# What is the connection between probability of getting $m$ heads from $n$ coins and getting $km$ heads from $kn$ coins?

Suppose we have a coin with $0 < p < 1$ probability of giving heads when tossed. Let $X_n$ be the random variable denoting the number of heads we obtained from flipping $n$ of the described coins independently.

For any positive integer $k$, does the following holds? $$\Pr[X_n = m] \le \Pr[X_{kn} = km]$$

• Inequality appears to depend on values of $n,m,k,p$.. – Arpan Mar 26 '15 at 12:26
• @ArpanBanerjee Hm, so it doesn't only depends on $p$? For my research, I have that $m = E[X_n]$, $2 \le k < n^{0.5}$ and $p = 1/k$. Is there any related strategy to solve this one? – Irvan Mar 26 '15 at 12:48

Hint/idea: you can check whether this is true of false explicitly (even get a hunch on the dependence on $p$) by using the explicit formulae for Binomial distributions. What you are trying to compare are the two quantities $$\binom{n}{m} p^m(1-p)^{n-m}, \qquad \binom{kn}{km} p^{km}(1-p)^{kn-km}$$ Your inequality is equivalent to checking whether $$1 \leq A(k, n,m, p) \stackrel{\rm def}{=} \frac{\binom{kn}{km}}{\binom{n}{m}} \left(p^{m}(1-p)^{n-m}\right)^{k-1}$$ If you are interested in the regime where $n\to\infty$ and $m=\mathbb{E}[X_n] = np$, you can approximate the RHS by Striling's approximation to get something easier to analyse as a function of $n,p,k$ only.
• Stirling approximation is an excellent suggestion! To clarify we don't really need $m = E[X_n]$ to do that, right? (i mean, as long as $m$ is known). – Irvan Mar 27 '15 at 4:55
• @Irvan: indeed, you just need to know something about $m$ (to take the asymptotics: i.e., how it compares to $n$.) – Clement C. Mar 27 '15 at 12:30