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# HMMT November 2021 Problem 6 General Round

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Hello,

Recently, I've been doing of past competitions to practice math and have stumbled upon this question.

Question:

Mario has a deck of seven pairs of matching number cards and two pairs of matching Jokers, for a total of 18 cards. He shuffles the deck, then draws the cards from the top one by one until he holds a pair of matching Jokers. The expected number of complete pairs that Mario holds at the end (including the Jokers) is m/n, where m, n are positive integers and gcd(m, n) = 1. Find 100m + n.

Considering ordering the nine pairs by the time they are first complete. Since the pairs are treated equally by the drawing process, this ordering is a uniform ordering. Therefore the problem becomes the following: consider ordering 7 N’s and 2 J’s randomly. What is the expected position of the first J?

We may solve this by linearity of expectation. Every N has exactly a 1/3 chance of being in front of the 2 J’s, so the expected number of N’s before the first J is 7/3. Thus the expected position of the first J is 7/3 + 1 = 10/3.

PDF:

https://hmmt-archive.s3.amazonaws.com/tournaments/2021/nov/gen/solutions.pdf

While I understand the second part of the solution using linearity of expectation, I am a bit confused on how the problem becomes ordering 7 N’s and 2 J’s randomly, since isn't that the equivalence of assuming that everytime of draw a card, you will also draw the paired card? Perhaps it has something to do with the number of jokers being halved too.

=^._.^=

Apr 30, 2022

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Please don't waste our time with these useless questions.

Apr 30, 2022
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I don't really understand why this question is useless.

=^._.^=

catmg  Apr 30, 2022
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These are not useless questions. These are great questions.

This comment comes from one of three (3) recently-new trolls, who post as guests. One is a low-life, asshole that targets CPhill with crude cocksucker jokes, posted via text images to escape the fourm’s auto censor.  These image texts are rendered on an #### server.

The first one caught me off-guard and I LMAO. But I’d still do an Irish step dance on the asshole’s face, if I knew where to find him. CPhill is like family. ...You are too.

GA

--. .-

Guest Apr 30, 2022
edited by Guest  Apr 30, 2022
edited by Guest  Apr 30, 2022
edited by Melody  Apr 30, 2022
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Thank you GA.

It seems quite stupid to be mean on a math forum to me.

=^._.^=

catmg  Apr 30, 2022
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GA is right catmg,

It's a good question.

It is best to just ignore rude trolls.

They get bored and go away when they are ignored.  They can always find another site where they can get 'good' reactions :)

Apr 30, 2022
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Aww, thank you.

=^._.^=

catmg  Apr 30, 2022
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I am sorry catmg, I have no idea.

I do not understand the given answer and even the question is open to interpretation.

I assume there are 2 pairs of jokers where the members of the pair are identical but the pairs are different from each other? I don't think any other interpretation is likely so I guess I am being pedantic.

I would also like someone knowledgable to discuss this question.

Apr 30, 2022
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I'm going to assume that the pairs of jokers are the same, and since the probability of drawing a pair is the same for all pairs, we can simply ignore that fact and simplify it to ordering 7 N’s and 2 J’s randomly.

I am not sure if this is the correct logic, but it's starting to make a bit more sense to me.

=^._.^=

catmg  Apr 30, 2022
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It is likely correct but entirely alien to me.

Melody  Apr 30, 2022
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I think that the key sentence in their answer is "Consider ordering the nine pairs by the time they are first complete." This is tricky to think about, but, in a combinatorial context, this does not skew the uniform randomness of the selection of cards. It's less that you gain the paired card with a draw of any card, but more so that the 7 N's and 2 J's serve as a model for the pairs that appear during Mario's selection. That is, the "position of the first J" is just analogous to the first point during Mario's drawing of cards whereby he has a pair of Jokers in front of him.

May 2, 2022
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Thanks Anthrax,

I expect you are right but your explanation is still lost on me.

That is fine, I always just move on to the next question.  Thanks for explaining.

Melody  May 2, 2022
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Thank you Anthrax, I am able to understand the problem now, it took a quite a while to process. :))

Dear Melody,

I'm not sure if this is helpful, but it's always nice to think about it in small cases.

Let's say I have a pair of jokers (j, j) and a pair of ones (1, 1). I randomly mix it up and draw cards until I get a pair, by symmetry, I will draw the pair of ones first 50% of the time, and will draw the pair of jokers first 50% of the time. Notice how this is the same as the probability of having two cards (1, j), and seeing which card I draw first. I will draw the one first 50% of the time, and will draw the joker first 50% of the time.This is because since both cards are pairs(1, 1 and j, j), the probability of which pair you draw first is still the same as is if you were to draw two non-paired cards (1, j).

Hope it makes a bit more sense.

=^._.^=

catmg  May 2, 2022
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Thanks Catmg,

I have thought about it and it is deliciously simple.

I will have to think on it some more though.

Thanks for encouraging me to continue thinking on it.

Melody  May 3, 2022