Calculate the probability and the number of ways

[mathmari]

Hey!

I am looking the following exercise:

1. At a tango course $12$ married couples participate. Find an appropriate measurable space and calculate the probability that exactly nine couples dance together.
2. $5$ women sit at a round table. $3$ men come later. With how many ways can the men sit between the women, if no man can sit next to an other man?

I have done the following:

1. The measurable space is the tuple $(\Omega, \mathcal{F})$, where $\Omega$ is a set and $\mathcal{F}$ a $\sigma$-Algebra over $\Omega$, right? Is $\Omega$ the set of $12$ men and $12$ women? We have $12$ couples, so $12$ men and $12$ women. We want that $9$ couples dance together.
   
   The first man has the probability $\frac{1}{12}$ to choose the right woman.
   The second man has the probability $\frac{1}{11}$ to choose the right woman.
   The third man has the probability $\frac{1}{10}$ to choose the right woman.
   ...
   The $i$-th man has the probability $\frac{1}{12-i+1}$ to choose the right woman.
   ...
   The $9$-th man has the probability $\frac{1}{12-9+1}=\frac{1}{4}$ to choose the right woman. The remaining $3$ men have to choose the wrong woman.
   
   The first man has the probability $\frac{2}{3}$ to choose a wrong woman.
   The second man has the probability $\frac{1}{2}$ to choose a wrong woman.
   The third man has the probability $1$ to choose a wrong woman. So, the probability that exactly nine couples dance together is equal to $$\frac{1}{12}\cdot \frac{1}{11}\cdot \frac{1}{10}\cdot \ldots \cdot \frac{1}{12-i+1}\cdot \ldots \cdot \frac{1}{4}\cdot \frac{2}{3}\cdot \frac{1}{2}\cdot 1$$
2. We have the following:
   
   View attachment 7484
   
   right?
   
   The first man has $5$ possibilities to sit between two women.
   The second man has $4$ possibilities to sit between two women.
   The third man has $3$ possibilities to sit between two women.
   
   Therefore, we get that there are $5\cdot 4\cdot 3$ ways so that the men sit between the women, if no man can sit next to an other man.

Is everything correct?

If yes, are the justifications correct and complete? Could we improve that or make that more formally?

(Wondering)

 

[I like Serena]

1. At a tango course $12$ married couples participate. Find an appropriate measurable space and calculate the probability that exactly nine couples dance together. The measurable space is the tuple $(\Omega, \mathcal{F})$, where $\Omega$ is a set and $\mathcal{F}$ a $\sigma$-Algebra over $\Omega$, right? Is $\Omega$ the set of $12$ men and $12$ women?

Hey mathmari!

$\Omega$ is the set of outcomes.
A single man is not an outcome of a tango course is it? (Wondering)

Either way, it seems we have to make a couple of assumptions.

Normally I'd expect that if we have married couples, that each couple would participate, and would dance with each other.
If that were the case, there's only 1 outcome, which is that each married couple dances with each other in the tango course.
But then we can't have exactly nine (married) couples dancing together.
Or rather, this is possible, and it fits the problem statement, and the resulting probability is 0.

To make it more interesting we might for instance assume that:
1. all couples do actually participate,
2. we only have opposite sex couples,
3. each man draws a lot to see who he dances with (uniformly distributed).

How does that sound?

A single outcome could then be a set of the 12 pairings of men with women.
What would $\Omega$ be then? (Wondering)

Our sigma algebra could for instance be $\mathcal F=2^\Omega$.

So, the probability that exactly nine couples dance together is equal to $$\frac{1}{12}\cdot \frac{1}{11}\cdot \frac{1}{10}\cdot \ldots \cdot \frac{1}{12-i+1}\cdot \ldots \cdot \frac{1}{4}\cdot \frac{2}{3}\cdot \frac{1}{2}\cdot 1$$

What if the first man picks a wrong woman, but the last man picks the right woman.
That's also possible isn't it? (Wondering)

2. $5$ women sit at a round table. $3$ men come later. With how many ways can the men sit between the women, if no man can sit next to an other man?

