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There are 30 people. Each of them will choose randomly two distinct numbers from the set $\{1,2,\cdots,20\}$.

What is the probability that exactly 3 of them will choose the same two numbers?

Let A denote the event that exactly 3 of them will choose the same two numbers.

Note that we have $\binom{20}{2}=190$ different ways to choose 2 distinct numbers from $\{1,2,\cdots,20\}$.

I see that there are basically two cases: all the remaining 27 choose distinct numbers or there might be groups of 2 people with the same numbers. I have difficulty with the second case!

Therefore, $P(A)=\frac{\binom{190}{1} \binom{30}{3}\binom{189}{27}27!+(???)}{190^{30}}$

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  • $\begingroup$ What do you mean by "exactly three"? Do you mean like there is a fixed pair for which there are exactly 3 people who have chosen that fixed pair? $\endgroup$ – Aryaaaaan Feb 3 at 23:27
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    $\begingroup$ For example, if three people choose the same pair, and other three people choose the (different) same pair, like (1,2),(1,2),(1,2),(3,4),(3,4),(3,4) is that also a "yes"? What if three people choose the same pair, and other four people choose the (different) same pair, like (1,2),(1,2),(1,2),(3,4),(3,4),(3,4),(3,4)? $\endgroup$ – Viliam Búr Feb 3 at 23:31
  • $\begingroup$ No. These cases are not considered. Should I say "there will be only one group of 3 people who choose the same two numbers"? $\endgroup$ – Probability student Feb 3 at 23:35
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    $\begingroup$ same as : we have an alphabet of 190 characters, and compose words of length 30; what is the probability of having exactly one max repetition of 3 of a certain character (other repeated 2,1,0) ? $\endgroup$ – G Cab Feb 3 at 23:59
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If you have $189$ selections that can be made by $27$ people with $p$ selections each chosen by two people and $27-2p$ each chosen by one person, there are $\frac{189!\,27!}{p!\,(27-2p)!\,(162+p)!\,2^p}$ ways of doing that.

Sum that over $0\le p \le 13$ and you get about $2.7025\times 10^{61}$

Multiply that by your $\binom{190}{1} \binom{30}{3}$ and you get about $2.0847\times 10^{67}$

Divide that by $190^{30}$ and you get a probability of about $0.09046$. Simulation suggests this is sensible.

Doing something similar with $190$ selections and $30$ people with no triples and you would get a probability of a less extreme result of about $0.90257$, leaving about $0.00697$ for a more extreme result

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