A game is played using one die. If the die is rolled and shows 5 5, the player wins $ 10 $10. If the die shows any number other

Question

oee [108]

3 years ago

10

A game is played using one die. If the die is rolled and shows 5 5, the player wins $ 10 $10. If the die shows any number other

than 5 5, the player wins nothing. If there is a charge of $ 2 $2 to play the game, what is the game's expected value?

Mathematics

2 answers:

Verizon [17] · Answer 1 · 2022-06-20T06:45:53+03:00

The expected value is -$0.33.

The probability of winning is 1/6. The probability of losing is 5/6.

If you win, your prize is 10-2 = 8, since you paid $2 to pay the game. If you lose, you lose the $2 you spent.

The expected value would be the probability of winning, 1/6, multiplied by the winnings, 8, added to the probability of losing, 5/6, multiplied by the loss, -2:
1/6*8 + 5/6(-2) = 8/6 - 10/6 = -2/6 = -0.33

jekas [21] · Answer 2 · 2022-06-20T08:55:56+03:00

Answer: The expected value is -$0.33

Step-by-step explanation: The expected value is calculated as:

E = ∑pₙxₙ

where xₙ is a given event (in this case winning 8$ or losing 2$) and pₙ is the probability for the event.

So in this you only win with one result in the dice, so the probability of winning is 1/6, and the probability of losing must be 5/6.

Then the expected value is:

E = 1/6*$8 - 5/6*$2 = $-0.33

Where I used the profit of $8 instead of $10, because you must pay $2 for play, and you win $10, the profit is ($10 - $2) = $8

A game is played using one die. If the die is rolled and shows 5 5​, the player wins $ 10 $10. If the die shows any number other

A game is played using one die. If the die is rolled and shows 5 5, the player wins $ 10 $10. If the die shows any number other