Basic Probability Question—Rolling Die

  • Thread starter bizboy1
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  • #1
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Basic Probability Question—Rolling Die!!!

Homework Statement


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.


Homework Equations





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!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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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
2,967
5


Why isn't p3 = 5/36?
 
  • #4
11
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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
2,967
5


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.
 

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