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A quick weekend riddle from the Riddler about the probability of getting a sequence of increasing numbers from dice with an increasing number of faces, eg 4-, 6-, and 8-face dice. Which happens to be 1/4. By sheer calculation (à la Gauss) or simple enumération (à la R):

> for(i in 1:4)for(j in (i+1):6)F=F+(8-j)
> F/4/6/8
[1] 0.25


The less-express riddle is an optimisation problem related with stick breaking: given a stick of length one, propose a fraction a and win (1-a) if a Uniform x is less than one. Since the gain is a(1-a) the maximal average gain is associated with a=½. Now, if the remaining stick (1-a) can be divided when x>a, what is the sequence of fractions one should use when the gain is the length of the remaining stick? With two attempts only, the optimal gain is still ¼. And a simulation experiment with three attempts again returns ¼.