imagine a betting game where we observe $N$ independent coin flips $x_1,...,x_n$ (where each $x_i \in {H,T}$) from the same coin, whose true weight is $\theta$. the task is to predict how many Heads we will get in the next $M$ flips of the same coin. the closer you are to the true number of heads, the higher your payoff will be. as example, the reward may be $r/(abs(Guess_H - True_H) + 1)$ where $Guess_H$ is your guess for number of Heads on next $M$ flips, $True_H$ is actual number of Heads in the next $M$ flips, and $r$ is some value $r > 1$.

what is the optimal strategy? how do you show formally which of the two following strategies is better?

strategy 1: estimate coin weight $\theta$ from $N$ first flips. if $\theta > 0.5$, predict all Heads for next $M$ flips ($M$ Heads), if $\theta < 0.5$, predict all Tails for next $M$ flips (0 Heads).

strategy 2: estimate coin weight $\theta$ from $N$ first flips. predict $\theta*M$ many heads for next $M$ flips.

which strategy is better? can this be shown formally? attempt:

expected reward for strategy 1:

assume $\theta = 0.75, M = 10$

$E(reward|strategy 1) = Binomial(10; 10, 0.75)*r$

expected reward for strategy 2:

$E(reward|strategy 2) = Binomial(0.75 * 10; 10, 0.75)*r$

this shows that for this case, strategy 2 is better, since:

$Binomial(0.75 * 10; 10, 0.75) > Binomial(10; 10, 0.75)$

how can this be shown analytically and in general, not assuming particular values for $M$ and the true $\theta$?

  • $\begingroup$ What are your thoughts? If the probability of heads is $0.75$, what would you expect? $\endgroup$ Nov 30, 2014 at 2:20
  • $\begingroup$ @RossMillikan: clarified q and added answer for 0.75 $\endgroup$
    – mvd
    Nov 30, 2014 at 2:44

1 Answer 1


If $N$ and $M$ are large, you expect the first part to be a good measure of $\theta$, then you expect a fraction $\theta$ of the tosses to be heads, so your prediction should be $\theta M$, strategy 2. If they are large enough that the normal approximation is good, you can use that. All heads will then be several standard deviations away. If $N,M$ are smaller, you have a lot of algebra ahead of you.

  • $\begingroup$ if strategy 2 is better does this not contradict the case that probability matching is flawed as in: en.wikipedia.org/wiki/Probability_matching - can you relate it to class conditional example, where if a class appears 60% of time then you maximize gains by predicting that class 100% of the time (not 60%)? see also: heim.ifi.uio.no/~inf3300/bilder/dudahart_chap2.pdf $\endgroup$
    – mvd
    Nov 30, 2014 at 15:03
  • $\begingroup$ There are two different predictions here. Your question asks about predicting the total number of heads over a long run. The link to Wikipedia is about making an individual prediction for each toss. In that case you always want to predict the more likely face. $\endgroup$ Nov 30, 2014 at 15:11
  • $\begingroup$ intuitively i see the difference, but can you explain formally in probability terms how the two predictions are different? hard to see formally why that distinction makes a big difference in which strategy maximizes reward $\endgroup$
    – mvd
    Nov 30, 2014 at 15:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .