optimality stands in the eye of the riddler

When looking at US elections on FiveThirtyEight, I came across this riddle:

Two players go (…) into separate booths (…) and a random number between zero and one appears on a screen (…) chosen from a standard uniform distribution. They can choose to keep that first number, or to (…) get a second random number, which they must keep (…) The prize (…) is awarded to the player who kept the higher number (…) Which number is the optimal cutoff for players to discard their first number and choose another? [The Riddler, Mar 4, 2016]

While the solution is now available, I wonder at the wording of this riddle, where “optimality” is not spelled out. Unless I missed some key entry, as it often happens with puzzles… Assuming both players use the same “optimal” cut-off C to make their decision to keep or drop the original uniform, the probability that one does better than the other is exactly ½ since both outcomes are iid. When considering the expected value of the number kept by a player, a simple integral shows that this value is

½(1-C²+C),

which is maximal for C=½. If one considers instead the median of the number Z kept by a player, a bit more computation leads to the function med(Z) = 1/2C when C²>1/2 and (½+C)/(1+C) when C²<1/2, which is maximal for C²=½.

“…using the golden ratio gives the best chance to win the gold bullion!” [The Riddler, Mar 4, 2016]

Incidentally (or not), the solution on 538 is also difficult to understand in that it starts with the event that the first draw is C, which is an event of probability zero. However, when trying to optimise the choice of C for one player, given that the other player has a known cuttoff of D, I found that the only case when C=D coincides with the solution proposed on Riddler, namely ½(√5-1). To check whether or not this derivation was correct, I also plotted the theoretical (right) versus the empirical (left) versions of the winning probability: There is no difference between the two. But the diagonal is exactly at the .5 level, as expected: with an interesting second curve at the .5 probability level. These two level sets happen to meet at the “golden number” solution, ½(√5-1), which comes as a confirmation of my earlier remark. [Another question connected with this riddle would be to figure out the value of D used by the other player from a sequence of games.]

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