Question

Throw two dice. If the sum of the two dice is 55 or more, you win $30. If not, you pay me $161.

Step 1 of 2:

Find the expected value of the proposition. Round your answer to two decimal places. Losses must be expressed as negative values.

Step 2 of 2:

If you played this game 794794 times how much would you expect to win or lose? Round your answer to two decimal places. Losses must be entered as negative.

Answer 1

If two dice are thrown the maximum possible sum of the number on both the dice is 12 that is when both the dice comes up with a six on the upper face.

So,

let X: Sum of the two dice

P(X>=55)=0

(As the sum ie X would take a maximum value as 12)

P(X<55)=1

Step 1

Therefore expected value of proposition is:-

=0*30+1*(-161)

= -161$

Hence if he plays it would be a loss of 161$.

Step 2

If the game is played 794794 times

let N:Number of times the game is played.

So expected loss= N*(loss per throw)

= 794794*(-161)

= -127961834

Therefore is 794794 games are played he would be having a loss of 127961834$.