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(This is a follow-up to Revealing the least helpful of three numbers.)

The game is like last time, but with one new step.

  • Andrew is secretly shown three real numbers, all independently, randomly, uniformly chosen from the interval [0, 1].
  • He must pick one to go in a transparent box. The other two go into opaque boxes.
  • Then Bndrew enters the room, and sees the number in the transparent box.
  • New Step: Andrew must tell Bndrew one of the numbers in the opaque boxes. Andrew chooses which of the two hidden numbers to divulge. He doesn't have to tell Bndrew which opaque box contains this number.
  • Bndrew guesses which of the three boxes contains the highest number. If he gets it right, he wins; otherwise Andrew wins.

Each player plays optimally (by which I mean they choose a strategy which maximises their win percentage under the assumption that the other player chooses the best possible counter-strategy). How likely is Bndrew to win?

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Standing on the shoulder of giants - ie the solvers of the previous iterations of the puzzle, I propose this solution:

Bndrew (still) wins $1/3$ of the games.

The strategies:

Andrew evaluates the distance function d(x) given by $x - 1/3$ if $x > 1/3$ and $2 * (1/3 - x)$ otherwise, for each number. They choose the two numbers with the lowest d, and put the numerically higher of these in the transparent box, and reveal the other one.

When Andrew follows this strategy, Bndrew can do no better than picking randomly.

The proof:

With the given construction, the remaining number will be uniformly distributed on the interval $[0;1/3-d_max/2] U [1/3+d_max;1]$, where d_max is the highest of the two d-values revealed. Since the upper interval is twice as large as the lower one, the chance of the unrevealed number to be the largest is exactly 2/3. Thus the largest number is in an opaque box 2/3 of the time, and in the transparent box 1/3 of the time - giving Bndrew a 1/3 chance to win for any numbers revaled, whether they choose the transparent or an opaque box

Since Bndrew can always achieve 1/3 by a random guess, this is an optimal solution for Andrew.

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  • $\begingroup$ Does it matter which opaque number is divulged, as long as it's lower than the transparent box? If we never put the lowest number in the transparent box, it's a foregone conclusion that the divulged number will be lower than what's in the transparent box - does it give any information at all? $\endgroup$ Commented Oct 28 at 15:41
  • $\begingroup$ @NuclearHoagie: What is your suggested alternative procedure? I am pretty sure that randomly divulging the other number 50 percent of the time iff they are both lower, will leak information. $\endgroup$ Commented Oct 28 at 17:38
  • $\begingroup$ Choose the two numbers with the lowest d, put the numerically higher one in the transparent box, and divulge any number that is lower (even if it has the highest d). Say the numbers are 0.1, 0.2, and 0.3 - Andrew puts 0.3 in the transparent box, but it's not clear to me why it matters if he divulges 0.1 or 0.2. Upon seeing the 0.3, Bndrew already knew the divulged number would be <0.3, and the numerical value of a not-highest number seems irrelevant. But maybe the distribution of the "pick either" divulged number shifts downward if both are less than the transparent number versus just one? $\endgroup$ Commented Oct 28 at 18:11
  • $\begingroup$ @NuclearHoagie Yes, your method skews the results (one way or the other). In your example with my method, when I hear 0.2 divulged, the last number is evenly distributed on [0-0.2] U [0.6-1]. Thus the chance of it being higher is 2/3. With yours, it is distributed on [0-0.3] U [0.6-1], but with only half the density on the lower interval. Thus the chance of it being in the upper interval is 8/11, and we win 4/11 by choosing an opaque box. $\endgroup$ Commented Oct 28 at 19:16

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