Coin flips and Dice rolls
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.
$\endgroup$ 81 Answer
$\begingroup$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 $$P(X-Y>0)=1-P(X-Y\leq0)=1-P(\frac{X-Y-50}{\sqrt{\frac{1325}{3}}}\leq0)=1-\Phi(-2.38)$$.
$\endgroup$ 0
