Counting confusion in probability

Question

Here is the quesiton:
According to the Guiness Book of World Records, Barbara Zula of Barberton, South Africa, bore six sets of twins in only seven years, between 1967 and 1973. Suppose Zula’s 12 children are divided randomly into pairs for a doubles tennis tournament, with the choice of pairs being completely uniform.
i) What is the probability that the oldest three sets of twins are all paired together?
ii) What is the probability that some set of twins is paired together?
iii) What is the probability that no set of twins is paired together?

I've been able to get part i and have an idea of how to get the other two. I'm having problems understanding the counting necessary to obtain the answer.
The first idea that came to me to solve part ii is to find the total number of teams where at least one set of twins plays together. The sample space is $$\Omega=\{(a,b):a,b\in\{1,2,3,4,5,6\}\} \therefore \#\Omega=6^2=36.$$ The ordered pair $(1,1),(2,2),\dots,(6,6)$ are considered to be the sets of twins. To choose all sets of twins to play together we use $\binom{6}{6}$ so the probability is $\Bbb P(\text{all sets of twins are together}) = \frac{1}{36}$. To choose only one set to play together we use $\binom{6}{1}$ and the probability is $\Bbb P(\text{only one set of twins play together}) = \frac{6}{36}$. I don't really understand how to count and arrange the elements to get part ii and iii. This is all I'm able to put together.

Answer 1

** The answer to part (iii) has been corrected only on $14$ Sept.**

The entire problem can be greatly simplified using the concept of the double factorial
For all possible pairs, if you have $2n$ total objects, the simplest count of total pairs is given by the double factorial $(2n-1)!!$, where for $2n = 6,$ e.g., it would be $5!3!1! = 15$

To come now to the problems, tackling them in order (i), (iii), (ii) for convenience

(i) There is only one way to pair the three oldest twins together, and using the double factorial formula, there are $(6-1)!! = 15$ ways to pair the remaining six, thus $Pr = \frac{15}{(12-1)!!} = \frac1{693}$

(iii) Applying inclusion-exclusion, we need to compute

There are a total of $11!!$ ways of pairing as explained in (i), and if one set of twins are paired, the remaining can be paired in $9!!$ ways, and son, so applying inclusion-exclusion, we get the probability as

$[(11!! - \binom619!! +\binom62*7!! - \binom63*5!! +\binom64*3!! - \binom65*1!! +\binom66]/11!! \approx 0.58$

(ii) This will just be the balance probability, $\approx 0.42$

Answer 2

*** Edited Sep. 11 2022: I had an $i$ in one spot below where I meant $j$ The correction is in red. ***

Here is a solution of part iii) by inclusion/exclusion.

Say a tournament has "property $i$" if the twins of set $i$ are paired together, for $1 \le i \le 6$, and define $S_j$ to be the total of the probabilities that a tournament has $\color{red}{j}$ of the properties, for $1 \le j \le 6$. Then $$S_j = \binom{6}{j} \prod_{i=1}^j \frac{1}{13-2j}$$ The probability that a tournament has none of the properties, i.e. no set of twins is paired together, is then $$1-S_1+S_2-S_3 + \dots + S_6 = \boxed{0.581049}$$

Having solved part iii), part ii) is now easy.

Answer 3

Per the comment of Daniel Mathias, following my answer, I made an analytical error.

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For part (ii), I label the $12$ children as $A_1,A_2, \cdots, A_6$ and $B_1, B_2, \cdots, B_6$. Here, I am assuming that $A_k,B_k ~: ~k \in \{1,2,\cdots,6\}$ represent a pair of twins.

What is the probability that some set of twins is paired together?

It seems to me that this wording is ambiguous and can intend either:

• What is the probability that $A_1,B_1$ are paired together?

• What is the complement of the probability that none of the $(6)$ pairs are twins?

Given that part (iii) is virtually the same as the 2nd bulletpoint above, I will assume that part (ii) intends the 1st bulletpoint above.

