# Geometric Law of Probability with Dice

• AllRelative
In summary: 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?
AllRelative

## 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 until the first type appears.
Let E2: Number of rolls until the second type appears.
Let E3: Number of rolls until 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.

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:
AllRelative
AllRelative said:

## 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 until the first type appears.
Let E2: Number of rolls until the second type appears.
Let E3: Number of rolls until 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:
Ray Vickson said:
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. Let's juste say that the dice has two sides with an A on it, two sides with a B and two sides with a C.

## 1. What is the Geometric Law of Probability with Dice?

The Geometric Law of Probability with Dice is a mathematical principle that states the probability of obtaining a specific outcome on a dice is equal to the number of ways that outcome can occur divided by the total number of possible outcomes.

## 2. How is the Geometric Law of Probability applied to dice?

The Geometric Law of Probability is applied to dice by calculating the probability of obtaining a specific number or combination of numbers on a dice, using the formula: P(E) = 1 / n, where n is the number of possible outcomes.

## 3. Can the Geometric Law of Probability be used for any type of dice game?

Yes, the Geometric Law of Probability can be applied to any type of dice game as long as the dice used are fair and have equally likely outcomes. The law can also be applied to multiple dice, by multiplying the individual probabilities together.

## 4. How is the Geometric Law of Probability useful in real life?

The Geometric Law of Probability can be useful in real life situations such as gambling, where understanding the likelihood of certain outcomes can help make informed decisions. It is also used in various industries, such as finance and insurance, to calculate risk and make predictions.

## 5. What are some limitations of the Geometric Law of Probability with Dice?

One limitation of the Geometric Law of Probability is that it assumes the dice are fair and have equally likely outcomes. In reality, this may not always be the case. Additionally, the law does not take into account external factors that may affect the outcomes, such as the force of the dice roll or the surface it lands on.

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