The following is a modified Jane Street interview question.

Question: Given $100$ fair coins. For each head obtained, we get $\$1$. If we can re-flip any number of coin once, what is the expected value of the game?

By 're-flip any number of coin once', I mean that we can flip those coins which do not give tails. For example, if we have $4$ coins and we obtain $HTHT$, then we can flip second and fourth coin again to increase our gains.

I know how to solve the problem for $4$ fair coins:

Without re-flipping, the expected value of the game is $\$2.$ With re-flipping, we can calculate the additional gain in the following manner:

$$\frac{1}{16}\times 0 + \frac{4}{16}\times 0.5 + \frac{6}{16}\times 1 + \frac{4}{16}\times 1.5 + \frac{1}{16}\times 2 = 1.$$

So, the expected value with re-flipping for $4$ fair coins is $\$3$.

I am able to do the above calculations in my head and get the answer without using pen and papers. However, if I am given $100$ coins, then I am not able to calculate the additional gain in my head as it is quite tedious.

I am wondering whether there is a shorter way to solve the $100$ coins problem without using pen and paper.

  • 2
    $\begingroup$ Hint: imagine that you flip every coin twice and the losses are exactly the $TT$ coins. $\endgroup$
    – lulu
    Commented Jan 11, 2020 at 15:32

1 Answer 1


Each coin has an expected value of $\$.75$ since the only outcome which does not yield $\$1$ is $TT$, a probability $\frac 14$ event.

By linearity the expected value of the $100$ coins is then $$.75\times 100=\boxed {$ 75}$$

Sanity check: with $4$ coins instead of $100$ this method would give $.75 \times 4=3$, in agreement with your calculation.

  • $\begingroup$ I see. Thanks for your fast and lucid explanations. $\endgroup$
    – Idonknow
    Commented Jan 11, 2020 at 15:37

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