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A blind man is on a strange island and he has 2 red pills and 2 white pills, completely identical and has kept in his pockets, he needs to take 1 red pill and 1 white pill order doesn't matter. If he takes 2 pills of the same color he dies. How does he survive. There is a standard solution for this and I was wondering if there could be a combinatorial way to solve this?

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    $\begingroup$ "A blind man is on a strange island" So... Honshu? (The people over there are a little strange...) $\endgroup$ – Akiva Weinberger Dec 31 '14 at 5:52
  • $\begingroup$ How would a combinatorial method exist, he has no information on the pills, if he can't break them all he can do is take a pill at random. $\endgroup$ – Jorge Fernández Hidalgo Dec 31 '14 at 5:54
  • $\begingroup$ I am just thinking if I was crazy to solve it this way. He pulls out one from the pocket and puts it in his left hand and one in his right hand and one in his other pocket. Then he swaps one in the left hand with the one in his right pocket. Now what he has in this right pocket and left pocket are always a different colored pill and thus he survives. Let me know if there is any flaw in the reasoning $\endgroup$ – Satish Ramanathan Dec 31 '14 at 5:55
  • $\begingroup$ Well that works if you know the pills in the same pocket have the same color. $\endgroup$ – Jorge Fernández Hidalgo Dec 31 '14 at 5:56
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    $\begingroup$ Ok, here is the scenario. He takes out a red pill from his left pocket and puts it in his left hand, and takes out a black pill and puts it in his right hand, and takes out a black pill and puts it in his right pocket, after this he switches his left hand and right pocket, so that the card in his right pocket is red the same color as the pill in his left pocket. $\endgroup$ – Jorge Fernández Hidalgo Dec 31 '14 at 6:04
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Just take a glass of water, fill it to the brim. Put all the pills inside, they should dissolve well, all pills will dissolve well, drink half the glass of water, even a blind man can tell, problem solved.

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Break each pill in half, and take half of each pill.

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  • $\begingroup$ This is the standard solution that I was talking about. I am just thinking if I there could be anything else $\endgroup$ – Satish Ramanathan Dec 31 '14 at 5:49
  • $\begingroup$ @satishramanathan This is the sort of riddle that requires an "Aha!" moment. I don't think there's a "combinatorial way" to solve this, because I don't think you can solve it without that "Aha!" moment. $\endgroup$ – Akiva Weinberger Dec 31 '14 at 5:51
  • $\begingroup$ I am just thinking if I was crazy to solve it this way. He pulls out one from the pocket and puts it in his left hand and one in his right hand and one in his other pocket. Then he swaps one in the left hand with the one in his right pocket. Now what he has in this right pocket and left pocket are always a different colored pill and thus he survives. Let me know if there is any flaw in the reasoning $\endgroup$ – Satish Ramanathan Dec 31 '14 at 5:53
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    $\begingroup$ Why would they be different colored? $\endgroup$ – Akiva Weinberger Dec 31 '14 at 5:56
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Take half of each pill. Solved!!

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HINT: $4\cdot\frac12=2$ and this is not forbidden in the formulation of the problem.

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  • $\begingroup$ This is the standard solution that I was talking about. Wondering if there could be anything else? $\endgroup$ – Satish Ramanathan Dec 31 '14 at 5:50
  • $\begingroup$ @satishramanathan I didn't know he standard solution. However it seems to be a natural (well, for mathematicians) trick, and combinatorics, using natural numbers, rather cannot help in surviving. :-) $\endgroup$ – Przemysław Scherwentke Dec 31 '14 at 5:56
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Chances that he survives $=1-$Chances that he dies

$1-\dfrac{2}{^4C_2}=1-\dfrac{2}{6}=\dfrac{2}{3}=66.67$% Chances

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  • $\begingroup$ This is the probability that he survives and would like a method that he survives $\endgroup$ – Satish Ramanathan Dec 31 '14 at 5:51
  • $\begingroup$ @satishramanathan So its a puzzle, and not a combinatorics problem. $\endgroup$ – Dheeraj Kumar Dec 31 '14 at 5:52
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I'm not sure what you would consider combinatorics, and what you wouldn't. The standard solution mentioned in other comments seems combinatorial to me. The insight necessary is that the pills are not "atomic", they can be divided.

One non-standard way to solve, which is kind of the same, is to smash the pills into infinitesimally small bits and then blend them until the particle distribution is uniform and well mixed, then eat half of the pile. with probability 1, you win.

The fact that doing this subdivision and recombination in the most coarse possible way ( break in two and recombine any way you like, then eat half ) actually works with probability 1 as well seems like a numerical miracle, which is kind of what I like to think combinatorics is.

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