Distribution of the non-absolute value difference of two identically distributed binomial

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You, your parents, your sister, go to visit grandma for her birthday. Grandma made a cake for the party. If she puts 20 raisins in the cake at random in the cake, and she divides the cake into 5 equal pieces, what's the probability that you get at least as many raisins as your sister?

Let $X_1$ be the number of raisins in your piece and $X_2$ the number of raisins in her piece.

$X_i \sim Bin(n=20, \theta=1/5)$

$X_1,X_2$ are identically distributed but not independent.

Find $P(X_1 \geq X_2)$.

$P(X_1 \geq X_2)=P(X_1 - X_2 \geq 0)$.

I don't know how to generate the cdf of $X_1-X_2$ and use the complement rule.

I know that the sum of two binomial variables is binomial but I don't know what distribution one binomial minus another is.

From the various answers at the linked question, it looks as if $$P(|X-Y|=0)={2n \choose n} \frac{1}{2^{2n}}$$ while for positive $z$ $$P(|X-Y|=z)={2n \choose n+z} \frac{1}{2^{2n-1}}$$

That isn't useful as I think I'm not looking for $|X-Y|$ but rather $X-Y$.

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In that specific case, you don't need to have more information to solve the problem : The problem is symmetric, i.e. you can reverse your situation with your sister without changing the problem.

Thus, $P(X>Y)=P(Y>X)$. Also, $P(X>Y)+P(Y>X)+P(X=Y)=1$, which gives you $P(X>Y)=\frac{1-P(X=Y)}{2}$ Thus $P(X>=Y)= P(X>Y)+P(X=Y) = \frac{1+P(X=Y)}{2}$

Since you already have $P(X=Y) = P(X-Y =0)$, you should be good assuming the expression you found in the linked problem fit your case.

However, they do not, because one required assumption is not met : $X_1$ and $X_2$ should be independent, which is not the case : you and your sister both have a non-zero probability of getting the 20 raisins, but both event cannot happen at the same time.

So you need to find another way to determine $P(X=Y)$.

Since there are not that many raisins, you can just go through all the possible numbers of raisins that both you and your sister would get : $P(X=Y)=\Sigma_{k=0}^{10}P(X=Y=k)=\Sigma_{k=0}^{10}\binom{20}{k}\binom{20-k}{k}0.2^{2k}0.6^{20-2k}$ which is about $0.1406$.

Thus the answer to the problem is around 0.5703