Solving Probability Problem: Flip a Coin n Times (X) and n+1 Times (Y)

[jamcc09]

As you may notice from a previous post of mine, I am working through the Blitzstein & Hwang probability book. I have run into a problem, which I cannot quite solve. It goes like this:

Alice flips a fair coin n times and Bob flips another fair coin n + 1 times, resulting in independent $X \sim Bin(n, 1/2)$ and $Y \sim Bin(n+1, 1/2)$.

(a) Let $V = min(X,Y)$ and $W = max(X,Y)$. Find $E(V) + E(W)$. If $X=Y$ then $X=Y=V=W$.

(b) Show that $P(X<Y) = P(n-X<n+1-Y)$.

(c) Compute $P(X<Y)$. Hint: Use (b) and the fact that $X$ and $Y$ are integer-valued.

In case anyone else is interested (a) is answered simply by seeing that $X+Y = V+W$ and therefore $E(V+W)=E(X+Y)=E(X)+E(Y)=E(V)+E(W)$. This simply uses linearity of expectation.

For (b), I believe that $P(X<Y) = P(n-X<n+1-Y)$ by symmetry, since both $X$ and $Y$ are binomially-distributed with $p=0.5$. So, my first question is: is this correct? I am not really sure how to justify this mathematically, but seems to be intuitively true because $P(X=k) = P(X=n-k)$ (although maybe this is wrong ;)).

For (c), the question suggests using part (b). So, I took that to mean $1 = P(X<Y) + P(Y>X) + P(X=Y)$. So, I took that to mean $$1 = P(X<Y) + P(Y>X) + P(X=Y) = P(X<Y) + P(n-X<n+1-Y) + P(X=Y) = 2P(X<Y) + P(X=Y).$$Therefore, $P(X<Y) = \frac{1-P(X=Y)}{2}$. I think then $P(X=Y) = \sum_{k=0}^{n} P(X=k)P(Y=k)$. I feel like I am now using that $X$ and $Y$ are integer-valued, but I am thinking there is a further simplification of this formula. Can someone help me get on the right track here? Just an additional note, is it possible to use part (a) at all in (b) and (c)? I don't really see how these are connected.

Thanks,
- J

 

[mmwave]

Dear J,

Thank you for sharing your progress and questions with us. It seems like you are making good progress in understanding the material in the Blitzstein & Hwang probability book. I can offer some insights and guidance on your questions.

(a) Your solution to this part is correct and well-reasoned. As you noted, this is a simple application of the linearity of expectation, which states that the expected value of a sum of random variables is equal to the sum of their expected values.

(b) Your intuition is correct. The symmetry of the binomial distribution with $p=0.5$ implies that $P(X=k) = P(X=n-k)$, and similarly for $Y$. This can be justified mathematically by the symmetry of the binomial probability mass function, which is $P(X=k) = P(X=n-k)$ for all integer values of $k$. Therefore, $P(X<k) = P(X>n-k) = P(n-X<k)$, and similarly for $Y$. This leads to the desired result $P(X<Y) = P(n-X<n+1-Y)$.

(c) Your approach is correct, but there is a simpler way to compute $P(X=Y)$ using the binomial distribution. Since $X$ and $Y$ are independent, the joint probability of $X$ and $Y$ is given by the product of their individual probabilities. Therefore, $P(X=Y) = \sum_{k=0}^{n} P(X=k)P(Y=k) = \sum_{k=0}^{n} P(X=k)P(X=k) = \sum_{k=0}^{n} P(X=k)^2$. Since $X$ is binomially distributed with $p=0.5$, we have $P(X=k) = \binom{n}{k}0.5^k0.5^{n-k} = \binom{n}{k}0.5^n$. Substituting this into the sum, we get $P(X=Y) = \sum_{k=0}^{n} \binom{n}{k}0.5^{2n} = 0.5^{2n}\sum_{k=0}^{n} \binom{n}{k} = 0.5^{2n}(2^n) = 0.5^n$. Therefore, $P(X