# Thread: Probability of dice and Chevalier de Mere's problem

1. ## Probability of dice and Chevalier de Mere's problem

Hi, I have a few questions, I think I got pats a & b for question1, but questions 2 and 3 I'm not sure of.

1a.) Calculate the probability of at least one "6" in four rolls
1b.) Calculate the probability of obtaining at least one "double 6" in 24 rolls

2.)Two dice are rolled. Given that the faces show different numbers, what is the probability that one face is a "4"

3.) A coin comes up heads 3 times as often as tails. The coin is tossed three times. Let the random variable X= number of heads that appear. Write out the Probability Distribution Function of X.

For 1a I got 1-(5/6)^4 which equals a probability of 0.5177
and for 1b. I got 1-(35/36)^24 equals 0.4914

For 2, I drew a table of all possible outcomes when 2 dice are rolled. I pretty sure the answer would be 5/30 because out of all 36 possible outcomes, only 30 of the rolls involve non-repeated numbers, and only 5 of the 30 rolls contain a 4, but I'm not 100% positive.

as for 3, I'm not really sure where to begin

2. ## Re: Probability of dice and Chevalier de Mere's problem

Originally Posted by crownvicman
Hi, I have a few questions, I think I got pats a & b for question1, but questions 2 and 3 I'm not sure of.

1a.) Calculate the probability of at least one "6" in four rolls
1b.) Calculate the probability of obtaining at least one "double 6" in 24 rolls

2.)Two dice are rolled. Given that the faces show different numbers, what is the probability that one face is a "4"

3.) A coin comes up heads 3 times as often as tails. The coin is tossed three times. Let the random variable X= number of heads that appear. Write out the Probability Distribution Function of X.

For 1a I got 1-(5/6)^4 which equals a probability of 0.5177
and for 1b. I got 1-(35/36)^24 equals 0.4914

For 2, I drew a table of all possible outcomes when 2 dice are rolled. I pretty sure the answer would be 5/30 because out of all 36 possible outcomes, only 30 of the rolls involve non-repeated numbers, and only 5 of the 30 rolls contain a 4, but I'm not 100% positive.

as for 3, I'm not really sure where to begin

1 is correct

2 is correct but you should think about it a bit to see if you can come up with a better way to find the solution.

To begin 3 solve the following

$p_H=3 p_T$ and $p_H+p_T=1$

From there it's just a binomial distribution with an unfair coin.

3. ## Re: Probability of dice and Chevalier de Mere's problem

I'm still a little lost on 3, this is what I have so far but I don't know if its correct I'm sorry I keep trying to post the photo but on my screen it appears sideways

4. ## Re: Probability of dice and Chevalier de Mere's problem

You did not solve the system of equations that romsek gave you. $p_H = 3p_T$ and $p_H + p_T = 1$. Solve for $p_H$ and $p_T$. You have two equations and two variables. The first equation says the probability of flipping heads is three times the probability of flipping tails. The second equation says the probability of flipping heads plus the probability of flipping tails equals 1 (in other words, a coin can only flip heads or tails, and there is a 100% chance that either one outcome or the other will occur).

Hint: if $p_H = 3p_T$ then in the equation $p_H + p_T = 1$, you know that $p_H = 3p_T$, so you can rewrite that equation as $(3p_T) + p_T = 1$. Then $3p_T + p_T = 4p_T$.

Next, the probability distribution function is:

$P(X = 0) =$??
$P(X = 1) =$??
$P(X = 2) =$??
$P(X = 3) =$??

Those are the only possible inputs, so to represent the function, you need to calculate the result of each input.

5. ## Re: Probability of dice and Chevalier de Mere's problem

Originally Posted by crownvicman
Hi, I have a few questions, I think I got pats a & b for question1, but questions 2 and 3 I'm not sure of.

1a.) Calculate the probability of at least one "6" in four rolls
1b.) Calculate the probability of obtaining at least one "double 6" in 24 rolls

2.)Two dice are rolled. Given that the faces show different numbers, what is the probability that one face is a "4"

3.) A coin comes up heads 3 times as often as tails. The coin is tossed three times. Let the random variable X= number of heads that appear. Write out the Probability Distribution Function of X.

For 1a I got 1-(5/6)^4 which equals a probability of 0.5177
and for 1b. I got 1-(35/36)^24 equals 0.4914

For 2, I drew a table of all possible outcomes when 2 dice are rolled. I pretty sure the answer would be 5/30 because out of all 36 possible outcomes, only 30 of the rolls involve non-repeated numbers, and only 5 of the 30 rolls contain a 4, but I'm not 100% positive.

as for 3, I'm not really sure where to begin
This is a poorly written question. In #1 it should say if one die is rolled or a pair of dice taking the sum.

In one reply you are told that your solution for #2 is correct. I disagree .
You have correctly understood that there are only thirty pairs with different faces.
However. there are ten pairs that contain a four:
$(4,1),~(4,2),~(4,3),~(4,5),~(4,6),~(6,4)~,(5,4)~, (3,4)~,(2,4)~,(1,4)$
So what is the correct answer for #2?

For #3. BTW. a couple of engineering professors at a CalState have proven that if a coin is filliped and caught on the fly, it is impossible for it to be unfair.

