1
$\begingroup$

I've been trying to work out a probability question but the solution has me a bit confused. This question is from 40 Puzzles and Problems in Probability and Mathematical Statistics by Wolfgang Schwarz:

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. 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)?

The solution states:

Peter’s sequence will usually consist of a certain number of 1s, each of which is followed by a number different from 1 with p = 5/6. If, after obtaining a 1, he fails to achieve the desired 1 − 1 run, then the number thrown was necessarily different from 1 and, therefore, cannot constitute the potential beginning of a potential 1 − 1 run. This is different for Paula: if she fails to throw a 2 after an initial 1, then she may do so by throwing another 1, which in turn could be the start of a potential 1 − 2 run. Therefore, the expected waiting time will be somewhat shorter for Paula.

I don't really understand this solution, perhaps I'm misunderstanding the question/solution. From what I understand, the suggestion is that Paula could throw 1 followed by another 1, which would not be a win, but would then form the potential start of a 1-2 run. The bit that confuses me is that this sequence (1-1) forms the win for Peter, so surely Peter has an advantage in this scenario?

Improve this question
$\endgroup$
3
  • $\begingroup$ Peter and Paula are not using the same die; each has their own. So Paula can not use Peter's result.. $\endgroup$
    – jginestet
    Commented Sep 6 at 23:20
  • $\begingroup$ For more on this problem (or very closely related version of it) please see stats.stackexchange.com/questions/12174 and stats.stackexchange.com/questions/544811. $\endgroup$
    – whuber ♦
    Commented Sep 7 at 16:29
  • $\begingroup$ If they were looking at the results of the same die, the probability of Peter's pattern coming before Paula's is the same as the probability of Paula's coming before Peter's so each $\frac12$ (first throw until you get a $1$, then if you immediately throw as $1$ or $2$ then stop with your winner, and otherwise start again). But that race is not the question, which is instead about expectations of numbers of throws. $\endgroup$
    – Henry
    Commented Sep 7 at 23:22

1 Answer 1

Reset to default
3
$\begingroup$

The question refers to the minimum number of die rolls and/or coin fills.

If Peter fails on the first two rolls, he must roll at least 2 more times for a minimum of 4 rolls. Such as: 1 - 2 - 1 - 1

If Paula fails on the first two rolls, it is then possible only need to one additional roll needed such as 1 - 1 - 2.

Thus Paula is likely to have the shorter wait time.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.