Where my logic went wrong while solving this probability problem from Sheldon Ross' book?

Question

I was solving following problem from the book "A first course in probability" by Sheldon Ross

A group of 6 men and 6 women is randomly divided into 2 groups of size 6 each. What is the probability that both groups will have the same number of men?

I solved it as follows:

It was clear to me that this division can be done in $\binom{6}{3}\times\binom{6}{3}$ ways. However I was a bit confused of denominator against which probability is to be calculated, that is, how many number of ways are there to divide 12 people into two groups. I thought it as follows:

• One group contains all men, other group contains all women. This can be done in only $1$ way.
• One group contains 1 man and 5 women, and the other group contains all remaining (5 men and 1 woman). This can be done in $\binom{6}{1}\times\binom{6}{5}$ ways.
• One group contains 2 men and 4 women, and the other group contains all remaining (4 men and 2 women). This can be done in $\binom{6}{2}\times\binom{6}{4}$ ways.
• One group contains 3 men and 3 women, and the other group contains all remaining (3 women and 3 men). This can be done in $\binom{6}{3}\times\binom{6}{3}$ ways. So I felt that the final probability would be $\frac{\binom{6}{3}\binom{6}{3}}{1+\binom{6}{1}\binom{6}{5}+\binom{6}{2}\binom{6}{4}+\binom{6}{3}\binom{6}{3}}=\frac{400}{662}\approx 0.6042$

But the answer given in the book is $0.4329$ which is apparently calculated as (I realised this when I googled) $$\frac{\binom{6}{3}\binom{6}{3}}{\frac{12!}{6!6!}}=\frac{400}{924}$$

My doubts

1. Where did I made mistake in my logic while forming sample space of 662?
2. Somehow, I still feel mine is correct. Is it so?

Some update

I tried forming sample space for smaller number of people: 2 men and 2 women divided in 2 groups of 2 people each:

By books formula, it will be $\frac{4!}{2!2!}=6$, which can me cross checked as follows:

m1 m2 w1 w2    Two groups     Division ID
-----------  --------------   -----------
g1 g1 g2 g2  {m1,m2}{w1,w2}       D1
g1 g2 g1 g2  {m1,w1}{m2,w2}       D2
g1 g2 g2 g1  {m1,w1}{m2,w1}       D3
g2 g2 g1 g1  {w1,w2}{m1,m2}       D4
g2 g1 g2 g1  {m2,w2}{m1,w1}       D5    
g2 g1 g1 g2  {m2,w1}{m1,w2}       D6

By my logic, it would have been $\binom{2}{1}\binom{2}{1}+\binom{2}{2}\binom{2}{0} = 4+1=5$. When I compared this to above table, I realised that my logic considers divisions D1 and D6 as same and I felt that its essentially correct as these divisions simply forms one group of all men and other group of all women. Books logic is counting them twice by once calling men's group as "group 1" and women's group as "group 2" and then again by calling men's group as "group 2" and women's group as "group 1". However both divisions are same. However, then I felt that

• D2 and D5 are also same
  Both groups m1 and w1 together and m2 and w2 together
• D3 and D6 are also same
  Both groups m1 and w2 together and m2 and w1 together

So total distinct divisions would have been just 3. That means even my logic would have been wrong!!!

Whats going wrong? Am I thinking unnecessarily too much?

Answer 1

One group contains $3$ men and $3$ women and the other group contains all remaining ($3$ women and $3$ men). This can be done in $\binom{6}{3}\binom{6}{3}$ ways.

This claim is true if the groups are labeled. For instance, if one group is assigned to do computer programming and the other is assigned to software testing, then it matters which person is in which group. Under that assumption, there are $\binom{12}{6}$ ways to select the two groups since it matters which six people are placed in the computer programming group. Under this assumption, your sample space was too small since you did not take into account the possibilities that there are more men than women in the group you are selecting. The number of ways of selecting exactly $\color{blue}{k}$ men and $\color{red}{6 - k}$ women for the computer programming group is $$\color{blue}{\binom{6}{k}}\color{red}{\binom{6}{6 - k}}$$ so the number of ways of selecting the computer programming group is $$\color{blue}{\binom{6}{0}}\color{red}{\binom{6}{6}} + \color{blue}{\binom{6}{1}}\color{red}{\binom{6}{5}} + \color{blue}{\binom{6}{2}}\color{red}{\binom{6}{4}} + \color{blue}{\binom{6}{3}}\color{red}{\binom{6}{3}} + \color{blue}{\binom{6}{4}}\color{red}{\binom{6}{2}} + \color{blue}{\binom{6}{5}}\color{red}{\binom{6}{1}} + \color{blue}{\binom{6}{6}}\color{red}{\binom{6}{0}} = \binom{12}{6}$$ Then the probability that both groups have the same number of men is $$\frac{\dbinom{6}{3}\dbinom{6}{3}}{\dbinom{12}{6}}$$

