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A Riddler puzzle on the three binary and sequential questions one should ask three players hiding their respective U(0,1) realisation, U, V, and W, to best guess which player holds the largest number, max{U,V,W}. Assuming questions of the type Is U, &tc., the challenge boils down to selecting seven bounds (one for U, two for V, and four for W) in order to optimise the probability of picking the right player. Which means for a given collection of such bounds to learn this probability from the three binary answers. These can be turned into eight (2³) binary variables and I used them as entries in a logistic regression to predict that W was larger than max(U,V), itself predicted by the first two answers. The optimisation of the bounds can then be achieved by simulated annealing (or otherwise) and the approach returns (random) outputs like the following bounds (one on U, two on V, and four on W)

`0.616 0.434 0.830 0.350 0.736 0.913 0.796 0.827`

for an estimated probability of 0.827. This is a somewhat coherent sequence of bounds when considering the simpler case of two players. Indeed, with three bounds, the probability of winning can be readily derived as

$b^2(b^1-b^2)+1/2+b^3(1-b^2+b^1)-b^1b^1$

logically optimised by (b¹,b²,b³)=(1/2,1/4,3/4) for a success probability of 0.875. And it coïncides with the solution posted by The Riddler, although there is no intuition behind the figures, contrary to the two player situation. In fact, I am surprised that the bound on W does not equate the expectation of max{U,V} under the current conditions:

```> x=runif(1e6,0,.616);y=runif(1e6,0,.434)
> mean(y+(x-y>0)*(x-y))
[1] 0.3591648
```