The probability that the sum of the numbers on both the dice is divisible by 10

In spite of all great answers, given here, I say, why not give another proof, from another point of view. The problem is we have 10 random variables $X_i$ for $i=1,\dots,10$, defined over $[6]=\{1,\dots,6\}$, and we are interested in distribution of $Z$ defined as $$ Z=X_1\oplus X_2\oplus \dots \oplus X_{10} $$ where $\oplus$ is addition modulo $6$. We can go on by two different, yet similar proofs.


First proof: If $X_1$ and $X_2$ are two random variables over $[6]$, and $X_1$ is uniformly distributed, sheer calculation can show that $X_1\oplus X_2$ is also uniformly distributed. Same logic yields that $Z$ is uniformly distributed over $[6]$.

Remark: This proves a more general problem. It says that even if only one of the dices is fair dice, i.e. each side appearing with probability $\frac 16$, the distribution of $Z$ will be uniform and hence $\mathbb P(Z=0)=\frac 16$.


Second proof: This proof draws on (simple) information theoretic tools and assumes its background. The random variable $Z$ is output of an additive noisy channel and it is known that the worst case is uniformly distributed noise. In other word if $X_i$ is uniform for only one $i$, $Z$ will be uniform. To see this, suppose that $X_1$ is uniformly distributed. Then consider the following mutual information $I(X_2,X_3,\dots,X_6;Z)$ which can be written as $H(Z)-H(Z|X_2,\dots,X_6)$. But we have: $$ H(Z|X_2,\dots,X_6)=H(X_1|X_2,\dots,X_6)=H(X_1) $$
where the first equality is resulted from the fact that knowing $X_2,\dots,X_6$ the only uncertainty in $Z$ is due to $X_1$. The second equality is because $X_1$ is independent of others. Know see that:

  • Mutual information is positive: $H(Z)\geq H(X_1)$
  • Entropy of $Z$ is always less that or equal to the entropy of uniformly distributed random variable over $[6]$: $H(Z)\leq H(X_1)$
  • From the last two $H(Z)=H(X_1)$ and $Z$ is uniformly distributed and the proof is complete.

Similarly here, only one fair dice is enough. Moreover the same proof can be used for an arbitrary set $[n]$. As long as one of the $X_i$'s is uniform, then their finite sum modulo $n$ will be uniformly distributed.

What is the probability that the sum of two dice is 10?

The probability that the sum of the number appearing is more than 10 is. ... .

What is the probability that the sum of the dice is a 10?

So that's three ways of making a 10 out of 36 ways of rolling the two dice (6 for the first times 6 for the second). That means the probability is 3/36 which reduces to 1/12 or 0.083.

What is the probability of rolling a 10 with two dice?

Two (6-sided) dice roll probability table.

What is the probability that the sum of the numbers on both the dice is divisible by4 or 6?

Required probability = P(S=4 or 6)P(Total)=1436=718.