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You are kidnapped and your offender plays a game with you. In front of you are 2 boxes, containing a total of 50 white balls and 50 black balls. The kidnapper will pull out a ball from one of the two boxes, at random. If the ball is white you survive, if not, you are shot.

Rules:

  • Every ball must be in one of the boxes.
  • No balls can be left out.
  • Each box must have at least one ball.

Example:

Box 1: 25 black balls and 25 white balls Box 2: 25 black balls and 25 white balls.

In this example the chance to live is obviously $50 \% $. How can you distribute the balls so as to increase that chance?

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    $\begingroup$ This one is generally presented as a lateral thinking problem, because the hidden assumption is that each container must have exactly 50 balls. $\endgroup$ – Joe Z. Jul 30 '14 at 1:55
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Box 1: 50 black balls, 49 white balls Box 2: 0 black balls, 1 white ball.
In this example the chance to live is obviously $50\%+50\%*49/99 \approx 75\%$. There is no way to increase that chance.

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  • $\begingroup$ You are absolutely right! $\endgroup$ – hetzi Jul 24 '14 at 10:46
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    $\begingroup$ Precisely it's about $74.7474...\%$ $\endgroup$ – justhalf Jul 25 '14 at 2:23
  • $\begingroup$ It's "obvious" that this solution is optimal, but by any chance do you or @hetzi know of a nice proof? $\endgroup$ – Mike Earnest Feb 12 '16 at 2:52
  • $\begingroup$ @MikeEarnest, well, the proof is straightforward, just write down the probability vs NBBalls & NWBalls formula and take an derivative. I don't think there is another way ,and for me it really doesn't worth work, since the dependency is clearly monotonic. $\endgroup$ – klm123 Feb 12 '16 at 7:27
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I think this can be slightly improved.

Box 1 is inside Box 2.

Box 1 contains 50 white balls. Box 2 contains Box 1 (which contains 50 white balls) and 50 black balls.

Upon choosing a random box, you get either Box 1 (100% survival) or Box 2 (50% survival).

Therefore your chance of survival is 75%, which beats the current accepted answer.

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  • $\begingroup$ In my opinion the change of survival is $50\%$, because if box 2 with box 1 inside is chosen, a choice has been made and therefore one ball from box 2 is selected. $\endgroup$ – hetzi Jan 26 '15 at 22:58
  • $\begingroup$ @hetzi As with all lateral thinking questions, it depends on the assumptions that you violate. You can either violate the "same number of balls" implicit in "distribute", or violate the topology of the boxes. IMO this answer is superior (due to its greater value and greater lateral-thinking level), but YMMV. $\endgroup$ – March Ho Jan 26 '15 at 23:02
  • $\begingroup$ Your answer could be even improved if you combine it with the other one. So there would be one white ball in box 2 and box 1 contains 49 white and 50 black balls. $\endgroup$ – hetzi Jan 26 '15 at 23:04
  • $\begingroup$ @hetzi Combining it doesn't change anything. Box 2 still contains 50 black and 50 white balls for a total of 50% survival rate for Box 2, while the rate of 100% for Box 1 is unchanged. $\endgroup$ – March Ho Jan 26 '15 at 23:07
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    $\begingroup$ @Geobits I wouldn't have given that answer if it had been phrased "one and only one". $\endgroup$ – March Ho Jan 27 '15 at 7:22

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