Question

In a game, 3 independent fair coins are tossed. If they all land the same way up, you win $10. If they do not, you re-toss any number or choice of coins (all at once). After this re-toss, if they are all the same way up you win $10. Assuming you act rationally, what is the expected value of the game?

Answer 1

The expected value of the coin toss game is $4.375. A player tossing three independent fair coins can expect to make a profit over many plays of the game, assuming they act rationally and re-toss accordingly.

Step-by-step explanation:

In the first phase of the game, the probability of winning $10 by having all coins showing the same side (either all heads or all tails) is ⅓ (for heads) plus ⅓ (for tails), which equals ⅔. The probability of winning nothing on the first toss is hence ¾. If you didn't win on the first toss, you should act rationally by re-tossing all three coins.

The probability of winning $10 on the re-toss is the same as the first toss, which is ⅔. The expected value of the game is thus calculated by: $10 times ⅔ (for the first toss) plus $10 times ¾ times ⅔ (for the re-toss, since you only get to re-toss ¾ of the time).

The formula for the expected value (EV) would be:

EV = ($10 times ⅔) + ($10 times ¾ times ⅔) = $10 times (⅔ + (¾ times ⅔))

By simplifying, we find that the EV of this game is:

EV = $2.50 + $1.875 = $4.375

Since the expected value is positive, over many plays of the game a rational player can expect to make a profit.