Thursday, October 24, 2019

Hopium Croissants, Part II

Reader Trumpeter commented that he put TWO Hopium Croissants into the shipment of 6561 croissants shipped from Abitibi, Quebec to Sault Ste. Marie, Michigan.


An approximation for our hero's chances of identifying one of the two Hopium Croissants follows.

There are fifteen ways (numbers down the left side of the table) two Hopium Croissants can present themselves in three, large lots.

There are four ways (highlighted in yellow) one Hopium Croissant can show up on each side of the balance beam and both get tossed on the assumption that the Hopium Croissant is in the unweighed lot.

4/15 = 0.27

There are eight ways (shown in pink) where one Hopium Croissant is unknowingly released to the wild and one is kept in the pool. At that point, the selection process goes as described in the previous post and our hero nails the Trois-Fils in court.

8/15 = 0.53

The remaining three-of-fifteen cases are where both Hopium Croissants are in the lot that is kept.

3/15 = 0.20

Definition of success
From the standpoint of the detective, success is being able to deliver to the Prosecuting Attorney the strongest possible evidence of wrong-doing while staying within the rules established by the court.

Suppose it was possible for him to isolate both Hopium Croissants. If he delivered both of them to the Forensics Lab it would taint the evidence and the case would be thrown out. He cannot deliver crumbs. He cannot mash ten of them together and weigh out the equivalent of a croissant. The court said he could send one croissant.

Success, under the rules the detective must follow, involves finding one Hopium Croissant in the lot of 6561 and delivering it to the lab.

Looking at rounds two-through-eight
The odds of both Hopium Croissants being lost in the second round are 0.20 (the chances of both Hopium Croissants being in the sample) * 0.27

The odds of losing both Hopium Croissants in the third round is 0.20 * 0.20 (the odds of both HCs being in the sample) * 0.27

The pattern is (0.20^(n-1))*(4/15)

The last round has different odds because it is not possible to have two Hopium Croissants in a single lot of one item. The odds of losing both Hopium Croisants in that round is 0.20^7 * 0.33

Adding up the chances that both Hopium Croissants will be lost in any of the eight "tests" yields 27% + 5% + 0.2% and then a bunch of very tiny numbers.

The bottom line is that there is a 32% chance both Hopium Croissants will be lost and a 68% chance our hero will successfully deliver one of the Hopium Croissants to the Forensics Lab.

Pretty good jump in odds from 0.015% to 68% chance of success.


  1. I haven't had enough coffee to math today... :-)

    1. It made my head hurt, even after two, stiff mugs of coffee.

      I was surprised at how well the sort worked even with the extra Hopium Croissant.

      The key to figuring it out was to make a table as shown in the first illustration. Then the odds just drop out for the important outcomes.

  2. Pretty damn awesome for an answer to a Smart Aleck.

    I wanted to thank you for yorylong running series post eebola. For being thought provoking, excellent writing, and incredible daily output.

    It is obvious you are doing the work you were meant for.


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