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Currently my guess for a solution is

[(6C3 x 3) * (3C2 x 2)] / 3^6 = 360/729

First bracket is considering 3 out of the 6 people picking the same hand to throw out (i.e rock paper or scissors) and the second bracket is considering the remaining 3 people, since they cannot throw out matching hands, 2 have to throw the same thing, and the last person throws out the last remaining option, in which this case there is two type of hands left to throw.

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  • $\begingroup$ Would $4$ or more out of $6$ choosing scissors count as $3$ out of $6$? Would $3$ choosing rock and $3$ choosing paper count as $3$ out of $6$? If the answer to both is yes then you can calculate the probability of $2$ rock, $2$ paper and $2$ scissors, and subtract that from $1$ to give $\frac{61}{81} \approx 0.753$ $\endgroup$ – Henry Jan 23 '15 at 7:09
  • $\begingroup$ Hmm, I am looking for exactly 3 picking the same hand. The remaining 3 can pick whatever they want. Rock rock rock paper paper paper is considered as a possible case, but there will be double counting which you must consider $\endgroup$ – Kasit Jan 23 '15 at 8:20
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The probability that 3 of 6 people, who all throw one of rocks, paper, or scissors at the same time, throw the same thing is $$ \frac{\binom{3}{1}\binom{6}{3}2^3}{3^6} \approx 0.685 $$ because you pick which of rocks, paper, scissors all three throwing the same thing throw, and you pick which three people throw that same thing. Picking the people who throw the same also picks the people who throw differently, and each of those remaining people can throw one of the two remaining possibilities, giving you the factor of $2^3$. And of course the denominator is all possible outcomes.


Removing the possibility of double counting: $$ \frac{\binom{3}{1}\binom{6}{3}\binom{2}{1}\binom{3}{1}}{3^6} \approx 0.494 $$ where the difference is that you want only one group of three people to have thrown the same thing. Of the remaining three people, one must throw differently that the other two, so you pick who it is and what he/she throws.

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  • $\begingroup$ Would this double count the situations where 3 pick the same choice and the other 3 also all pick the same? Like RRRSSS? $\endgroup$ – turkeyhundt Jan 23 '15 at 3:37
  • $\begingroup$ Yes (ten more characters) $\endgroup$ – CactusHouse Jan 23 '15 at 3:38
  • $\begingroup$ Hey cactus, I just realised you actually double counted in your first solution. You consider rock rock rock paper paper paper but 2³ considers paper paper paper rock rock rock which is the same scenario again. $\endgroup$ – Kasit Jan 23 '15 at 8:21
  • $\begingroup$ Yes, hence my response 'yes' to the person asking if that is true, and my mention of it in the second part of my post $\endgroup$ – CactusHouse Jan 23 '15 at 13:36

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