I'm looking for some different ways to solve a problem I thought of in my head while watching basketball.

The problem is essentially: on an NBA team, there are 15 players, 5 of which can play on the court simultaneously. What is the probability that 2 specific players are chosen, assuming that each player has equal probability of being chosen?

One way to approach this problem is with combinations. There are 15 choose 5 players total unique possibilities. Now fix 2 players to be the 2 in question, we are left with 13 possibly players to choose from and 3 spots to fill.

So the probability is $\frac{13 \choose 3}{15 \choose 5}$.

What are some other ways to approach this problem. I'm trying to come up with a solution that doesn't involve combinations, and only involves the 1/15 probability that a player is chosen.

After writing this, I attempted to reason with just probabilities, but I think I'm think, but along the right path. I was thinking that the probability of the first player being 1 of the 2 players is 2/15, and the 2nd player being the 2nd player is 1/14.


So the probability of the 1st and 2nd player chosen being the 2 players in question is 2/15*1/14. This result is exactly an order of magnitude less than the combinations approach. I want to say that there are 5 choose 2 = 10 ways to arrange the 2 players among the 5 players, and that would give the right answer $(2/15*1/14*5C2)$, but this doesn't seem right to me intuitively.


Now that I'm thinking about this again, I feel like both of the above ideas are wrong. We draw each of the 5 players from a pool of 15 players (each time we draw, we don't replace the player).

There are 10 different ways we could draw the 2 players in question:

1st and 2nd draw. probability = (2/15*1/14)

1st and 3rd draw. probability = (2/15*1/13)

1st and 4th draw. probability = (2/15*1/12)

1st and 5th draw. probability = (2/15*1/11)

2nd and 3rd draw. probability = (2/14*1/13)

2nd and 4th draw. probability = (2/14*1/12)

2nd and 5th draw. probability = (2/14*1/11)

3rd and 4th draw. probability = (2/13*1/12)

3rd and 5th draw. probability = (2/13*1/11)

4th and 5th draw. probability = (2/12*1/11)

Totaling these probabilities gives me 0.12085137085, whereas the above 2 answers gives me 0.095. I think this question is correct.

  • $\begingroup$ The problem is underspecified. Different distributions are compatible with all players being chosen with the same probability. From your solution with combinations, it seems that you're assuming that each combination of players is chosen with the same probability. $\endgroup$
    – joriki
    Commented Mar 2, 2020 at 4:42
  • $\begingroup$ @joriki. Hmm I'm trying to reason with this. Is the problem fully specified, if I instead say "INITIALLY, each of the 15 players has a probability of 1/15 of being drawn. After the first draw, the remaining 14 players each has a probability of 1/14 of being drawn, and so on..." If so, does the combinations approach match this statement? $\endgroup$
    – 24n8
    Commented Mar 2, 2020 at 4:45
  • $\begingroup$ Yes, those are equivalent (because there's nothing in that process that distinguishes one combination from another, so they must be equiprobable). $\endgroup$
    – joriki
    Commented Mar 2, 2020 at 4:52
  • $\begingroup$ @joriki Does that mean my pure probabilistic approach (see Edit 2) is wrong, since the combination approach resulted in a probability of ~ 0.095, where my Edit 2 resulted in ~0.12. If so, I'm not seeing how this my thought process was wrong on that one. $\endgroup$
    – 24n8
    Commented Mar 2, 2020 at 4:58
  • 1
    $\begingroup$ Yes, that's wrong because you didn't include the probabilities for not drawing the players on the previous draws. E.g. for the $1$st and $3$rd draw, the probability should be $\frac2{15}\cdot\frac{13}{14}\cdot\frac1{13}=\frac2{14\cdot15}$, the same as on the first two draws or any other pair of draws, as there's full symmetry among the draws and thus the probability of drawing two specific players is the same on any pair of draws (and thus your approach in "Edit 1" is correct). $\endgroup$
    – joriki
    Commented Mar 2, 2020 at 5:03

1 Answer 1


Your 10 cases are actually equally probable with $P=1/105$

You will get this if you include the probabilities of choosing the other players.

e.g. $$P_{1,3} = 2/15*13/14*1/13$$ $$P_{4,5} = 13/15*12/14*11/13*2/12*1/11$$

So the total probability is $10/105$ which agrees with the combinatoric solution.


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