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A die is rolled 100 times, and the sum of the numbers that are rolled is recorded as X (for example, if a 6 is rolled every time, X = 600). A coin is tossed 600 times, and the number of heads is recorded as Y. Find P(X > Y).

I know E[X] = 350 and E[Y] = 300, but I am not able to find the probability of X > Y. Any help is appreciated.

David G. Stork
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Harry P
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  • Given that the dice roll and coin flip sample requirements are huge, what do you think about a normal approximation?(I mean, without it you are not going to be able to do this on paper, I presume. Even ten coins and five dice rolls and you are not going to do it on paper!) Do you know how to approximate the random variables $X$ and $Y$ by normals? – Sarvesh Ravichandran Iyer Sep 07 '19 at 04:45
  • @астонвіллаолофмэллбэрг Since E[X] = 350 and Var(X) = 875/3, X ~ N(350,875/3). Similarly, Y ~ N(300,150). Am I correct? – Harry P Sep 07 '19 at 05:05
  • Yes, you are correct. Now, when is $N(350,\frac{875}{3}) > N(300,150)$? Think about writing both these random variables in terms of the standard normal $N(0,1)$ (You know the scaling, right?), then you will get an inequation in terms of the standard normal, whose solution set will give you the approximate probability. – Sarvesh Ravichandran Iyer Sep 07 '19 at 05:08
  • @астонвіллаолофмэллбэрг So X - Y is normally distributed with mean 50 and variance $1325/3$. Then $P(X>Y) = 1 - P(X-Y \leq 0) = 1 - P (\frac{X-Y-50}{1325/3} \leq \frac{0-50}{1325/3} )= 1 - \Phi (-0.11) = 1 - 0.13 = 0.87.$ Does this look right? – Harry P Sep 07 '19 at 05:34
  • Yes, this looks right to me, you should post it as an answer and then accept it. – Sarvesh Ravichandran Iyer Sep 07 '19 at 05:38
  • Thanks for the help. – Harry P Sep 07 '19 at 05:41
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    When you standardize the normal variable $X-Y$, should you not divide by the standard deviation, not the variance? – Harry Reed Aug 07 '20 at 20:14
  • How do you derive the variance as 875/3? – Trajan Dec 28 '21 at 14:09

1 Answers1

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Try to collect all the intelligent ideas from the comment above. And hope I didn't mess it up. First, rolling dice i.i.d. 100 times follows a multi-nomial distribution with mean $$E[x] = 350$$ and variance $$Var(X) = \frac{875}{3}$$. Then, flipping coins i.i.d. 600 times follows a binomial distribution with mean $$E[x] = 300$$ and variance $$Var(x) = 150$$. Considering that those tests are sample for a large number of times, we can assume that they both follow normal distributions with $$X \sim N(350, \frac{875}{3})$$ and $$Y \sim N(300, 150)$$. Then, the result we want $$P(X>Y) = P(X-Y>0)$$, where $$X-Y \sim N(50, \frac{1325}{3})$$. Then, it's simple to know that $$P(X-Y>0)=1-P(X-Y\leq0)=1-P(\frac{X-Y-50}{\sqrt{\frac{1325}{3}}}\leq0)=1-\Phi(-2.38)$$.

Chason Young
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