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A fair red die and a fair blue die are rolled $2$ times each. What is the probability of the sum of numbers on the red die less than the sum of numbers on the blue die?

My attempt:

Two dice are thrown. Then there are $36$ possibilities for the red die as well as the blue die.

The condition is the sum of numbers on the red die less than the sum of numbers on the blue die.

Is that possibilities are $35$? I am a bit confused. Please help me

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    $\begingroup$ Hint: it’s the same as the probability of the sum’s being greater. $\endgroup$ – amd Sep 8 '17 at 1:05
  • $\begingroup$ @amd,,can you expalin plz how many are there $\endgroup$ – Inverse Problem Sep 8 '17 at 1:12
  • $\begingroup$ @amd..is that 30 possibilites for that $\endgroup$ – Inverse Problem Sep 8 '17 at 1:15
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    $\begingroup$ The hint of amd is $1=P(X>Y)+P(Y>X)+P(X=Y)$, where $P(X>Y)=P(Y>X)$. Thus $P(X>Y)=\frac12-\frac12\cdot P(X=Y)$ $\endgroup$ – callculus Sep 8 '17 at 1:22
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    $\begingroup$ What is your answer to my question ? More specific. What is $P(X=Y=2)$ ? $\endgroup$ – callculus Sep 8 '17 at 1:30
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Let's consider just the red dice. Of the 36 ways in which they could land, the sum can be 2 in just 1 way, can be 3 in 2 ways (1+2 or 2+1), ..., can be 7 in 6 ways, can be 11 in 2 ways (6+5 or 5+6) and can be 12 in just 1 way.

Let X represent the total from the red dice. P(X=2) = 1/36,...P(X=)=6/36..., P(X=11)=2/36 etc.

Now, letting Y represent the total from the blue dice, which is distributed just like X, we can find the probability that X=Y.

$$P(X=Y) = P(X=Y=2) + ... + P(X=Y=12) = \frac{1^2+...6^2+...1^2}{36^2}$$, which I get to be $\frac{146}{36^2}$

But $$P(X<Y) = P(Y<X) = \frac{1-P(X=Y)}{2} = 0.4436728$$

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