A game is played using one die. If the die is rolled and shows a 2, the player wins $7. If the die shows any number other than 2, the player wins nothing. If there is a charge of $1 to play the game, what is the game's expected value? Round to the nearest cent.

CHOOSE ONE

A
$0.17

B
–$0.17

C
$6.00

D
–$1.00

Question

A game is played using one die. If the die is rolled and shows a 2, the player wins $7. If the die shows any number other than 2, the player wins nothing. If there is a charge of $1 to play the game, what is the game's expected value? Round to the nearest cent.

CHOOSE ONE

A
$0.17

B
–$0.17

C
$6.00

D
–$1.00

anonymous · Answer

You have a chance of 1/6 to win the game, right? So the expected money that you will win will be 1/6 * 7 = 1,17. But you pay 1, so you paied 0,17.

I think that's it.

kropot72 · Answer

Let the dollar profit be X. For a win, the profit is $6. For a loss the profit is -$1.

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Setting up a probability distribution variable gives
$$E(X)=-1\times \frac{5}{6}+6\times \frac{1}{6}=\frac{1}{6}$$
So, on average, the player would expect to win about 17 cents per game.