C3_W2_Assignment: Exercise 12

Course Q&A Math for Machine Learning Probability & Statistics for Machine Learning &...
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I am trying to solve Exercise 12 of C3_W2_Assignment:

Exercise 12:
Given a n-sided dice (could be fair or not). You throw it twice and record the sum (there is no dependance between the throws). If you are only given the histogram of the sums can you use it to know which are the probabilities of the dice landing on each side?

My Solution:
For simplicity to demonstration, assume n = 3.
And given the histogram of the sum, I should know value of:
P(sum=2), P(sum=3), P(sum=4), P(sum=5), P(sum=6)

And because there is no dependance between the throws:
P(X=1)=P(Y=1) =: P1
P(X=2)=P(Y=2) =: P2
P(X=3)=P(Y=3) =: P3

And I can have equations such as:
P(sum=2) = P(X=1)P(Y=1) = P1^2
P(sum=3) = P(X=1)P(Y=2) + P(X=2)P(Y=1) = 2* P1 * P2
P(sum=4) = P(X=1)P(Y=3) + P(X=2)P(Y=2) + P(X=3)P(Y=1) = 2* P1 * P3 + P2^2
P1 + P2 + P3 = 1 (sum of all probabilities should be 1)

So, I believe from the equations above I should able to solve the value for P1, P2, P3 (each side probability)

And I am not care about the dice is fair or not in the solution.

So my answer is “yes, regardless if the die is fair or not”, but its actually wrong.

Please help to indication which step or assumption is wrong of my solution, thank you very much for your time!

2 Likes
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Hello @yusun
I think it’s the step where you assume independence between the throws is incorrect. When you throw the dice twice and record the sum, the outcome of the second throw should be dependent on the outcome of the first throw.

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Hi @Isaak_Kamau
Thanks for replaying, but the question has statement “(there is no dependance between the throws)”, thats the reason I am assume independence between the throws.

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I still don’t have access to the course but if the question explicitly states that there is no dependence between the throws your approach and equations should be valid.

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I think this might be a semantic issue.

I read the no dependence as one literally does not affect the other, not that it denies the conditionality. In a sense you are always throwing the second die but only using it’s result based on the outcome of the first. In this way it would change the answer, at least as I understand it.

Just my two cents.

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Hi yusun,

You’re absolutely correct. The assessment goes wrong for this one.

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I think you are right. I chose the same answer and I was very surprised to see that it was marked as wrong.

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Hi all, this is indeed a bug in the grader, it will be fixed asap! Sorry for this

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The fix is now live! Thanks @yusun for flagging this :slight_smile:

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Thanks for fixing the issue :slight_smile:

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