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A fair coin is flipped 100 times in a row. For each flip, if it comes up heads you win \$2, if it comes up tails you lose \$1. You start with \$50, if you run out of money you must stop prematurely. If you don't run out of money you stop after 100 flips.

What is the expected value of this game?

So for the case where you have no stopping condition before 100 flips you get that $E(100)=50$, and I assume the possibility of stopping will reduce this expectation somewhat. But how exactly is it changed?

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Try it with a smaller number. Assume you are allowed eight flips and start with $4$. As you say, without the constraint of the starting value, the expected value is $4$. You run out of money if you throw TTTT, HTTTTTT, THTTTTT, TTHTTTT, TTTHTTT. Throwing four tails deprives you of four flips worth $2$ and has a chance of $\frac 1{16}$. Each of the rest deprives you of one throw worth $\frac 12$ and has a chance of $\frac 1{128}$. The expected value is then $$4-2\cdot \frac1{16}-4 \cdot \frac12 \cdot \frac 1{128}=3\frac {55}{64}$$ Similarly you lose chances if you throw $50$ tails in a row, or one head and $25$ tails among the first $53$ throws (where the head is among the first $50$), or so on. Each of these is a very small chance so the reduction in expected value is very small.

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  • $\begingroup$ Hi Ross, I just had a quick question. I see that the expectation is reduced by all of the games that could have been played that make money e.g. TTTTHHHH which makes \$4 with probability $2^{-8}$. However, are there not also games that would lose money that you also do not get to play, which would increase the expectation? e.g. TTTTTTTH which occurs with probability $2^{-8}$ and loses \$5? By this I mean that the expectation should be larger than $3\frac{55}{64}$? $\endgroup$
    – Ellya
    Aug 13, 2020 at 12:28
  • $\begingroup$ Is it then the case that the games you do not get to play that increase expected value outweigh the games where you lose expected value (much like the game with no stopping criteria), and so the expectation is always reduced. In short, we remove any game which follows a loss - this creates a subgame of positive expectation, and this expectation is subtracted from the original one, thus the expectation always decreases. $\endgroup$
    – Ellya
    Aug 13, 2020 at 12:37
  • $\begingroup$ @Ellya: don't confuse the outcome of a particular set of throws with the expected value over all the possible sets of throws. It is true that by stopping after four tails you delete possibilities that would have made you lose money. That was counted in the expectation drop because you stopped early. Let's imagine a game with one flip. The expectation is $\frac 12$. If some event prevents you from playing, you have lost the expectation of $\frac 12$. $\endgroup$
    – Ross Millikan
    Aug 13, 2020 at 14:25
  • $\begingroup$ @Ellya: True, one of the filps (tails) would have led to a loss of $1$, but we have counted that as well as the one that leads to a gain of $2$ in the calculation. $\endgroup$
    – Ross Millikan
    Aug 13, 2020 at 14:25
  • $\begingroup$ Of course, by a "subgame" I meant really that it is a game of fewer moves, but the properties are similar to the full game, kind of how you took the problem from 100 flips with \$50 to 8 flips with \$4 - the key part is that winning a flip outweighs the loss of one flip. I did wonder exactly where the value $2$ came from - so in this you have included both the missed "positive" turns and the "negative" turns? $\endgroup$
    – Ellya
    Aug 13, 2020 at 15:19

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