The first man has $5$ possibilities to sit between two women.
The second man has $4$ possibilities to sit between two women.
The third man has $3$ possibilities to sit between two women.

Therefore, we get that there are $5\cdot 4\cdot 3$ ways so that the men sit between the women, if no man can sit next to an other man.

Yep. All correct. (Nod)

 

[mathmari]

A single outcome could then be a set of the 12 pairings of men with women.
What would $\Omega$ be then? (Wondering)

Do you mean that $\Omega$ will be the set of each 2-tupels $(i,j)$, where $i,j\in \{1,2,\ldots , 12\}$ ? (Wondering)

What if the first man picks a wrong woman, but the last man picks the right woman.
That's also possible isn't it? (Wondering)

Oh yes (Thinking)

There are $\binom{12}{9}$ ways to permute the position of the men.

So, the probability that exactly nine couples dance together is equal to
$$\binom{12}{9}\cdot \frac{1}{12}\cdot \frac{1}{11}\cdot \frac{1}{10}\cdot \ldots \cdot \frac{1}{12-i+1}\cdot \ldots \cdot \frac{1}{4}\cdot \frac{2}{3}\cdot \frac{1}{2}\cdot 1 $$

Is this correct? (Wondering)

Yep. All correct. (Nod)

Do we not have to take here also into consideration the position of the men? (Wondering)

 

[I like Serena]

Do you mean that $\Omega$ will be the set of each 2-tupels $(i,j)$, where $i,j\in \{1,2,\ldots , 12\}$ ? (Wondering)

I don't think so. Just a single pair doesn't sound like an outcome, just like a single man is not an outcome...
A single pair could be an event though, where nothing is known about the other pairs. (Thinking)

Oh yes (Thinking)

There are $\binom{12}{9}$ ways to permute the position of the men.

So, the probability that exactly nine couples dance together is equal to
$$\binom{12}{9}\cdot \frac{1}{12}\cdot \frac{1}{11}\cdot \frac{1}{10}\cdot \ldots \cdot \frac{1}{12-i+1}\cdot \ldots \cdot \frac{1}{4}\cdot \frac{2}{3}\cdot \frac{1}{2}\cdot 1 $$

Is this correct? (Wondering)

It is correct in this case, but generally we can't treat the last 3 like that.

Suppose we would have 4 pairs left, with people numbered 1-4.
Then we could have (1,3), (2,4) as the the first 2 couples, but 3 and 4 now both have probability $1$ to get a wrong woman instead of probabilities $\frac 12$ and $1$. (Thinking)

Do we not have to take here also into consideration the position of the men? (Wondering)

Are we not covering all possible positions for the men? (Wondering)

 

[mathmari]

I don't think so. Just a single pair doesn't sound like an outcome, just like a single man is not an outcome...
A single pair could be an event though, where nothing is known about the other pairs. (Thinking)

I haven't really understood what $\Omega$ should be. Could you explain it further to me?

Should it maybe be $\Omega=\{1,2,3\ldots,12\}^2$ ?

(Wondering)

It is correct in this case, but generally we can't treat the last 3 like that.

Suppose we would have 4 pairs left, with people numbered 1-4.
Then we could have (1,3), (2,4) as the the first 2 couples, but 3 and 4 now both have probability $1$ to get a wrong woman instead of probabilities $\frac 12$ and $1$. (Thinking)Are we not covering all possible positions for the men? (Wondering)

I got stuck right now.

We consider first a permutation of men.
We want that the first $9$ men choose the right women.
The probability that these $9$ men choose the right woman is equal to $\displaystyle{\frac{1}{12}\cdot \frac{1}{11}\cdot \frac{1}{10}\cdot \ldots \cdot \frac{1}{12-i+1}\cdot \ldots \cdot \frac{1}{4}=\prod_{i=1}^9\frac{1}{12-i+1}}$.
There are $\binom{12}{9}$ permutations to get $9$ of $12$ men.
So, the probability that $9$ men choose the right woman is equal to $\displaystyle{ \binom{12}{9}\cdot \prod_{i=1}^9\frac{1}{12-i+1}}$.

The remaining $3$ men have to choose the wrong woman.
The probability that these $3$ men choose the wrong woman is equal to $\frac{1}{2}\cdot 1\cdot 1$.

