# probability of exactly 3 people choosing the same two numbers from n numbers

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

• 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? – Aryaaaaan Feb 3 at 23:27
• 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)? – Viliam Búr Feb 3 at 23:31
• No. These cases are not considered. Should I say "there will be only one group of 3 people who choose the same two numbers"? – Probability student Feb 3 at 23:35
• 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) ? – G Cab Feb 3 at 23:59

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