Problems

A coin is tossed repeatedly. The probability that a head appears is \(p\) and the probability that a tail appears is \(q = 1 - p\).

1. A and B play a game. The game ends if two successive heads appear, in which case A wins, or if two successive tails appear, in which case B wins. Show that the probability that the game never ends is \(0\). Given that the first toss is a head, show that the probability that A wins is \(\dfrac{p}{1 - pq}\). Find and simplify an expression for the probability that A wins.
2. A and B play another game. The game ends if three successive heads appear, in which case A wins, or if three successive tails appear, in which case B wins. Show that \[\mathrm{P}(\text{A wins} \mid \text{the first toss is a head}) = p^2 + (q + pq)\,\mathrm{P}(\text{A wins} \mid \text{the first toss is a tail})\] and give a similar result for \(\mathrm{P}(\text{A wins} \mid \text{the first toss is a tail})\). Show that \[\mathrm{P}(\text{A wins}) = \frac{p^2(1-q^3)}{1-(1-p^2)(1-q^2)}.\]
3. A and B play a third game. The game ends if \(a\) successive heads appear, in which case A wins, or if \(b\) successive tails appear, in which case B wins, where \(a\) and \(b\) are integers greater than \(1\). Find the probability that A wins this game. Verify that your result agrees with part (i) when \(a = b = 2\).

1. Two people adopt the following procedure for deciding where to go for a cup of tea: either to a hotel or to a tea shop. Each person has a coin which has a probability \(p\) of showing heads and \(q\) of showing tails (where \(p+q = 1\)). In each round of the procedure, both people toss their coins once. If both coins show heads, then both people go to the hotel; if both coins show tails, then both people go to the tea shop; otherwise, they continue to the next round. This process is repeated until a decision is made. Show that the probability that they make a decision on the \(n\)th round is $$(q^2 + p^2)(2qp)^{n-1}.$$ Show also that the probability that they make a decision on or before the \(n\)th round is at least $$1 - \frac{1}{2^n}$$ whatever the value of \(p\).
2. Three people adopt the following procedure for deciding where to go for a cup of tea: either to a hotel or to a tea shop. Each person has a coin which has a probability \(p\) of showing heads and \(q\) of showing tails (where \(p + q = 1\)). In the first round of the procedure, all three people toss their coins once. If all three coins show heads, then all three people go to the hotel; if all three coins show tails, then all three people go to the tea shop; otherwise, they continue to the next round. In the next round the two people whose coins showed the same face toss again, but the third person just turns over his or her coin. If all three coins show heads, then all three people go to the hotel; if all three coins show tails, then all three people go to the tea shop; otherwise, they go to the third round. Show that the probability that they make a decision on or before the second round is at least \(\frac{7}{16}\), whatever the value of \(p\).

In a game, a player tosses a biased coin repeatedly until two successive tails occur, when the game terminates. For each head which occurs the player wins \(\pounds 1\). If \(E\) is the expected number of tosses of the coin in the course of a game, and \(p\) is the probability of a head, explain why \[ E = p \l 1 + E \r + \l 1 - p \r p \l 2 + E \r + 2 \l 1 - p \r ^2\,, \] and hence determine \(E\) in terms of \(p\). Find also, in terms of \(p\), the expected winnings in the course of a game. A second game is played, with the same rules, except that the player continues to toss the coin until \(r\) successive tails occur. Show that the expected number of tosses in the course of a game is given by the expression \(\displaystyle {1 - q^r \over p q^r}\,\), where \(q = 1 - p\).

Calamity Jane sits down to play the game of craps with Buffalo Bill. In this game she rolls two fair dice. If, on the first throw, the sum of the dice is \(2,3\) or \(12\) she loses, while if it is \(7\) or \(11\) she wins. Otherwise Calamity continues to roll the dice until either the first sum is repeated, in which case she wins, or the sum is \(7\), in which case she loses. Find the probability that she wins on the first throw. Given that she throws more than once, show that the probability that she wins on the \(n\)th throw is \[ \frac{1}{48}\left(\frac{3}{4}\right)^{n-2}+\frac{1}{27}\left(\frac{13}{18}\right)^{n-2}+\frac{25}{432}\left(\frac{25}{36}\right)^{n-2}. \] Given that she throws more than \(m\) times, where \(m>1,\) what is the probability that she wins on the \(n\)th throw?

A patient arrives with blue thumbs at the doctor's surgery. With probability \(p\) the patient is suffering from Fenland fever and requires treatment costing \(\pounds 100.\) With probability \(1-p\) he is suffering from Steppe syndrome and will get better anyway. A test exists which infallibly gives positive results if the patient is suffering from Fenland fever but also has probability \(q\) of giving positive results if the patient is not. The test cost \(\pounds 10.\) The doctor decides to proceed as follows. She will give the test repeatedly until either the last test is negative, in which case she dismisses the patient with kind words, or she has given the test \(n\) times with positive results each time, in which case she gives the treatment. In the case \(n=0,\) she treats the patient at once. She wishes to minimise the expected cost \(\pounds E_{n}\) to the National Health Service.

1. Show that \[ E_{n+1}-E_{n}=10p-10(1-p)q^{n}(9-10q), \] and deduce that if \(p=10^{-4},q=10^{-2},\) she should choose \(n=3.\)
2. Show that if \(q\) is larger than some fixed value \(q_{0},\) to be determined explicitly, then whatever the value of \(p,\) she should choose \(n=0.\)