What is the expected value of this five-dice game?

[leumas614]

Homework Statement

Five dice are rolled. You win a dollar for each number other than a five that is rolled, but you don't win extra for duplicate numbers; for example if [3,5,3,2,1] is the result, then you win three dollars. What should you pay to play this game (what is expected value)?

Homework Equations

Can't think of any specific ones

The Attempt at a Solution

It's hard for me to explain what I did because I don't know how to draw a tree diagram on a computer.

First I did (5/6)*(4/6)*(3/6)*(2/6) but apparently that's wrong. Then I drew a tree diagram so I had 5/6 chance of rolling 1,2,3,4,6. If I rolled a 1, then I have a 2/3 chance of rolling a 2,3,4,6 but not 1,5. If I rolled a 3 let's say then 1/3 chance of rolling one of the remaining numbers namely 4,6. If I rolled a 4 then 1/6 chance of rolling a 6 and lastly multiplied by random variable 5 for five tosses. Thus my equation is this starting from bottom of the tree:

5(1/6)2(1/3)4(1/2)5(2/3)(1/6)=1.85 The 5,2,4,5 I added into account for the fact that I only considered certain numbers and left some out.

Is this anywhere near correct? Thanks

 

[tiny-tim]

Five dice are rolled. You win a dollar for each number other than a five that is rolled, but you don't win extra for duplicate numbers; for example if [3,5,3,2,1] is the result, then you win three dollars. What should you pay to play this game (what is expected value)?

Hi leumas614!

You need to be systematic …

how many ways are there of winning $1?
$2?
…
$5?

 

[leumas614]

# ways of rolling anything is 5^{5}

# of ways of winning $0: only 1 (rolling 5,5,5,5,5)

# of ways of winning $5:

1,2,3,4,6 in any order so 5!

# of ways of winning $1

any combination of 1 and 5, 2 and 5 ... 4 and 5. The prob is the binomial distribution:
4\sumfrom n=1 to 5(^{5}_{n})(1/6)^{n}(1/6)^{5-n}=4(1/6)^{5}(1+5+10+5)=0.01080

There has to be a way to solve this without enumerating all possible answers. What about this:

odds of rolling only two numbers out of 6 are 5[(2/6)^{5}-(1/6)^{5}] It's multiplied by 5 because you can make 5 pairs of two numbers (1,5 2,5...6,5). It's minus (1/6)^{5} to account for the overlap. You are counting one number twice. For the other numbers:

C(5,1)[(2/6)^5-(1/6)^5] +C(5,2)[(3/6)^5-(2/6)^5]+...+C(5,5)[(1-(5/6)^5]=3.2447

This includes rolling 3 numbers, 4 numbers, 5 numbers

 

[leumas614]

Never mind I found out the answer.

X = I₁+I₂+I₃+I₄+I₆


I₁ is the indicator of getting a 1. That is you either get a 1 or you don't. So X is getting a number other than 5 and no repeats which equals \sumI_n

Take expected value of both sides so E(X) = P(1)+P(2)+P(3)+P(4)+P(6) which is just 5P(1).

P(1) is the odds of getting at least one 1 which is just 1-(5/6)⁵

So E(X)=5P(1)=5(1-(5/6)⁵)=2.99