Hello kaitoli Originally Posted by

**kaitoli** Hi everyone

I have a question which I cannot put my finger on..

A symmetric coin with heads on one side and tails on the other side is tossed 491 times after one another. The total amount of times you get tails is either even or odd. Is the probability that you get an even amount of tails exactly 50%? And the question requires a strong mathematical evidence, e.g. a formula.

I'm thankful for all ideas!

The answer is Yes, it is 50%.

Proof

The probability that a given coin lands tails is $\displaystyle \frac12$, and the probability that it doesn't is $\displaystyle 1-\frac12=\frac12$. Therefore, using the Binomial Distribution Formula, the probability that we get exactly $\displaystyle r$ tails from $\displaystyle n$ tosses is:$\displaystyle P(r) =\binom n r \left(\frac12\right)^r\left(\frac12\right)^{n-r}=\binom n r \left(\frac12\right)^n$

We must prove that $\displaystyle P(0)+ P(2) + P(4) + ... = \frac12$. So, we need to prove that, if $\displaystyle n$ is odd (and therefore if $\displaystyle n=491$):$\displaystyle \binom n 0\left(\frac12\right)^n + \binom n 2\left(\frac12\right)^n+\binom n 4\left(\frac12\right)^n + ... +\binom{n}{n-1} \left(\frac12\right)^n = \frac12$

i.e. that:$\displaystyle \binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1} = 2^{n-1}=S$, say

Now it is well known that:$\displaystyle \binom n 0 + \binom n 1+\binom n 2 + ... +\binom{n}{n} = 2^{n}$ ...(1)

(The proof of this is to write down the Binomial Expansion of $\displaystyle (1+1)^n\;( = 2^n)$)

And also that:$\displaystyle \binom{n}{r} =\binom{n}{n-r}$ ...(2)

Now, adding $\displaystyle S$ to itself, we get:$\displaystyle 2S = \binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1}+\binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1}$$\displaystyle = \binom n 0 + \binom n 2+\binom n 4 + ... +\binom{n}{n-1}+\binom n n + \binom{ n}{n- 2}+\binom{ n}{n- 4} + ... +\binom{n}{1}$, using (2)

$\displaystyle = 2^n$, from (1)

$\displaystyle \Rightarrow S = 2^{n-1}$ which completes the proof.

Grandad