Peter and Paula play a simple game of dice, as follows. Peter keeps throwing the

Question

Peter and Paula play a simple game of dice, as follows. Peter keeps throwing
the (unbiased) die until he obtains the sequence 1 - 1 in two successive
throws. For Paula, the rules are similar, but she throws the die until she
obtains the sequence 1 - 2 in two successive throws.
a. On average, will both have to throw the die the same number of times?
If not, whose expected waiting time is shorter (no explicit calculations are
required)?
b. Derive the actual expected waiting times for Peter and Paula.

Explanation / Answer

a. Peter’s sequence will usually consist of a certain number of 1s, each ofwhich is followed by a number different from 1 with p = 5/6. If, after obtaininga 1, he fails to achieve the desired 1 1 run, then the number thrown wasnecessarily different from 1 and, therefore, cannot constitute the potentialbeginning of a potential 1 1 run. This is different for Paula: if she fails tothrow a 2 after an initial 1, then she may do so by throwing another 1, whichin turn could be the start of a potential 1 2 run. Therefore, the expectedwaiting time will be somewhat shorter for Paula.

b. As for Peter, let us assume that in each throw a 1 occurs with a fixedprobability of p. We are looking for the expected waiting time to see two 1sin a row for the first time; call this expectation W.

Consider the outcome of Peter’s first throw. With probability 1 p, theoutcome is different from 1, that is, one of the numbers from 2 to 6. In thiscase (shown as the upper branch in Figure 3.1) one trial is used up, and thestate of the game after the first trial is as it was before the game started. Thismeans that the additional expected waiting time — namely, in addition to the

first trial — is just as large as the original expected waiting time was; that is,it equals W. Therefore, if the first case applies (which has probability 1 p),then the conditional expectation of the total waiting time is equal to 1 +W.With probability p, Peter’s first trial results in a 1, in which case we needto consider the outcome of the second trial. If his second trial results in a 1too, then he has achieved two 1s in a row, and the game is over after onlytwo trials. This case is shown as the lowest branch in Figure 3.1, and itsoverall probability is equal to p2. Alternatively, his second trial might notyield another success, in which case he is again in the same state as he wasright before the start of the game, except that he has already spent two trials.This case is shown as the middle branch in Figure 3.1. The expected numberof throws in this case is 2+W, and its overall probability is equal to p· (1p).These three cases exhaust all opening possibilities, and they are mutuallyexclusive. Therefore, we can obtain an equation for W by conditioning on thesethree cases and weighing each conditional expectation of W by the probabilityof each case. That is,W = (1+W) · (1 p) + (2+W) · p(1 p) + 2· p2On isolating the terms containing W, the solution of this equation isW =1 + pp2Specifically, for p = 1/6, Peter’s mean waiting time until the first time twoconsecutive 1s show up equals 42.The corresponding computation for Paula’s expected number of throwsis slightly less straightforward. One of the easiest ways is to define a simpleMarkov chain describing the current state of the game using just three statesSi, as follows. Let S0 be the starting state, let S1 be the state when the lastthrow was a 1, and let S2 be the final (absorbing) state 1 2. The rules ofthe game then imply the following transition matrix:

S0 S1 S2S0S1S25616231616 1For example, a throw in state S0 (at start) will lead with probability 1/6 intostate S1 (namely, if a 1 is thrown), and with probability 5/6 will Paula stayin state S0 (with any other number thrown). Similarly, after a 1 has beenthrown (i.e., out of state S1), Paula will either stay in state S1 (if a further1 is thrown, probability 1/6), or she will enter the final state S2 (if a 2 isthrown, probability 1/6), or she will enter state S0 (any number other than 1or 2, probability 2/3).

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