Rolling a pair of dice has a 1/36 chance of getting a pair of ones. But if I have four dice, at first I assumed it will still be 1/36 since the other two dice won't matter, but then the other two dice both have the chance to get a 1, so my possible results should be [1,1,x,x],[1,x,x,1],[x,x,1,1],[x,1,1,x],[1,x,1,x],[x,1,x,1],[1,1,1,x],[1,1,x,1],[1,x,1,1],[x,1,1,1],[1,1,1,1], so its 11/1296? but why has it become smaller than 1/36? I'm doing something wrong? Also how would I compute it for 3 rolls
Sorry, if there is an obvious answer for this, but its been years since I last took basic probability
Best Answer
Your $x$s can have any one of five values, so for example $[1,1,x,x]$ is really $25$ options, not $1$. The total number of successful rolls of four dice is $$6\times25+4\times5+1=171$$ and the probability is $$\frac{171}{1296}\ .$$ The chance of failing on one roll of four dice is $$1-\frac{171}{1296}=\frac{1125}{1296}\ ,$$ the chance of failing on three rolls is $$\Bigl(\frac{1125}{1296}\Bigr)^3$$ and the chance of succeeding at least once in four rolls is $$1-\Bigl(\frac{1125}{1296}\Bigr)^3=0.3459\ .$$