Then, the answer to part (ii) must be $~\dfrac{1}{11},~$ because there are $(11)$ people that $A_1$ might be paired with, only one of which is $B_1$.

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In my opinion the best way to attack part (iii) is combinatorically, rather than as a probability of events. I will express the probability as

$$\frac{N}{D},$$

where $N$ represents the number of satisfying pairings (i.e. no twins paired together) and $D$ represents the total number of pairings.

Then

$$D = \binom{12}{2} \times \binom{10}{2} \times \cdots \times \binom{2}{2} = \frac{(12)!}{2^6}. \tag1 $$

$\color{red}{\text{Actually, the above computation is wrong.}}$
I overlooked that I am overcounting the pairings by a factor of $(6!)$. That is, I am counting the pairings of $A_1,B_1$, then $A_2,B_2$ as distinct from first pairing $A_2,B_2$, and then pairing $A_1,B_1$.

Therefore,

$$D = \binom{12}{2} \times \binom{10}{2} \times \cdots \times \binom{2}{2} = \frac{(12)!}{6! \times 2^6}. ~~~\color{red}{(1)} $$

Therefore, the problem reduces to enumerating $N$. Here, I advocate dispensing with any (elegant) attempt at a direct approach and instead using the simpler (and much less elegant) approach of Inclusion-Exclusion.

See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

Using the syntax of the second Inclusion Exclusion link, let $S$ denote the set of all pairings, (satisfactory or not), and $S_k$ denote the subset of $S$ that has $A_k,B_k$ paired together, where $k \in \{1,2,\cdots,6\}$.

Then

$$|S| = D = \frac{(12)!}{2^6 \times 6!}$$

and

$$N = |S| - |S_1 \cup S_2 \cup \cdots \cup S_6| \tag2 $$

Let $T_0$ denote $|S|$.

For $k \in \{1,2,\cdots,6\}$, let $T_k$ denote the sum of the $\displaystyle \binom{6}{k}$ terms, given by

$$\sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq 6} |S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_k}|.$$

Then, Inclusion Exclusion theory indicates that

$$N = \sum_{k=0}^6 (-1)^k T_k. \tag3 $$

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$\underline{\text{Computation of} ~T_1}$

If $A_1,B_1$ are paired together, there are then $~\displaystyle \frac{(10)!}{2^5 \times 5!}~$ ways of pairing the other $(10)$ people. This equals $|S_1|$.

By symmetrical considerations, $|S_k| = |S_1| ~: k \in \{2,3,4,5,6\}$.

Therefore,

$$T_1 = |S_1| + \cdots + |S_6| = \frac{6 \times (10)!}{2^5 \times 5!}. \tag4 $$

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$\underline{\text{Computation of} ~T_2}$

If $A_1,B_1$ are paired together, and $A_2,B_2$ are paired together, then there are then $~\displaystyle \frac{(8)!}{2^4 \times 4!}~$ ways of pairing the other $(8)$ people. This equals $|S_1 \cap S_2|$.

By symmetrical considerations, the other $\displaystyle ~\left[\binom{6}{2} - 1\right]~$ terms in the computation of $T_2$ will be identical.

Therefore,

$$T_2 = \frac{\binom{6}{2} \times (8)!}{2^4 \times 4!}. \tag5 $$

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$\underline{\text{Computation of} ~T_k ~: k \in \{3,4,5,6\}}$

From the previous two sections, the pattern is clear:

$$T_k = \frac{\binom{6}{k} \times [12 - (2\times k)]!}{2^{[6-k]} \times [6-k]!}. \tag 6$$

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$\underline{\text{Final Computation For Part (iii)}}$

$$T_0 = D = \frac{(12)!}{2^6 \times 6!}.$$

$$T_k = \frac{\binom{6}{k} \times [12 - (2\times k)]!}{2^{[6-k]} \times [6-k]!} ~: ~k \in \{1,2,\cdots,6\}.$$

$$N = \sum_{k=0}^6 (-1)^k T_k.$$

For part (iii), the probability that none of the twins are paired together is

$$\frac{N}{D}.$$