Q: I throw a die until the running total exceeds 12. What is the
most likely final total?
[Puzzle Panel, BBC Radio 4, 2 Aug 1998]
A: 13. To see this:
The way to get a final total of 13 is to build up some total between
7 and 12 inclusive, then make a single throw of the appropriate value.
The way to get a final total of 14 is to build up some total between 8 and 12 inclusive, then make a single throw of the appropriate value.
Thus if we take the list of sequences producing 14, then subtract 1 from the final throw of each sequence, we will have part but not all of the list of sequences producing 13. Moreover, corresponding sequences are equally likely to occur, because they contain the same number of throws. Thus 13 is strictly more likely than 14.
A similar argument shows that 14 is strictly more likely than 15, and so on. Hence 13 is the most likely total.
I must admit I thought of this argument after I'd started finding the probabilities explicitly by counting with a computer, so I carried on to get the following probabilities (to 10dp):
|13||3647371105 / 13060694016||0.2792631923|
|14||3095329537 / 13060694016||0.2369957931|
|15||2511747217 / 13060694016||0.1923134570|
|16||1901445049 / 13060694016||0.1455852994|
|17||1271727037 / 13060694016||0.0973705559|
|18||633074071 / 13060694016||0.0484717022|
Note that the odds are better than 50:50 that the final total will be
13 or 14.
[Emailed to programme, 3 Aug 1998]
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