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A fair die bearing the numbers 1, 2, 3, 4, 5, 6 is repeatedly thrown
until the running total first exceeds 12. What’s the most likely total
that will be obtained?
This must be the most likely total!
Here is the question!
Just consider the the next-to-last throw where last throw would make total exceeding 12.
Till then total must be either of 12,11,10,9,8 or 7. And next throw will make total greater than 12.
At this point -
If total is 12 then the first total exceeding 12 could be any from 13,14,15,16,17,18.
If total is 11 then the first total exceeding 12 could be any from 12,13,14,15,16,17,
If total is 10 then the first total exceeding 12 could be any from 12,13,14,15,16.
If total is 9 then the first total exceeding 12 could be any from 12,13,14,15.
If total is 8 then the first total exceeding 12 could be any from 12,13,14.
If total is 7 then the first total exceeding 12 could be any from 12,13.
It's quite evident that the number 13 appears in every above case.
Hence, 13 is the most likely total.