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I see from this link that if we have a fair coin and toss it 100 times, the probability that we get odd number of heads is equal to even number of heads. How about the converse? Given a coin with unknown bias, but we know when tossing 100 times, the probability that we get odd number of heads is equal to even number of heads, is it true that it must be fair? I write out the explicit formula for 4 and 6, and plot it on desmos, it is a monotone decreasing function from 0 to 1/2 and monotone increasing function from 1/2 to 1, so I guess it is also true for 100 coins. Is it the case?

vyjtkbyykyhuk
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  • Yes, if we know that the probability of getting an odd number of heads is equal to the probability of getting an even number of heads, then the coin must be fair. But the problem is that in a real situation you can't know these probabilities by tossing the coin just 100 times. Otherwise, in a hypothetical situation where you repeat the 100 tosses an infinite number of times, yes, you can know for sure that the coin is fair. – Amogh Mar 06 '23 at 06:22

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Let $O(n)$ and $E(n)$ be the probability to get odd number and even number of heads from $n$ throws. Also let $p$ be the probability to get head from a throw. We have the following relations:

$$ \begin{aligned} O(n) &= O(n-1)\cdot(1-p)+E(n-1)\cdot p \\ E(n) &= E(n-1)\cdot (1-p) + O(n-1)\cdot p \end{aligned} $$

Thus we have the following:

$$ O(n)=E(n) \implies (1-2p)\cdot O(n-1) =(1-2p)\cdot E(n-1) $$

The implication trivially holds for $p=\frac{1}{2}$. Now let's assume $p\neq\frac{1}{2}$ to see if any other $p$ satisfy the implication. From this assumption, $1-2p\neq 0$ so the implication becomes the following:

$$ O(n)=E(n) \implies O(n-1)=E(n-1) \implies ... \implies O(1)=E(1) $$

However, $O(1)=p$ and $E(1)=1-p$ so in the end we again end up with $p=\frac{1}{2}$.

acat3
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