# Basic Probability Question—Rolling Die

1. Apr 5, 2012

### bizboy1

Basic Probability Question—Rolling Die!!!

1. The problem statement, all variables and given/known data
Suppose you roll a fair six-sided die repeatedly until the first time you roll a number that you have rolled before. A) for each r=1,2,... calculate the probability pr that you roll exactly r times.

2. Relevant equations

3. The attempt at a solution
p8=p9=...=pinfinity=0
p1=0
p2=1/6
p3=5/18
I think I'm doing this the long way and the wrong way. I know these answers are right because they make sense. But it's getting to hard to count. I need insight on how I should think about this problem. It is basic probability from pitman section 1.6 problem 6. This is not homework. I am just reviewing. Please help! Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 5, 2012

### bizboy1

Re: Basic Probability Question—Rolling Die!!!

Actually I just figured it out. It's similar to the birthday problem. Just create trees for multiplication rule for n events and you will get it.

p4=(5/6)*(4/6)*(3/6)
p5=(5/6)*(4/6)*(3/6)*(4/6)
p6=(5/6)*(4/6)*(3/6)*(2/6)*(5/6)
p7=(5/6)*(4/6)*(3/6)*(2/6)(1/6)*(6/6)

3. Apr 5, 2012

### Dickfore

Re: Basic Probability Question—Rolling Die!!!

Why isn't p3 = 5/36?

4. Apr 5, 2012

### bizboy1

Re: Basic Probability Question—Rolling Die!!!

There's two ways of doing it. Short way: given that you rolled a number, the probability of getting that number on the second roll is 1/6. But since we didn't get that number, we have 5/6. Now, the probability of getting one of those two numbers is 2/6. So we multiply 5/6*2/6=10/36.

The way I did it first was the dumb way. p(1|12)+...+P(1|16),...P(6|61)+...+P(6|65);P(1|21)+...+P(1|61),...,P(6|16)+...+P(6|56).
We notice that p3=(5x6+5X6)/6^3=5/18. You're just counting all possible combinations!

5. Apr 6, 2012

### Dickfore

Re: Basic Probability Question—Rolling Die!!!

Oh, it's until you rolled ANY number that was already rolled. I misread it needs to be the same number as the first one rolled, so I erroneously attributed it a probability of 1/6.