Is everything correct so far?

I haven't understood if we have in that way all the possible permutations of the $3$ men. Do we not have to multiply the probability $\frac{1}{2}\cdot 1\cdot 1$ by a binomial coefficient?

(Wondering)

 

[I like Serena]

I haven't really understood what $\Omega$ should be. Could you explain it further to me?

Should it maybe be $\Omega=\{1,2,3\ldots,12\}^2$ ?

(Wondering)

How about $\Omega = S_{12}$, the permutations of 12 women? (Wondering)

I got stuck right now.

We consider first a permutation of men.
We want that the first $9$ men choose the right women.
The probability that these $9$ men choose the right woman is equal to $\displaystyle{\frac{1}{12}\cdot \frac{1}{11}\cdot \frac{1}{10}\cdot \ldots \cdot \frac{1}{12-i+1}\cdot \ldots \cdot \frac{1}{4}=\prod_{i=1}^9\frac{1}{12-i+1}}$.
There are $\binom{12}{9}$ permutations to get $9$ of $12$ men.
So, the probability that $9$ men choose the right woman is equal to $\displaystyle{ \binom{12}{9}\cdot \prod_{i=1}^9\frac{1}{12-i+1}}$.

The remaining $3$ men have to choose the wrong woman.
The probability that these $3$ men choose the wrong woman is equal to $\frac{1}{2}\cdot 1\cdot 1$.

Is everything correct so far?

We still need to factor in that the remaining 3 have to pick a wrong woman, but otherwise this is correct.

I haven't understood if we have in that way all the possible permutations of the $3$ men. Do we not have to multiply the probability $\frac{1}{2}\cdot 1\cdot 1$ by a binomial coefficient?

No, instead we have to count the number of permutations of 3 persons that have no invariant points, and divide by the total number of permutations.
That is:
$$P(\text{3 men pick a wrong woman}) = \frac{\#\{(123),(132)\}}{\#S_3}$$
(Thinking)

Note that formally (assuming uniform probabilities when picking) we have:
$$P(\text{exactly 9 couples}) = \frac{\text{#permutations of 12 women with 9 invariant points}}{\text{#permutations of 12 women}}$$
And in the denominator we can see what $\Omega$ is supposed to be.

 

[mathmari]

How about $\Omega = S_{12}$, the permutations of 12 women? (Wondering)

So, we consider the $12$ men.
We consider the dance-couple assignment as a permutation, so $\Omega$ is the set of all possible permutations of the $12$ women, i.e., $S_{12}$. So, $\Omega$ is the set of all possible dance-couples?

Have I understood that correctly? (Wondering)

Note that formally (assuming uniform probabilities when picking) we have:
$$P(\text{exactly 9 couples}) = \frac{\text{#permutations of 12 women with 9 invariant points}}{\text{#permutations of 12 women}}$$
And in the denominator we can see what $\Omega$ is supposed to be.

How can we calculate the number of permutations of 12 women with 9 invariant points? Do we use the product of post #5 ? (Wondering)

 

[I like Serena]

So, we consider the $12$ men.
We consider the dance-couple assignment as a permutation, so $\Omega$ is the set of all possible permutations of the $12$ women, i.e., $S_{12}$.

That's what I'm suggesting yes. (Nod)

So, $\Omega$ is the set of all possible dance-couples?

Have I understood that correctly?

Not quite, $\Omega$ is the set of all possible sets of partner assignments.
Not just the set of possible dance-couples.
That is:
$$\begin{aligned}
\Omega&= S_{12} \\
&= \{ (1), (12), ... \} \\
&\cong \{ \{ (1,1), (2,2), ..., (12,12) \},\quad\{ (1,2), (2,1), (3,3), ..., (12,12) \},\quad ... \}
\end{aligned}$$
(Thinking)

How can we calculate the number of permutations of 12 women with 9 invariant points? Do we use the product of post #5 ? (Wondering)

Yep.
For the 9 invariant points that works fine, since there is only 1 choice - each represents a married couple.
However, I believe that for the 3 points that not invariant, we need to consider the different possible assignments by enumerating all of them. (Thinking)

