C3_W2_Assignment Issue

Course Q&A Math for Machine Learning Probability & Statistics for Machine Learning &...
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Exercise 7:

Given a 6-sided fair dice you try a new game. You only throw the dice a second time if the result of the first throw is lower or equal to 3. Which of the following probability mass functions will be the one you should get given this new constraint?
Hints:

  • You can simulate the second throws as a numpy array and then make the values that met a certain criteria equal to 0 by using np.where

My solution code:

{moderator edit - solution code removed}

Actual Output

image

Expected answer

image
image997Ă—267 22.1 KB

Correct Coursera answer was supposed to be right-center. Can anyone explain how does this work?

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Hi @abdallahheidar!

Please do not include the solution code in your post as stated in the guideline: “include relevant info but please do not post solution code or your entire notebook”.

On my end, it seems like the expected answer should be the left-center graph, the one in blue.

Keep in mind that if the first throw is greater than 3, we do not throw the die a second time, but we still keep the sum of just the first throw’s value. So you don’t want to apply the filter to include the first throw when they are only <= 3. Instead, you want to apply the filter on the second throws, on the condition of the first throw being <= 3, since we only throw the die a second time when that condition is true.

To explain the expected answer, we can count up the possible sums of the throws.

  • If the first throw is 1, there are 6 possibilities for the second throw: 1, 2, 3, 4, 5, or 6
  • If the first throw is 2, there are 6 possibilities for the second throw: 1, 2, 3, 4, 5, or 6
  • If the first throw is 3, there are 6 possibilities for the second throw: 1, 2, 3, 4, 5, or 6
  • If the first throw is 4, 5 or 6, we do not throw the die a second time.

As a result, all the possible sums are:

  • when first throw is 1, possible sums: 2, 3, 4, 5, 6, 7
  • when first throw is 2, possible sums: 3, 4, 5, 6, 7, 8
  • when first throw is 3, possible sums: 4, 5, 6, 7, 8, 9
  • when first throw is 4, possible sums: 4
  • when first throw is 5, possible sums: 5
  • when first throw is 6, possible sums: 6

The counts for each sum are then:
sum of 2: 1
sum of 3: 2
sum of 4: 4
sum of 5: 4
sum of 6: 4
sum of 7: 3
sum of 8: 2
sum of 9: 1

Based on this outcome, the closest graph would be the left-center one.

I hope this helps!

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Thanks for the reply, I changed the logic according to your instructions and got this output. Does it match yours?
image

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Hi @abdallahheidar, glad I could help.

Yes, I think that’s looking closer to the expected output. Good job!

I noticed there are only 7 bars in your plot, but there should be 8, one for each possible sum.

In order to get the expected plot, I would suggest referring back to the notebook for the Lab “Simulating Dice Rolls with Numpy”. The notebook should have some calls to sns.histplot() which you can try to follow using the same parameter values, such as discrete.

Best of luck!

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