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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 $X-Y$.

1 Answers1

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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

WNG
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  • The final answer is about $1/2$? How did you come up with the formula $\Sigma_{k=0}^{10}\binom{20}{k}\binom{20-k}{k}0.2^{2k}0.6^{20-2k}$ I don't understand why there's the product of binomial coefficients and why it is $20-2k$ rather than $20-k$ –  Jan 31 '20 at 20:58
  • Why is the summation up to 10? –  Jan 31 '20 at 22:08
  • Cause you cannot have both 11 raisins or more. The binomial coefficients would give 0 anyway in that case – WNG Jan 31 '20 at 22:09
  • Is the reason we have $0.2^{2k}$ rather than $0.2^k$ because we're picking $2k$ things at a time rather than $k$ things at a time? –  Jan 31 '20 at 22:59
  • @Numbers: the probability must be just over $\frac12$ since either you get the same number as your sister, or you get more than her, or you get fewer than her. The second and third are equally probable by symmetry and the fairness of your grandma, and the three probabilities add up to $1$. – Henry Feb 01 '20 at 22:02