Probability Question: No Couples in Randomly Split Groups of Four Couples

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Homework Help Overview

The problem involves determining the probability that, when four couples are randomly split into groups of two, no group contains a couple that came together. The context is rooted in combinatorial probability and derangements.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various methods for calculating total combinations and the probability of avoiding couples in pairs. Some explore the idea of calculating the complement probability, while others question the feasibility of certain approaches, such as handling cases with multiple couples.

Discussion Status

The discussion is ongoing, with participants offering different perspectives on how to approach the problem. Some suggest using derangements as a potential solution, while others express uncertainty about the complement method and its implications.

Contextual Notes

Participants note the complexity of calculating probabilities involving multiple couples and the challenges of defining independent events in this context. There is also mention of a reference to derangements, indicating a specific combinatorial concept relevant to the problem.

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Homework Statement


Four couples go to a party. They decide to split randomly into groups of 2. What is the probability that no group has a couple that came together.


Homework Equations


I actually have an MS in math; feel free to use mathematics at any level. The answer is supposed to be 37.5% or 1-37.5 I have forgotten.


The Attempt at a Solution


I've tried several different approaches.

First: Total ways possible is 8! / 2!2!2!2! OR you can do: 8C2 * 6C2 * 4C2 * 2C2 (which is the same thing); I believe this is correct.

Total ways of getting no couples (this part is incorrect). 7P6 * 5P4 * 3P2 * 1P1 = 7*5*3*1.

Second: There are 8 people, label them 1-8. 1,2 = couple, 3,4 = couple, 5,6 = couple, 7,8 = couple. This gives us 28 different combinations of couples that are possible. 7+6+5+4+3+2+1(because 1 can be paired with 2-8 = 7 ways, 2 can be paired with 3-8 = 6 etc).

But I can't figure out all the different cases; mostly because I don't want to do it this way. I feel like there should be an easier way to solve this that I'm over looking. The book solves this problem by setting up an experiment.
 
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Perhaps you can solve for the probability of the complement instead more easily.
 
vela said:
Perhaps you can solve for the probability of the complement instead more easily.

The problem with looking at the complement is that it makes it even harder. The complement of none is at least 1. In this case, you'd need 1-P(One couple)+P(Two Couples)+P(Three Couples) + P(4 Couples).

But ouch, P(3 Couples).. what's that mean? Is it possible to have 3 couples and the 4th set not being a couple? nope cannot. So how do we handle that.
 
That just means P(3) and P(4) aren't independent, so you can't just add the individual probabilities to get P(3 or 4).
 

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