 

[mathmari]

$\Omega$ is the set of all possible sets of partner assignments.
Not just the set of possible dance-couples.
That is:
$$\begin{aligned}
\Omega&= S_{12} \\
&= \{ (1), (12), ... \} \\
&\cong \{ \{ (1,1), (2,2), ..., (12,12) \},\quad\{ (1,2), (2,1), (3,3), ..., (12,12) \},\quad ... \}
\end{aligned}$$
(Thinking)

Ah ok! (Thinking)

Yep.
For the 9 invariant points that works fine, since there is only 1 choice - each represents a married couple.
However, I believe that for the 3 points that not invariant, we need to consider the different possible assignments by enumerating all of them. (Thinking)

We have arranged all men in order from 1-12, and now there are 1-12 free positions for women. So there are 12! arrangements possible.
For exactly 9 couples, we select 9 positions out of 12 that contain women with their respective partner, and other 3 not with their partner.
The number of ways to select 9 positions out of 12 that contain women with their respective partner, i.e., the number of ways to select 9 couples out of 12. is equal to $\binom{12}{9}$.
The number of ways so that the other 3 are not with their partner is equal to $\#\{(2,3,1), (3,1,2)\}=2$.

So, we get that $\text{#permutations of 12 women with 9 invariant points}=\binom{12}{9}\cdot 2$.

Therefore, we get the following $$P(\text{exactly 9 couples}) = \frac{\text{#permutations of 12 women with 9 invariant points}}{\text{#permutations of 12 women}}=\frac{\binom{12}{9}\cdot 2}{12!}$$

Is this correct? (Wondering)

 

[I like Serena]

Yep. All correct. (Nod)

 

[mathmari]

Yep. All correct. (Nod)

Great! (Yes)

2. $5$ women sit at a round table. $3$ men come later. With how many ways can the men sit between the women, if no man can sit next to an other man?I have done the following:

2. We have the following:
right?
The first man has $5$ possibilities to sit between two women.
The second man has $4$ possibilities to sit between two women.
The third man has $3$ possibilities to sit between two women.
Therefore, we get that there are $5\cdot 4\cdot 3$ ways so that the men sit between the women, if no man can sit next to an other man.Is everything correct?
If yes, are the justifications correct and complete? Could we improve that or make that more formally?
(Wondering)

We have that $5$ women are at the table. Between these women there are $5$ positions for the men.

We have $3$ men.

The first man has $5$ possibilities to sit between $2$ women.
The second man has $4$ possibilities to sit between $2$ women.
The third man has $3$ possibilities to sit between $2$ women.

So, there are $5\cdot 4\cdot 3$ possibilities so that the men sit betweem $2$ women and no man sits next to a man. Or: We have $5$ free positions. We want to choose $3$ out of $5$ positions. That can be done in $\binom{5}{3}$ different ways.

The men can be arranged in $3!$ ways.

So, there are $\binom{5}{3} \cdot 3!$ possibilities so that the men sit betweem $2$ women and no man sits next to a man. Is everything correct? (Wondering)
I have an other question about the measurable space. If we have to give an appropriate measurable space, do we take always the 2-tuple $(\Omega,\mathcal{F})$, where $\mathcal{F}$ is the power set of $\Omega$, i.e. $\mathcal{F}=2^{\Omega}$ ?

For example at the following:

We throw a dice twice and we add the numbers that we get. I want to give an appropriate measurable space $M$.

Is it as follows?

The universal set is $\Omega=\{1,2,3,4,5,6\}^2$, since from each dice we get the numbers from $1$ to $6$.

The measurable space is $M=(\Omega, \mathcal{F})$, where $\mathcal{F}$ is a $\sigma$-Algebra on $\Omega$, for example $\mathcal{F}=2^{\Omega}$, i.e. the power set of $\Omega$.

(Wondering)

 

[I like Serena]

We have that $5$ women are at the table. Between these women there are $5$ positions for the men.
...
Is everything correct? (Wondering)

Yep. Both solutions are correct and give the same result. (Happy)

 

[mathmari]

Ok! Thank you so much! (Mmm)