# Math Help - Probability - married couples paired correctly at a dance

1. ## Probability - married couples paired correctly at a dance

A ballroom contains four married couples (four men and their four wives). The gentlemen are to be paired up randomly with the ladies, one man to each woman. After one dance, the eight people are to be separated, except for the married couples that happen to be together by chance. The separated people will then be paired up again randomly for the second dance, again one man to each woman.

(a) Find the probability that exactly two married couples will be dancing together on the second dance.
(b) Suppose that exactly one married couple are dancing together on the second dance. What is the probability that no man was dancing with his wife on the first dance?

2. Hello Robb
Originally Posted by Robb
A ballroom contains four married couples (four men and their four wives). The gentlemen are to be paired up randomly with the ladies, one man to each woman. After one dance, the eight people are to be separated, except for the married couples that happen to be together by chance. The separated people will then be paired up again randomly for the second dance, again one man to each woman.

(a) Find the probability that exactly two married couples will be dancing together on the second dance.
(b) Suppose that exactly one married couple are dancing together on the second dance. What is the probability that no man was dancing with his wife on the first dance?
This problem is equivalent to looking at the re-arrangements (permutations) of objects A, B, C, ... and finding out how many times a particular object comes in its 'natural' place: A in 1st place, B in 2nd place, and so on.

Let's start with two objects A and B. There are obviously 2! (=2) arrangements of them. In one arrangement, both objects are in their natural place, and in the other neither is. So if we use p(n) to denote the probability that n objects are in their natural place, then for two objects:

• p(2) = $\tfrac12$
• p(1) = 0
• p(0) = $\tfrac12$

With three objects, there are 3! (=6) arrangements.

• In one of these all three objects are in their natural place; so p(3) = $\tfrac16$.
• It's impossible to have exactly 2 objects in their natural place; so p(2) = 0
• There are three ways of choosing one of the objects to be in its natural place, and then just one way of arranging the remaining two so that neither is in its natural place; so p(1) = $\tfrac36=\tfrac12$
• There are just two arrangements left from the six possible ones; so p(0) = $\tfrac26 = \tfrac13$

With four objects, there are 4! (=24) arrangements. Using similar arguments:

• p(4) = $\tfrac{1}{24}$
• p(3) = 0
• For p(2), there are $^4C_2=6$ ways of choosing the two to be 'correct', and then just one way of making the remaining two 'incorrect'. So p(2) = $\tfrac{6}{24}=\tfrac14$
• For p(1), there are 4 ways of choosing the 'correct' one, and then 2 ways of arranging the remaining 3 where none is correct (see above, for 3 objects). So p(1) = $\tfrac{8}{24}=\tfrac13$
• This accounts for 1+6+8 = 15 of the 24 arrangements. So in the remaining 9, none is in the correct place. So p(0) = $\tfrac{9}{24} = \tfrac38$.

Now that we have established all the relevant individual probabilities, we can start to answer the question. I'll get you started:

(a) This will happen in one of three ways:

• 2 'correct' from 4 followed by 0 correct from 2. The probability of this is ... x ... (using the above results)
• 1 'correct' from 4, followed by 1 from 3. The probability of this is ...
• 0 correct from 4, followed by 2 from 4.

Work out these three probabilities and add them together.

(b) Use Bayes' Theorem. See here

Can you complete it now?

3. Originally Posted by Robb
A ballroom contains four married couples (four men and their four wives). The gentlemen are to be paired up randomly with the ladies, one man to each woman. After one dance, the eight people are to be separated, except for the married couples that happen to be together by chance. The separated people will then be paired up again randomly for the second dance, again one man to each woman.

(a) Find the probability that exactly two married couples will be dancing together on the second dance.
(b) Suppose that exactly one married couple are dancing together on the second dance. What is the probability that no man was dancing with his wife on the first dance?
Part (a)

One can consider this problem as a problem of placing n balls in n boxes. Let's take balls as men and boxes as women. There are 4! unique possibilities of pairing 4 men with 4 women, such that no woman is paired with more than one man. Assuming each of such arrangements as equally likely, there are 4C2 possible arrangements in which one gets to have 2 married couples paired up. Please note that there's only one possible arrangement of pairing the remaining men and women who aren't married to each other against each possible arrangement of exactly 2 married couples.

So, the probability of having exactly two married couples at second dance is 6/4!.

Part (b)

The probability of such happening is zero as we have already assumed that there's a single married couple dancing in the second dance.

I hope my reasoning is correct.

4. Originally Posted by Steppenwolf
Part (a)

One can consider this problem as a problem of placing n balls in n boxes. Let's take balls as men and boxes as women. There are 4! unique possibilities of pairing 4 men with 4 women, such that no woman is paired with more than one man. Assuming each of such arrangements as equally likely, there are 4C2 possible arrangements in which one gets to have 2 married couples paired up. Please note that there's only one possible arrangement of pairing the remaining men and women who aren't married to each other against each possible arrangement of exactly 2 married couples.

So, the probability of having exactly two married couples at second dance is 6/4!.

Part (b)

The probability of such happening is zero as we have already assumed that there's a single married couple dancing in the second dance.

I hope my reasoning is correct.
Sorry, but this is just plain wrong. See my detailed analysis above.

Sorry, but this is just plain wrong. See my detailed analysis above.

Can you please elaborate a bit as to how my reasoning is wrong? I am still not able to figure this out..

