Each of the two persons makes a single throw with a pair of unbiased dice.What is the probability that the throws are equal?

Since the same throws can result if either both of them get 1 or both of them get 2 or both of them get 3 or both of them get 4 or both of them 5 or both of them get 6.

So I calculated probability as $\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}=\frac{1}{6}$

But my answer is wrong and the correct answer is $\frac{73}{648}$.Please help me with the correct approach to solve it.Thanks.

  • $\begingroup$ First person throws any number with probability $1$ and the second person has to throw the same number with probability $1/6$. So, the probability that the two throws are equal is $1 \cdot 1/6 = 1/6$. It seems you are indeed correct! Edit: It seems each person throws two dice. You and I misinterpreted the question. $\endgroup$
    – Ritz
    Oct 2, 2015 at 9:55
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    $\begingroup$ Will you reproduce the exact question ? As it stands, I can't make out what it is asking for ! $\endgroup$ Oct 2, 2015 at 9:56
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    $\begingroup$ I think the question means that each player throws two dice. $1/6$ would indeed be correct if each threw only one. $\endgroup$
    – kviiri
    Oct 2, 2015 at 9:56
  • $\begingroup$ I have edited the question,i left some words,typing mistake!@trueblueanil $\endgroup$
    – diya
    Oct 2, 2015 at 10:07
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    $\begingroup$ Are the dice the same color? In short, what is your equals test -- add up the values? $\endgroup$
    – Yakk
    Oct 2, 2015 at 14:11

2 Answers 2


If I'm interpreting this correctly...

There are a total of $36$ combinations you can get when rolling a pair of $6$-sided dice. There is $1$ way to get a $2$, $2$ ways to get a $3$, $3$ ways to get a $4$, etc.

You have a total of $36\cdot 36= 1296$ events. Now, we do a bit of counting. There is $1$ way both people can get a $2$, $2 \cdot 2$ ways both people can get a $3$, $3 \cdot 3$ ways both people can get a $4$, etc.

Thus, the number of successes is (by symmetry) $2(1+4+9+16+25)+36 = 146$.

The answer you are looking for is $\frac{146}{1296} = \frac{73}{648}$.


Deducing, like Sherlock Holmes, that the question is:
Two people throw two dice each. What is the probability that both get the same sum ?

Ways of throwing sums of $2 - 12$ with $2$ dice follows the pattern

So P(both get the same sum) = $\dfrac{(1^2+2^2+3^2+ ....+6^2+5^2+4^2... +1^2)}{36^2}= \dfrac{146}{1296}=\dfrac{73}{648}$


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