Pairs of dice are tossed 10 times (10 experiments). We know that first dice rolled 4 times with some specific number (let's say number one) out of all 10 experiments. We know that the second dice rolled 2 times with the same number one. How to calculate probability of both 2 dice NOT rolling number one in any of this 10 experiments simultaneously?
from first dice perceptive there is (10-2)/10 = 0.8 probability that second dice is not number one in the same experiment. And we know first dice rolled 4 times total. So that I'd calculate total probability as 0.8 ^ 4 = 0.4096.
However if you look from second dice perspective you will see that it's (10-4)/10 = 0.6 chances first dice is not number one in the same single experiment. Then for 2 experiments where we have number one for second dice we'll get 0.6 ^ 2 = 0.36 outcome.
why do I get different results here? What is the right way to solve this task?
Best Answer
wlog we may assume that the first dice had a 1 on the first four rolls and no others.
There are $6^{10}$ equally likely outcomes for the second dice, of which just $S={6\choose 2}5^8$ had a 1 on just two of the last six rolls and no others, but a total of ${10\choose 2}5^8$ for which the second dice had a total of just two 1s.
Hence the probability of no rolls on which both dice had a 1 is $\frac{6\choose2}{10\choose2}=\frac{1}{3}$.