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I am interested in knowing how often (in terms of percentage) you would expect the total of 3 fair rolled dice to exceed the total of 2 fair rolled dice.

Thanks

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2 Answers

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Numerically, it happens about 77,85% of the times. Precisely, 6054 times every 7776. Below the C++ code (I just happened to have it written for calculating probabilities in Risiko).

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#include<stdio.h>
#include<stdlib.h>
using namespace std;
void main(){ int i,j,k,a,b; int win=0; int h=0; for(i=1;i<=6;i++) for(j=1;j<=6;j++) for(k=1;k<=6;k++) for(a=1;a<=6;a++) for(b=1;b<=6;b++) { if(i+j+k>a+b) win++; h++; } printf("\n\n We have %d\n victories out of %d\n possibilities.",win,h); return; }

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I agree with Lorenzo's answer having both simulated the experiment in Python and calculated the probabilities as follows.

If A = total score on 2 dice then the probability distribution for A is simply:

and if B = total score on 3 dice then the probability distribution for B is:

We want P(B>A).

Clearly, $ P( B > A ) = P( B > A \cap A=2) + P( B > A \cap A=3 ) + ............... + P( B > A \cap A=12) $

We then need to calculate each of these 11 probabilities. The first three are as follows:

$ P(B>A\cap A=2) = P(B\geqslant3 \cap A=2) = 1/36 \times 1 $

$ P(B>A \cap A=3) = P(B\geqslant4 \cap A=3) = 2/36 \times 215/216 $

$ P(B>A \cap A=4) = P(B\geqslant5 \cap A=4) = 3/36 \times 212/216 $

Once we have worked out all 11 probabilities we add them up and get$ P(B>A) = 6054/7776 $

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