Who is more likely to win the game?

Question

You meet your friend again, and this time she decides to play the following game with you. She rolls 6n dice and wins if she gets at least n sixes. She offers you 6(n+1) dice, and declares that you need at least n + 1 sixes to win. Who is more likely to win the game?

Answer 1

Denote by $X$ the number of sixes that appears in $6n$ rolls of your friend. And let $Y$ be the number of sixes in your first $6n$ rolls, and $Z$ be the number of sixes in your last $6$ rolls. Variables $X$ and $Y$ has the same distribution, $Y$ and $Z$ are independent, and total number of sixes you get is $Y+Z$.

We need to compare two probabilities: $$ \mathbb P(X\geq n) \vee \mathbb P(Y+Z\geq n+1). $$

We can replace $Y$ by $X$ in the second probability and deal with $\mathbb P(X+Z\geq n+1)$.

Simplify both probabilities: $$\tag{1}\label{1} \mathbb P(X\geq n)=\mathbb P(X=n)+\color{red}{\mathbb P(X\geq n+1)}. $$ $$\begin{align} \mathbb P(X+Z\geq n+1) &=\color{red}{\mathbb P(X\geq n+1)}\ +\\ & \mathbb P(X = n, Z\geq 1)\ +\\ & \mathbb P(X = n-1, Z\geq 2)\ +\\ & \mathbb P(X = n-2, Z\geq 3)\ +\\ & \mathbb P(X = n-3, Z\geq 4)\ +\\ & \mathbb P(X = n-4, Z\geq 5)\ +\\ & \mathbb P(X = n-5, Z=6)\end{align}\tag{2}\label{2} $$ Two probabilities marked red are the same, and we need to compare the rest. Note that $X$ is the number of successes in $6n$ Bernoulli trials with success probability $1/6$, and $n=\lfloor(6n+1)\frac16\rfloor$ is the most probable number of successes. Therefore $\mathbb P(X = n-k) < \mathbb P(X = n)$ for any integer $k\geq 1$ and $$ \mathbb P(X = n-1, Z\geq 2) = \mathbb P(X = n-1) \mathbb P(Z\geq 2) <\mathbb P(X = n) \mathbb P(Z\geq 2), $$ $$ \mathbb P(X = n-2, Z\geq 3) = \mathbb P(X = n-2) \mathbb P(Z\geq 3) <\mathbb P(X = n) \mathbb P(Z\geq 3), $$ and the same is valid for all black summands in (\ref{2}).

We have: $$\begin{align} \mathbb P(X+Z\geq n+1) & < \color{red}{\mathbb P(X\geq n+1)}\ + \cr & \mathbb P(X=n)\bigl( \mathbb P(Z\geq 1)+ \mathbb P(Z\geq 2)+\ldots +\mathbb P(Z=6)\bigr)= \cr & \color{red}{\mathbb P(X\geq n+1)} + \mathbb P(X=n) = \mathbb P(X\geq n) \end{align} \tag{3}\label{3} $$ The sum in (\ref{3}) disappears since it is equal to 1: $$ \mathbb P(Z\geq 1)+ \mathbb P(Z\geq 2)+\ldots +\mathbb P(Z=6) = \mathbb EZ = 1. $$

Conclude that your friend is more likely to win the game.