6. Hello Steppenwolf
Originally Posted by Steppenwolf

Can you please elaborate a bit as to how my reasoning is wrong? I am still not able to figure this out..
You haven't taken into account the fact that there are two separate selections: once before the first dance, and again before the second. Please read the original questions and my first reply carefully.

Hello SteppenwolfYou haven't taken into account the fact that there are two separate selections: once before the first dance, and again before the second. Please read the original questions and my first reply carefully.

Thanks.

My mistake, I mis-interpreted the problem.

I think i have part (a) sorted, with
$P(\mbox{2 couples dancing in the second})=\frac{6}{24}\cdot \frac{1}{2}+\frac{8}{24}\cdot \frac{2}{6}+\frac{9}{24}\cdot \frac{6}{24}=\frac{37}{96}$

But for (b) i am having problems with using Baye's rule...

I have defined the event A=1 pair dancing in the second
event B=no pair dancing in the first.

So
$P(B)=\frac{9}{24}$ and $P(A)=P(\mbox{1 pair in first from 4} )*$ $P( \mbox{0 pair in second dance from 3})$ $+P(\mbox{0 pair in first from 4})*P(\mbox{1 pair in second dance from 4})$ $=\frac{8}{24}\cdot \frac{1}{3}+\frac{9}{24}\cdot \frac{8}{24}=\frac{17}{72}$

So for bayes rule, $P(B|A)=\frac{P(B) \cdot P(A|B)}{P(A)}$

with $P(A|B)=\frac{P(A \cap B)}{P(B)}$

So I am unsure what $P(A \cap B)$ should equal. I think from the definition of finding the probability of A (Using the law of total probability?) $P(A)=P(A \cap B)+P(A \cap \bar{B})$ in which case $P(A \cap B)=\frac{9}{24}\cdot \frac{8}{24}=\frac{1}{8}$
Which means that $P(A|B)=\frac{1/8}{9/24}=\frac{1}{3}$

So $P(B|A)=\frac{9/24 \cdot 1/3}{17/72}=\frac{9}{17}$

I really feel like i have done something wrong with the intercept of A and B...

9. Hello Robb
Originally Posted by Robb
I think i have part (a) sorted, with
$P(\mbox{2 couples dancing in the second})=\frac{6}{24}\cdot \frac{1}{2}+\frac{8}{24}\cdot \frac{2}{6}+\frac{9}{24}\cdot \frac{6}{24}=\frac{37}{96}$
Correct!
But for (b) i am having problems with using Baye's rule...

I have defined the event A=1 pair dancing in the second
event B=no pair dancing in the first.

So
$P(B)=\frac{9}{24}$ and $P(A)=P(\mbox{1 pair in first from 4} )*$ $P( \mbox{0 pair in second dance from 3})$ $+P(\mbox{0 pair in first from 4})*P(\mbox{1 pair in second dance from 4})$ $=\frac{8}{24}\cdot \frac{1}{3}+\frac{9}{24}\cdot \frac{8}{24}=\frac{17}{72}$

So for bayes rule, $P(B|A)=\frac{P(B) \cdot P(A|B)}{P(A)}$

with $P(A|B)=\frac{P(A \cap B)}{P(B)}$

So I am unsure what $P(A \cap B)$ should equal. I think from the definition of finding the probability of A (Using the law of total probability?) $P(A)=P(A \cap B)+P(A \cap \bar{B})$ in which case $P(A \cap B)=\frac{9}{24}\cdot \frac{8}{24}=\frac{1}{8}$
Which means that $P(A|B)=\frac{1/8}{9/24}=\frac{1}{3}$

So $P(B|A)=\frac{9/24 \cdot 1/3}{17/72}=\frac{9}{17}$

I really feel like i have done something wrong with the intercept of A and B...
This is also correct, but you've made it unnecessarily complicated. With $A$ and $B$ as you have defined them, then, as you correctly said:

$p(A) = \frac{17}{72}$ and $p(B)= \frac38$

Then $p(A|B)$ = probability that one pair is dancing in the second dance, given that no pair was dancing in the first = probability of getting one 'correct' from 4 $= \frac13$.

It's confusing to start talking about $p(A\cap B)$; you simply need to focus on the words: given that no pair was dancing in the first dance, what is the probability that one pair is dancing in the second? We are given that no pair was dancing in the first dance - so take that as read. What, then - after this has happened - is the probability that there is one pair dancing in the second dance? The answer is that, at the second selection from all four couples, just one pair is 'correct'. And the probability of this is $\frac13$.

Now we need to find the probability that there were no married couples dancing in the first dance, given there was just one married couple dancing in the second. In other words, we need $p(B|A)$. Which you have correctly found, using Bayes' Theorem.

I must confess, I often find Bayes' Theorem problems difficult to sort out, and in my experience, most students do as well. So here's another way you can look at it (which may make it clearer):

Note first that $p(A) = \frac18+\frac19 = \frac{17}{72}$, where the two fractions $\frac18$ and $\frac19$ arise through the two different 'routes' to get to the result denoted by $A$. Suppose, then, that this whole 'experiment' (that is, two dances with two selection processes) is repeated a large number of times. Then we have found that, in the long run, out of every 72 experiments, we can expect that in 17 there will be one married couple dancing in the second dance. Out of these 17 occasions, 9 will arise after there were no married couples in the first dance ( $= \tfrac18$th of 72) and 8 will arise after there was one married couple in the first dance ( $= \tfrac19$th of 72). The ratio of these different 'routes' to the final scenario is 9:8. So the probability that it has been arrived at via the first route (i.e. no married couple dancing in the first dance) is $\frac{9}{8+9}=\frac{9}{17}$.