# Two dice tossed 10 times. Probability of rolling same number knowing amount the number rolled?

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?

• Your calculation $(0.8)^4$ (and the other one) both assume independence. Apr 27, 2016 at 20:21
• You really just want the probability that the 4 times you hit with the first die lie among the 8 times you missed with the second die. This probability is ${{8}\choose{4}} / {{10}\choose{4}}=1/3$. (Equivalently, the probability that the 2 times you hit with the second die lie among the 6 times you missed with the first die is ${{6}\choose{2}} / {{10}\choose{2}} = 1/3$ as well.) Apr 27, 2016 at 20:41

## 2 Answers

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 $\frac{6\choose2}{10\choose2}=\frac{1}{3}$.

Once you know exactly how many times the "one" showed up on each die, the $10$ rolls of the first die are no longer independent from each other, and likewise the $10$ rolls of the second die are no longer independent.

Consider this example: what is the probability that the first three rolls of the second die were all "ones"? Each roll has a $0.2$ probability ($2/10$) to have been a "one", so if we take the first three rolls as independent events, the probability would be $0.2^3 = 0.008$. But in fact the probability is zero, because we have already been told there were only two "ones" on the second die.

Instead of treating each roll of each die as a Bernoulli trial, try considering all the different ways the two "ones" on the second die could have occurred within the $10$ rolls. Are those different ways all equally likely? Given any particular way in which the four "ones" of the first die occurred, how likely is it that neither of the "ones" on the second die occurred at the same time as a "one" on the first die?