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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:

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

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?

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  • $\begingroup$ What about the probability of an even number of sixes minus the probability of an odd number of sixes? $\endgroup$ – Lord Shark the Unknown Nov 23 '18 at 20:23
  • $\begingroup$ Have you considered using induction? $\endgroup$ – James Nov 23 '18 at 20:27
  • $\begingroup$ I don't see how that helps - @LordSharktheUnknown $\endgroup$ – Alex Nov 23 '18 at 20:32
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$\sum_{k=0}^{[n/2]}{n\choose 2k}(1/6)^{2k}(5/6)^{n-2k}+\sum_{k=0}^{[n/2]}{n\choose 2k+1}(1/6)^{2k+1}(5/6)^{n-2k-1}=1$

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

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

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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, $$1+G(-1)=2(p_0+p_2+\cdots)=2P(X\text{ is even}).$$

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?

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

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