However, the assumption that the groups are labeled is not necessary. But we need to be careful when we divide a group of $12$ people into two groups of six people. Choosing six of the $12$ people to be in one group means the other six people are in the other group. Hence, we get the same two groups when we pick a particular group of people and when we pick its complement. Thus, in selecting six of twelve people to be in one group, we count every division into two groups of six twice. Therefore, the number of ways of dividing twelve people into two groups of six is $$\frac{1}{2}\binom{12}{6}$$ and the number of ways of selecting three men and three women to be in one group is $$\frac{1}{2}\binom{6}{3}\binom{6}{3}$$ so the probability that each group contains the same number of men is $$\frac{\dfrac{1}{2}\dbinom{6}{3}\dbinom{6}{3}}{\dfrac{1}{2}\dbinom{12}{6}} = \frac{\dbinom{6}{3}\dbinom{6}{3}}{\dbinom{12}{6}}$$

Alternate Solution: Suppose that Alexander is one of the men. He must be in one of the two groups. There are $\binom{11}{5}$ ways to select five of the other eleven people to be in his group. In a favorable case, two of the other five men and three of the six women must be in his group. The number of such selections is $\binom{5}{2}\binom{6}{3}$. Hence, the probability that both groups will have the same number of men is $$\frac{\dbinom{5}{2}\dbinom{6}{3}}{\dbinom{11}{5}}$$

Answer 2

You have considered the following cases:

• 6 men, 0 women
• 1 man, 5 women
• 2 men, 4 women
• 3 men, 3 women

You are missing the other three:

• 0 men, 6 women
• 4 men, 2 women
• 5 men, 1 woman

Update your denominator to $$ \left[1 + \binom 6 1 \binom 6 5 + \binom 6 2 \binom 6 4 + \binom 6 3 \binom 6 3 \right] + \binom 6 4 \binom 6 2 + \binom 6 5 \binom 6 1 + 1$$ to get the correct answer.

I think your intuition is that there is something fundamentally similar about, say, 4 men and 2 women versus 2 men and 4 women. That is of course correct; the probabilities are the same, but nevertheless they are separate cases that (in this formulation) must be accounted for separately. In short, your sample space is indeed too small.

EDITED TO ADD: Perhaps the real crux of the confusion is the distinguishability of the groups. Generally, in probability you want to regard all things as distinguishable, even if it feels like they may not be distinguishable in practice.

As an example, if you roll two identical dice, they may not be "distinguishable" to your eye, but they are nevertheless different dice and must be treated as distinguishable when you do the calculation; this is why you fix one to be "Die A" and the other to be "Die B," even if in practice you would not be able to tell which was which after rolling them. The situation is similar here with the groups. Although there may be no meaningful distinction between them, imagine that one must stand on the left side of a room and the other on the right, or that one group is lucky and receives a prize, etc. Clearly, such matters can't actually affect the computation, and distinguishing between the groups helps you to avoid issues like the $\binom 6 3$ repetition articulated by others in this thread.

Answer 3

Suppose we have to board 12 people in two canoes.

Actually there are three problems here :

i) the indistinguishable canoes having the solution:

$$\frac{ \frac{1}{2}\binom{6}{3}\binom{6}{3}}{1+\binom{6}{1}\binom{6}{5}+\binom{6}{2}\binom{6}{4}+\frac{1}{2}\binom{6}{3}\binom{6}{3}}= \frac{200}{1+36+225+200}=\frac{200}{462}$$

ii) indistinguishable canoes with Alexander and other five men $$\frac{\binom{5}{2}\binom{6}{3}}{\binom{11}{5}}=\frac{200}{462}$$

iii) the distinguishable canoes : $$\frac{\binom{6}{3}\binom{6}{3}}{\binom{12}{6}}=\frac{400}{924}$$

If the twelve people were to be distributed in three unlabeled canoes, there are five partition types :

$0+2+4 \to \binom{6}{2} \binom{6}{2} = 15×15 $

$1+1+4 \to \frac{1}{2}\binom{6}{2} \binom{6}{3} = 15×20 $

$0+3+3 \to \frac{1}{2}\binom{6}{3} \binom{6}{1} \binom{5}{1}= 10×30 $

$1+2+3 \to \binom{6}{1} \binom{5}{2}\binom{6}{1} \binom{5}{2} = 60×60$

$2+2+2 \to \frac{1}{6}\binom{6}{2} \binom{4}{2} \binom{2}{2}\binom{6}{2} \binom{4}{2} \binom{2}{2}= 15×90 = 1350$

$$P_{unlabeled} = \frac {1350}{5775} = \frac{18}{77} \approx 0.2337 $$

The labeled case is simpler :

$$P_{labeled} = \frac {\binom{6}{2} \binom{4}{2} \binom{2}{2}\binom{6}{2} \binom{4}{2} \binom{2}{2} } { \binom{12}{4} \binom{8}{4} \binom{4}{4} } = \frac{18}{77} \approx 0.2337 $$

Finally, I would like to add the species/e.g.f. approach:

$$ \frac {coeff \ 6,6} {coeff \ 6,6} \ \frac {(E_2[X]·E_2[Y])^3}{(E_4[X+Y])^3} \to \frac {8100}{34650} = \frac {18}{77} $$ We have to divide the coefficients of $\frac {x^6y^6}{6!.6!}$ in the two egf's of the associated species.

Answer 4

What went wrong with your logic has been pointed out in two answers, so I shan't dwell on it.

Re the question of groups being labelled or unlabelled, please understand that # of ways hasn't been asked for, and since probability is a ratio, you will get the same answer under either of the assumptions, and you can choose the assumption that is simplest to compute.

So I choose to consider the groups as labelled, and compute$\quad Pr =\dfrac{\binom63\binom63}{\binom{12}6}$