Betting the bias

frequentest approach == bayesian approach

Jun 29, 2025

A coin has an unknown head‐probability p∼Uniform(0,1)
The coin is flipped 100 times, independently
Before each flip, you may (optionally) bet $1 on heads or on tails
Bets pay 1:1 (you win $1 if your call is correct, or lose $1 if it isn’t)

What is the optimal strategy?
What is the EV of playing this game?

Solution (so spoilers…duh)

The optimal strategy is simply just to bet the side that has come up more times than the other. In the event of a tie of the frequency of both sides (which is guaranteed at the very beginning because 0==0), you do not wager anything as it is just as likely to be one side favorited as the other. Well actually the EV doesn’t change even if you bet on these tie scenarios, but you just increase variance for 0 EV so it’s silly to do so.

If it’s not clear why the side that has come up most frequently is most likely to be the side that is favored then go ask your buddy Claude or read my “Beta believe it” post1 or both.

Ok so now that we have established the optimal play style, we can now calculate the EV of this game. Assuming you bet on the favored side and we look at cases of p > 0.5, then your EV is just p - (1 - p) = 2p - 1. Likewise, if you bet on the wrong side, then your EV is (1 - p) - p = 1 - 2p. For every flip we have our EV equal to (2p - 1) * P(favored side came up more frequently) + (2p - 1) * P(unfavored side appeared more frequently) + 0 * P(both sides came up with the same frequency). Clearly the last term is just 0, so we can remove it. We do this across every flip and then integrate over all possible p values to get the EV of the game—well actually we’re just integrated from 0.5 to 1 and then multiplying by 2 because of symmetry.

I will refer to cases where the unfavored side came up more frequently as the ‘misleading’ cases, and cases where the favored side came up more frequently as the ‘expected’ cases.

$P_{\text{misleading}}(n, p) = P(H_n < n/2) = \sum_{h=0}^{\lfloor(n-1)/2\rfloor} \binom{n}{h} p^h (1-p)^{n-h}$

$P_{\mathrm{equal}}(n,p) = P\!\bigl(H_n = n/2\bigr) = \begin{cases} \displaystyle \binom{n}{\tfrac n2}\,p^{n/2}(1-p)^{n/2} & \text{if n is even}\\[6pt] 0 & \text{if n is odd} \end{cases} $

$P_{\text{expected}}(n, p) = P(H_n > n/2) = 1 - P_{\text{misleading}}(n, p) - P_{\text{equal}}(n, p)$

$E(p) = \sum_{n=0}^{99} \left[ P_{\text{misleading}}(n, p) \cdot (1-2p) + P_{\text{expected}}(n, p) \cdot (2p-1) \right]$

$\bar{E} = 2 \int_{0.5}^{1} E(p) \,dp$

And with a little bit of code to evaluate this integral, we get:

And this checks out with a quick sanity check right? Because on average we have 0.5 in EV for every flip (if we are correct with which one we bet on). This is because the expected absolute different between p and 1 - p is 0.5. To see why this is the case, let’s break this down into the case of p > 0.5 and note that the logic mirrors p < 0.5 cases by symmetry. Conditioned on p being greater than 0.5, it is uniform between 0.5 and 1 and therefore 0.75 in expectancy; in this case, 1 - p would be 0.25 and our difference is 0.5. So basically we have 0.5 in EV for every single flip if we have bet on the actual +EV side of this coin. The upper bound of EV is $49.50 (99 flips * 0.5) and we can subtract out a few flips to lower EV a bit because in expectation we have times where the unfavored side can take the ‘lead’ of most frequent side and also there are ties. Recall that the first time is guaranteed to be a tie (we don’t wager anything), hence why I used 99 * 0.5 as the upper rather than 100 * 0.5. It’s clear from the plot above that the EV is dragged down additionally when the variance is higher (p is closer to 0.5) as that statistically adds more cases of ‘ties’ and/or ‘misleading’; this drag down is non linear w.r.t. p.

Further Analysis

Suppose the initial distribution is not uniform, but rather comes from a normal distribution with mean 0.5 and variance sigma^2. Obviously you can just weight your integral by the pdf of the distribution, but I want to see if you can intuitively describe the relationship between EV and sigma^2. What would the curve of EV as a function of sigma^2 look like?
Suppose you had to market make this game, but you are able to make markets on the EV of the game while it is being played. How would you find the intra-game fair value?