A fair coin is tossed. If it comes up heads, you win £200, if it comes up tails, you lose £100. Find the expected profit from this game.
Here are my thoughts:
One way of doing this would be to deduce the probability distribution and take the expectation, forming an infinite sum. However, that seems messy given how much symmetry there is in the game. I feel that I can use conditional expectation to exploit this.
$E(X) = \frac{1}{2}E(X|H)$ [since you lose instantly if you toss a tails]
$E(X|H) = \frac{1}{2}E(X|HH) + \frac{1}{2}E(X|HT)$
I think that I can then somehow solve for $E(X|H)$ using the following symmetries:
$E(X) = E(X|HT) - 100$
[since once you flip a HT you are back to a very similar game but simply have an extra £100]
$E(X|HHT) = E(X|HTH)$
[since either route through the tree diagram gets you to the same game with £300 in the kitty]
When I try to use these I never quite end up being able to find $E(X|H)$.
Any help would be much appreciated. Thanks.
By the question it seems that coin is tossed once. If the coin is tossed once then Expectation $= P(Heads\ comes\ up)*£200 + P(Tails\ comes\ up)*(-£100)=\frac{1}{2}£200-\frac{1}{2}£100=£50$