Jtdylktuy

A fair die is thrown $n$ times. Show that the probability of getting an even number of sixes is $frac{1}{2}[ 1 + (frac{2}{3})^{n}]$, where $0$ is counted as even number.

My solution. I have probability of getting even number of sixes as:

$sum_{k=0}^{{lfloor n/2rfloor}} {n choose 2k}(frac{1}{6})^{2k}(frac{5}{6})^{n-2k}$

I also know that the probability of getting an even number of sixes plus an odd number of sixes is $1$, i.e.

$$sum_{k=0}^{{lfloor n/2 rfloor}} {n choose 2k}(frac{1}{6})^{2k}(frac{5}{6})^{n-2k}+ sum_{k=0}^{{lfloor n/2 rfloor}+1} {n choose 2k+1}(frac{1}{6})^{2k+1}(frac{5}{6})^{n-2k-1} = 1$$

However I am not sure how to extract the "even" part of the expression to obtain the answer required?

$sum_{k=0}^{[n/2]}{nchoose 2k}(1/6)^{2k}(5/6)^{n-2k}+sum_{k=0}^{[n/2]}{nchoose 2k+1}(1/6)^{2k+1}(5/6)^{n-2k-1}=1$

and $sum_{k=0}^{[n/2]}{nchoose 2k}(1/6)^{2k}(5/6)^{n-2k}-sum_{k=0}^{[n/2]}{nchoose 2k+1}(1/6)^{2k+1}(5/6)^{n-2k-1}=sum_{k=0}^{n}{nchoose k}(-1/6)^{k}(5/6)^{n-k}=(2/3)^n$

Hence $sum_{k=0}^{[n/2]}{nchoose 2k}(1/6)^{2k}(5/6)^{n-2k}=1/2(1+(2/3)^n).$

Let's suppose you have a discrete random variable $X$ taking on non-negative
integer values. Let $p_n=P(X=n)$. The generating function of $X$ is
$$G(s)=E(s^X)=sum_{n=0}^infty p_ns^n.$$
Then
$$G(1)=E(1^X)=sum_{n=0}^infty p_n=1$$
and
$$G(-1)=sum_{n=0}^infty (-1)^np_n.$$
Adding these,
$$1+G(-1)=2(p_0+p_2+cdots)=2P(Xtext{ is even}).$$

In your example, $X$ is the number of sixes in $n$ throws of a die.
Then $X$ is a binomial random variable with parameters $1/6$ and $n$.

So (i) what is the generating function of a binomial random variable,
and (ii) how do you apply that to the question at hand?

Let $e(n)$ be the probability of getting an even number of sixes in $n$ rolls.

It is easy to show directly that $e(0)=1$, so for $n=0$, $e(n)=frac{1}{2}[ 1 + (frac{2}{3})^{n}]$.

Let $n$ be a positive integer and assume that $e(n-1)=frac{1}{2}[ 1 + (frac{2}{3})^{n-1}]$. If this assumption implies that $e(n)=frac{1}{2}[ 1 + (frac{2}{3})^{n}]$, the desired result holds, thanks to the principle of mathematical induction.

After $n-1$ rolls of the die, the probability of an even number of sixes is $e(n-1)$ and the probability of an odd number of sixes is $1-e(n-1)$. After one more roll, the probability of an even number of sixes is ${5over6}e(n-1)+{1over6}(1-e(n-1))$. (There are an even number of sixes after $n$ rolls only if there were an even number after $n-1$ rolls and the $n$-th roll was not a six, or if there were an odd number after $n-1$ rolls and the $n$-th roll was a six.)

Therefore $e(n)={5over6}e(n-1)+{1over6}(1-e(n-1))={5over6}left(frac{1}{2}[ 1 + (frac{2}{3})^{n-1}]right)+{1over6}(1-left(frac{1}{2}[ 1 + (frac{2}{3})^{n-1}]right))$, which (it can be seen with a bit of algebra) equals $frac{1}{2}[ 1 + (frac{2}{3})^{n}]$.