after, so A will play in exactly n contests regardless of whether he wins or loses the n-th contest. In other words, the probability that A plays in exactly n contests is just the probability that A wins all contests up to and including contest [pic]. These statements give us [pic] as

[pic].

b) Let E denote the event that contestants A and B ever play each other. We can obtain this probability by conditioning on the events [pic]. In other words, we know that

[pic].

Note that [pic] means A faced i different contestants, and each of the [pic] other contestants have an equal chance of ever playing with A; hence, [pic].

Using what we determined for [pic] from (a), we have

[pic]

c) Let [pic] be the probability that A and B meet in a contest with [pic] players. Now, let F be the event that A and B play in the first contest; then [pic]. But since we know that the probability that A and B meet in the first contest is just [pic], we get that

[pic].

Now, let G be the event that both A and B advance to the second round of this contest (i.e., the event that both win their first match). Then we have that

[pic]

In other words, the probability that they meet given that they both advance to the second round and don’t meet in the first round is just the probability that out of [pic] possible contestants, we see A and B meet ([pic] because the number of contestants left after 1 round has been completed is half the original number). Hence, we have that

[pic].

To verify that this holds, let us plug in [pic] (and hence [pic]) to verify that this identity holds:

[pic]

(Checks out)

d) Note that every game played yields 1 loser. Also, with [pic] contestants and only 1 winner, there must be [pic] losers; hence, we require [pic] games in order to obtain [pic] losers.

e) Let [pic] denote the event that A and B play each other in game i, where [pic]. Note that there are [pic] possible pairings that can occur. Also, since players are equally skilled, the 2 players in game i are equally likely to be any of the [pic] pairs. Hence, it follows that [pic].

f) Since the events [pic] are mutually exclusive, then the event E is just the union of the events [pic]; hence,

[pic].

3.

Let [pic] denote the probability of having no two consecutive heads in a sequence of k tosses of a fair coin. Then let [pic] be the probability of having at least 2 heads in a row if there are k tosses remaining, i consecutive heads have already appeared, and there has not already been a run of at least 2 heads prior to the current run of i heads in a row. A recursive formula using [pic]’s will give the complement to [pic].

Adapting the FT shooting recursive formula from the addendum to the notes for a coin-tossing scenario, we have that

[pic]

with boundary conditions

[pic] and [pic].

Then the probability that we have no runs of at least 2 consecutive heads in a sequence of k tosses of a fair coin is just [pic].

The following R function will be used to calculate [pic] for k = 5, 10, and 25 by calculating [pic].

bmat ................
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