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I'm trying to work out the following problem:

In a study concerning a new treatment of a certain disease, two groups of $25$ participants in each were followed for five years. Those in one group took the old treatment and those in the other took the new treat- ment. The theoretical dropout rate for an individual was $50%$ in both groups over that $5$-year period. Let X be the number that dropped out in the first group and Y the number in the second group. Assuming independence where needed, give the sum that equals the probability that $Y \geq X + 2$. Hint: What is the distribution of $Y − X + 25$?

Apparently the distribution of $Y - X + 25$ is binomial with $p = .50$ and $n = 50$ but I am unsure how to arrive at this solution.

Here's what I've tried thus far: Let $Z = Y - X + 25$. Since $X$ and $Y$ are both binomial with $n = 25$ and $p = .5$, we have \begin{equation} M_Z(t) = E[e^{Zt}] = E[e^{(Y - X + 25)t}] = E[e^{Yt}]E[e^{-Xt}]E[e^{25t}] = (.5 + .5e^t)^{25}(.5 + .5 e^{-t})^{25}e^{25t}. \end{equation}

However, I'm not sure where to progress from here.

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Hint: $(0.5+0.5e^t)(0.5+0.5e^{-t}) e^t = (0.5+0.5e^t)^2$

So: $(0.5+0.5e^t)^{25}(0.5+0.5e^{-t})^{25} e^{25t} = (0.5+0.5e^{t})^{50}$

This is the MGF of what distribution?

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    I think he $e^{25}$ should be $e^{25t}$ but that is precisely the MGF of the binomial with $n = 50$ and $p = .5$.2017-02-13