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Solution to puzzle 24: Die: median throws

What is the minimum number of times a fair die must be thrown for there to be at least an even chance that all scores appear at least once?  (Computer assistance advisable.)


Let p(n,m) be the probability that, out of n throws, there are m distinct scores, where n greater than or equal to m greater than or equal to 1.
Then p(n,m) = (probability of no new score) · p(n−1,m) + (probability of new score) · P(n−1,m−1).
So p(n,m) = (m/6) · p(n−1,m) + ((7−m)/6) · p(n−1,m−1).

We seek the first p(n,6) > 1/2.

These values can be calculated by hand or with the aid of a computer, whereby:
p(12,6) = 0.438, p(13,6) = 0.514, to 3 significant figures.

Therefore the minimum number of times a fair die must be thrown for there to be at least an even chance that all scores appear at least once is 13.

Of course, this is also the answer to the question: "How many fair dice must be thrown, simultaneously, for there to be at least an even chance that all scores appear at least once?"

The result is characterized as the median number of throws by which all scores are represented.

Source: Extra Stuff: Gambling Ramblings, by Peter Griffin. See Chapter 4.

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