Assume it's a fair coin. What are the odds you roll two heads, two tails, and two feet on six rolls? I got the following:
$ 3 \times \binom{6}{2} \over 3^6 $
$3$ for the different sides, $6 \choose 2$ for all the ways you can get two events (e.g. heads, tails, feet) and $3^6$ for all possible outcomes.
My classmate thinks the answer is just $\frac{1}{3}$. Who's right, here? Or, rather, is either of us right?
$\endgroup$ 32 Answers
$\begingroup$Neither of you is right. You have $6 \choose 2$ ways to position the heads and $4 \choose 2$ ways to position the tails, so the chance is only $\frac {15 \cdot 6}{3^6}=\frac {10}{81}$
$\endgroup$ $\begingroup$Total number of possible outcomes is $3^6$.
And no. of favourable outcomes is $\frac{6!}{2!2!2!}$ since :
Consider there are $6$ such coins and all of them are rolled together only once and simultaneously. So the order comes into play, like, which coin has what result. So the results can be arranged in $6!$ ways and our desired result has $2$ hads, $2$ tails and $2$ feet, hence the divisions by $2!$ thrice.
The probability becomes $\frac{90}{3^6}=\frac{10}{81}$
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