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Geometric Law of Probability with Dice

  • #1

Homework Statement


We have a normal 6 sided dice marked from 1 to 6. There is an equal chance to get each number at every roll. Let's put 1&2 as A type, 3&4 as B type and 5&6 as C type.

We roll the dice over and over until we get a number of every type.
Let X be the number of rolls.
We are interested in the mean of X.

Homework Equations


I see three geometric distributions in there.
Let Y be a Geometric Random Variable and p the probability of success:
P(Y=x) = p(1-p)^(x-1)
Mean(Y) = 1/p

The Attempt at a Solution


I am not sure if my reasoning is legit. I split an experiment into 3 smaller ones and added their mean.

I split the problem into three steps.
Let E1: Number of rolls untill the first type appears.
Let E2: Number of rolls untill the second type appears.
Let E3: Number of rolls untill the third type appears.

1)
Here p = 1. The first roll to eliminates a first type.
Mean(E1) = 1

2)
Since a first type is out, The new p for either of the other two types of appeating are 4/6= 2/3.
Therefore,
Mean(E2) = 1 / p = 1 / (2/3) = 1.5

3)
The new p is now 2/6 = 1/3.
So
Mean(E3) = 1 / p = 1 / (1/3) = 3

We can say that the Mean(X) = 1 + 1.5 + 3 = 5.5

We wait on average 5.5 rolls before the three types apprears.
 

Answers and Replies

  • #2
RPinPA
Science Advisor
Homework Helper
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That looks exactly right. This is a classic probability calculation called the "coupon collector's problem". It is usually solved by defining exactly the random variables you did. You are correct that they are geometric and everything looks perfect in your calculation. Well done!
 
Last edited:
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
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Homework Statement


We have a normal 6 sided dice marked from 1 to 6. There is an equal chance to get each number at every roll. Let's put 1&2 as A type, 3&4 as B type and 5&6 as C type.

We roll the dice over and over until we get a number of every type.
Let X be the number of rolls.
We are interested in the mean of X.

Homework Equations


I see three geometric distributions in there.
Let Y be a Geometric Random Variable and p the probability of success:
P(Y=x) = p(1-p)^(x-1)
Mean(Y) = 1/p

The Attempt at a Solution


I am not sure if my reasoning is legit. I split an experiment into 3 smaller ones and added their mean.

I split the problem into three steps.
Let E1: Number of rolls untill the first type appears.
Let E2: Number of rolls untill the second type appears.
Let E3: Number of rolls untill the third type appears.

1)
Here p = 1. The first roll to eliminates a first type.
Mean(E1) = 1

2)
Since a first type is out, The new p for either of the other two types of appeating are 4/6= 2/3.
Therefore,
Mean(E2) = 1 / p = 1 / (2/3) = 1.5

3)
The new p is now 2/6 = 1/3.
So
Mean(E3) = 1 / p = 1 / (1/3) = 3

We can say that the Mean(X) = 1 + 1.5 + 3 = 5.5

We wait on average 5.5 rolls before the three types apprears.
The problem statement has some ambiguity. Does "and" (as 1 & 2) really mean "and" as in logic, so that in order to collect an A you need to have obtained both a 1 and a 2? Alternatively, does "and" really mean "or" in this problem (as 1 or 2), so that in order to collect an A you need to get either 1 or a 2?
 
Last edited:
  • #4
The problem statement has some ambiguity. Does "and" (as 1 & 2) really mean "and" as in logic, so that in order to collect an A you need to have obtained both a 1 and a 2? Alternatively, does "and" really mean "or" in this problem (as 1 or 2), so that in order to collect an A you need to get either 1 or a 2?
Sorry I wrote the question by memory and english is not my first language. Lets juste say that the dice has two sides with an A on it, two sides with a B and two sides with a C.
 

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