Question

In a certain dice game, three fair dice are rolled and the total on all three dice is counted. Let the random variable x be the total count on all three dice.
You pay one dollar to roll the dice. If you roll a 16 or greater, you will win $10. If you roll a 6 or lower, you win $5. If you roll anything else, you lose your money.
What is the expected value of this game?

Answer 1

we need to compute the probability of each of these events

Answer 2

shouldn't be that bad, 16 or higher means
a) (5 5 6), (5,6,5) ,(6,5,6)
b) 5 6 6 another three
c) 6 6 6 only one
for a total of 7 out of the possible \(6^3=216\) rolls, probability it
\[\frac{7}{216}\] that you win \(\$10\)

Answer 3

Good job dude.

Answer 4

six or lower, many more choices
@emah, thanks
there is lots more to do

Answer 5

I know, I just also know I couldn't explain it that well.

Answer 6

here is a picture of the probabilities

Answer 7

then for the sixes, looks like
1, 1, 1, 1 of these
1, 1, 2 (3 of these)
1, 1, 3 (3 of these)
1,1,4 (3 of these)
1 2 2 (3 of these)
1,2,3 (3 more)
2,2,2 1 of these

Answer 8

oooh NOW you tell me you have these listed!!
then it is real easy

Answer 9

sorry, i was trying to upload the pic asap lol

Answer 10

lol^

Answer 11

i get the second part but not the first one can you guys explain it to me

Answer 12

then the probability you get a number 6 or less is
\[\frac{20}{216}\]
probability you get a number 16 or greater is \[\frac{10}{216}\]

Answer 13

so are we looking for the expected value of rolling a specific number or how much money would be won?

Answer 14

and you can win \(\$9, \$4\) or \(-\$1\)

Answer 15

then the expected winnings is the average
it is
\[9\times P(X=9)+4\times P(X=4)-1\times P(X=-1)\]

Answer 16

that is
\[\frac{9\times 10+4\times 20-1\times 186}{216}\]

Answer 17

the reason it is 9 and 4, rather than 10 and 5, is presumably you don't win ten dollars plus get your one dollar back

Answer 18

you can think of it more simply this way
imagine you play exactly 216 times and that the probabilities are exact
you pay a total of $216
you win back \(10\times 10+5\times 5=100+25=125\)
for a loss of \(125-216=-91\)

Answer 19

averages over each play, you lose \(-\frac{91}{216}\) each time
that is your expected value

Answer 20

@Alessandr09 you here to right? because you had the same question
is it clear how to do it?

Answer 21

oh I see! so would I just leave it as a fraction or would it be better to put it as a decimal?

Answer 22

that is up to you i guess
since it is money, you might want a decimal

Answer 23

okay, thank you sooooo much!!!

Answer 24

yw

Answer 25

thank u i got it @satellite73 you saved my life

Answer 26

yw, glad to help (and that your life is saved)

Answer 27

what would the probability distribution look like?

Answer 28

@aprilmelody, I attached a photo in an earlier reply. It would basically look like that but only for the section with three dice

Answer 29

ok! could you explain why it goes up to 18?

Answer 30

or do I stop at 6 rolls?

Answer 31

You would only stop at 6 rolls if you were only dealing with one die. It goes up to 18 rolls because since there are three dice being rolled, there are a total of 18 outcomes (6+6+6=18).

Answer 32

So the first two columns, you can just ignore. All you wanna focus on is the third column (3 dice). :]

Answer 33

Thanks!:DD

Answer 34

You're welcome :]