For a game played using the dice in Problem #13, you make a 2$ bet on the outcome of a roll of the dice. If the roll is 5 or lower you win 3$. If the roll is 11 or 12, you win 10$. What is the expected value of this game?(note the winning amounts are net amounts.)
Here is the dice probability distribution for problem 13
Here is what i did
Is this correct?
Why .3225*5? Should it be *3 since thats how much money you get if you roll a 5 or lower?
Also why .3325?
.09+.09+.1425+.15= .4725 right?
THen 1-.4725+.0175 = .5100
so: .4725*3+.0175*10-.51*2= .5725
Also, doesnt the net value mean that you lose the 2$ you bet and any winnings are -2? So the 3$ win is actually 1$ net profit, and the 10$ win is actually 8$ net profit?
So should i times the probabilities by 1 and 8 instead?
If that is what you mean net amount. It is an odd way of expressing it but
if you have it defined that way OK.
yes, slip of the calculator
Also why .3325?
.09+.09+.1425+.15= .4725 right?
THen 1-.4725+.0175 = .5100
so: .4725*3+.0175*10-.51*2= .5725
If that is what you mean by net winnings (I would interpret net winningsAlso, doesnt the net value mean that you lose the 2$ you bet and any winnings are -2? So the 3$ win is actually 1$ net profit, and the 10$ win is actually 8$ net profit?
So should i times the probabilities by 1 and 8 instead?
to be how much better off you are after compared to before the bet), then:
0.4725*1 + 0.0175*8 - 0.51*2 = +0.4675
My experience is that when you bet your returned stake is not counted
part of your winnings. Just think of the fun the tax man would have if it did.
RonL
Hello, solace!
Rolling a pair of dice, there are 36 possible outcomes.You make a $2 bet on the outcome of a roll of the dice.
If the roll is 5 or lower you win $3.
If the roll is 11 or 12, you win $10.
What is the expected value of this game?
The totals and their respective probabilities are:
. .
. .
Hence: .
Therefore, you can expect to win an average of about 30½¢ per game.