We of course can design an experiment that mimics this question.
Let $X$ be the number of heads in three tosses, $X=0,1,2,3$.
$\mathcal{P}(X=n)=\dbinom{3}{n}(1/4)^n(3/4)^{3-n}$

6. ## Re: Probability of dice and Chevalier de Mere's problem

Ah thank you! I forgot to include the rolls when 4 was the second number rolled for number 2. As far as 3 is concerned, I calculated it out (.421875+.412875+.140625+.015625) and as expected the result of p(x=1) + p(x=2) + p(x=3) = 1. Is that what is meant when it asks for the distribution?

7. ## Re: Probability of dice and Chevalier de Mere's problem

Originally Posted by crownvicman
Ah thank you! I forgot to include the rolls when 4 was the second number rolled for number 2. As far as 3 is concerned, I calculated it out (.421875+.412875+.140625+.015625) and as expected the result of p(x=1) + p(x=2) + p(x=3) = 1. Is that what is meant when it asks for the distribution?
the distribution in this case is the set of the numbers

Pr[k heads in 3 tosses] = {0.015625, 0.140625, 0.421875, 0.421875} corresponding to k=0,3.

8. ## Re: Probability of dice and Chevalier de Mere's problem

Originally Posted by Plato
This is a poorly written question. In #1 it should say if one die is rolled or a pair of dice taking the sum.

In one reply you are told that your solution for #2 is correct. I disagree .
You have correctly understood that there are only thirty pairs with different faces.
However. there are ten pairs that contain a four:
$(4,1),~(4,2),~(4,3),~(4,5),~(4,6),~(6,4)~,(5,4)~, (3,4)~,(2,4)~,(1,4)$
So what is the correct answer for #2?

For #3. BTW. a couple of engineering professors at a CalState have proven that if a coin is filliped and caught on the fly, it is impossible for it to be unfair.

We of course can design an experiment that mimics this question.
Let $X$ be the number of heads in three tosses, $X=0,1,2,3$.
$\mathcal{P}(X=n)=\dbinom{3}{n}(1/4)^n(3/4)^{3-n}$
That is double counting because the 2 dice are independent. If they were dependent than yes you would have to take into consideration all those 10 but in this case we assume it is independent thus the number of different rolls that have a particular number on a die it is a combination not a permutation.

9. ## Re: Probability of dice and Chevalier de Mere's problem

Independent probability is where order doesn't matter. Combinations are the same way thus me saying you were double counting Plato.

10. ## Re: Probability of dice and Chevalier de Mere's problem

Originally Posted by caters
That is double counting because the 2 dice are independent. If they were dependent than yes you would have to take into consideration all those 10 but in this case we assume it is independent thus the number of different rolls that have a particular number on a die it is a combination not a permutation.
Originally Posted by caters
Independent probability is where order doesn't matter. Combinations are the same way thus me saying you were double counting Plato.
You are the one who is confused on how to count the outcome space.
Roll two a pair of dice. The outcome space contains 36 pairs that form the elementary events.
If we are interested is the sum of the faces there are five ways to get the sum of eight.
Look at this expansion. The term $5x^8$ tell us there are five ways to get an eight:
$(2,6),~(3,5),~(4,4),~(5,3),~(6,2)$.
So the probability of rolling a sum of eight is $\dfrac{5}{36}$.
The sums are not are combinations nor permutations, they are pairs.

11. ## Re: Probability of dice and Chevalier de Mere's problem

numbers of pairs of dice(assuming order doesn't matter) = 6C2

numbers of pairs of dice(assuming order does matter) = 6P2

12. ## Re: Probability of dice and Chevalier de Mere's problem

Originally Posted by caters
numbers of pairs of dice(assuming order doesn't matter) = 6C2
numbers of pairs of dice(assuming order does matter) = 6P2
You are very welcome to post. But you have been posting false informatiom.
All of that is false/incorrect information.

What is the probability of rolling two dice and getting a sum of six?

What is the probability of rolling two dice and getting a sum of six, given the faces are different?

13. ## Re: Probability of dice and Chevalier de Mere's problem

you are saying something for specific whereas I am saying it for overall. You are saying it for a specific sum whereas I am saying it for the whole probability space.

14. ## Re: Probability of dice and Chevalier de Mere's problem

Originally Posted by caters
you are saying something for specific whereas I am saying it for overall. You are saying it for a specific sum whereas I am saying it for the whole probability space.
I guess that you cannot answer either of two very very simple probability questions.
So why in the world would you presume to comment on the whole probability space?

15. ## Re: Probability of dice and Chevalier de Mere's problem

5/36 for both because on 2 dice there are only 5 ways to get a 6 without double counting.

As for commenting on the whole probability space it is because 6P2 = 6^2 = 36 and there are 36 different possibilities if you consider each one to be different. 6C2 is half of that which is 18 pairs if you consider (2,4) to be the same as (4,2) and so on. Thus 36 total possibilities 18 of which are different. Everything in probability has to do with combinations and/or permutations like the different binary numbers you get if you flip a coin 8 times and you have heads = 1 and tails = 0. That has to do with permutations. Different ways to have a 9 card hand has to do with combinations. Why does the binary have to do with permutations? Because there is only 1 way to get a particular number in binary. Why does the 9 card hand have to do with combinations? Because no matter how you order the cards in the 9 card hand the cards are going to be the same.