Find Probability that even numbered face occurs odd number of times

Question

A Die is tossed $2n+1$ times. Find Probability that even numbered face occurs odd number of times

First i assumed let $a,b,c,d,e,f$ be number of times $1$ occurred, $2$ occurred and so on $6$ occurred respectively. Then we have

$$a+b+c+d+e+f=2n+1$$ is a linear equation where variables are non negative integers.

So number of non negative integer solutions of above is $\binom{2n+6}{5}$

But $b$, $d$ and $f$ should be odd. So let $b=2x+1$, $d=2y+1$ and $f=2z+1$ where $x,y,z$ are non negative integers.

so we get

$$a+2x+c+2y+e+2z=2n-2$$ which is possible in the following two cases

Case $1.$ $a,c,e$ are all even so let $a=2m+1$, $c=2q+1$ and $e=2r+1$ we get

$$m+x+q+y+r+z=n-1$$ so number of solutions is $\binom{n+4}{5}$

Case $2.$ Exactly two of $a,c,e$ are odd and other is even which can be possible in 3 ways.

So let $a=2m+1$, $c=2q+1$, $e=2r$ we get

$$m+x+q+y+r+z=n-2$$

Number of solutions is $3 \times \binom{n+3}{5}$

Hence Probability is

$$P(A)=\frac{\binom{n+4}{5}+3\binom{n+3}{5}}{\binom{2n+6}{5}}$$

Is my approach fine?

Answer 1

First let us simplify the problem: the probability of an even face is 1/2, so is the probability of an odd face. The problem is therefore equivalent to a number of coin tosses of a fair coin, with even/odd faces corresponding to heads/tails.

Define the events $E$: the number of heads is even and $F$: the number of tails is even. On the one hand, heads and tails are perfectly interchangeable, therefore $$ P(E)=P(F). $$ Moreover, given an odd number of trials, either the number of heads is even or the number of tails is even, which implies $$ P(E)+P(F)=1. $$ You may then conclude that $$ P(E)=P(F)=1/2. $$

Therefore, the probability that even numbered face occurs odd number of times is 1/2.

Answer 2

Even and odd faces both occur with a proabability of $\frac{1}{2}$. The probability of attaining an odd number of evens is \begin{eqnarray*} \sum_{i=0}^{n} \binom{2n+1}{2i+1} \left( \frac{1}{2} \right)^{2n+1} \end{eqnarray*} and an even number of odd faces is \begin{eqnarray*} \sum_{i=0}^{n} \binom{2n+1}{2i} \left( \frac{1}{2} \right)^{2n+1} \end{eqnarray*} Now note that these two sums are equal by the symmetry of the binomial coefficients so they are both $\color{red}{\frac{1}{2}}$.

Answer 3

First of all, it should be noted that $0+3+0+1+0+1=5$ and $1+1+1+1+0+1=5$ are not equally likely. There are $\frac{5!}{3!}$ ways to get the first and $5!$ ways to get the second.

So counting these is not going to be a good approach.

Letting $P(x_2,x_4,x_6)=(3+x_2+x_4+x_6)^{2n+1}$, you then let: $$P_2(x_2,x_4,x_6)=\frac{1}{2}\left(P(x_2,x_4,x_6)-P(-x_2,x_4,x_6)\right),$$ this gives us the terms where $x_2$ has odd power.

Then $P_4(x_2,x_4,x_6)=\frac{1}{2}\left(P_2(x_2,x_4,x_6)-P_2(x_2,-x_4,x_6)\right)$ has the terms where $x_2$ and $x_4$ have odd powers.

Finally, $P_6(x_2,x_4,x_6)=\frac{1}{2}\left(P_4(x_2,x_4,x_6)-P_4(x_2,x_4,-x_6)\right)$ is the terms where all of $x_2,x_4,x_6$ have odd exponents.

Then the probability you want is:

$$\begin{align}\frac{P_6(1,1,1)}{6^{2n+1}}&=\frac{6^{2n+1}-3\cdot 4^{2n+1}+3\cdot 2^{2n+1}}{8\cdot 6^{2n+1}}\end{align}$$

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Another approach is to use Markov processes, with states $0$ to $3$, which is the number of $2,4,6$ which have occurred an odd number of times. Start in state $0$ and you get a transition matrix:

$$A=\begin{pmatrix}\frac{1}{2}&\frac{1}{2}&0&0\\ \frac{1}{6}&\frac{1}{2}&\frac{1}{3}&0\\ 0&\frac{1}{3}&\frac{1}{2}&\frac{1}{6}\\ 0&0&\frac{1}{2}&\frac{1}{2} \end{pmatrix}$$

And the probability you want is $$\begin{pmatrix}0&0&0&1\end{pmatrix}A^{2n+1}\begin{pmatrix}1\\0\\0\\0\end{pmatrix}$$

Diagonalizing gives eigenvalues of $A$ equal to $0,\frac{1}{3},\frac{2}{3},1$, and the same probability as above.