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The draw for the fifth round of the FA Cup is about to be made. There are 16 teams, leading to eight matches. Your task is to pair the teams off, in an attempt to guess as many as possible of the actual matches in the real Cup draw. You are not asked which teams will be drawn at home, just which pairs will be selected.

What is the probability of at least four correct guesses?

I thought about using a conditional probability tree, after the third level it became too large and confusing.

I split the tree into 8 different levels, at level one probability of selecting a correct pair is 1/15, then, level 2 depends on whether level one guess picked a correct pair or not. Previous correct pair will compute to 1/13, a previous wrong selection will compute to 6/91. As I developed d tree it became a little complex.

osita unegbe
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1 Answers1

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Yes, a tree diagram will be unwieldy.

First, we count the number of ways to pair up the teams. We can arrange the teams in a line in $16!$ ways. Then we pair the first team with the second, the third with the fourth, and so on. Of course, in each of the groups we get the same result if the teams come in the other order, so we have to divide by $2^8.$ Also, the order of the $8$ groups is irrelevant, so we have to divide by $8!$. This gives $${16!\over2^88!}=2,027,025$$ possible pairings. (So a tree diagram seems out of the question.)

To calculate the number of ways to get at least $4$ pairs correct, note that there are ${8\choose4}$ ways to choose the correct pairs, and then we can pairs off the remaining $8$ teams in any way, since we have assured ourselves of $4$ correct pairs already. By the same logic as above, this gives $${8\choose4}{8!\over2^44^!}$$ ways.

There is a problem with this calculation. Suppose $5$ pairs are chosen correctly. We have counted this case $5$ times -- once for each of the groups of $4$ pairs among the $5.$ So, we have to subtract $4$ times the number of ways to choose $5$ correct pairs.

But we still have a problem with the case where we choose $6$ correct pairs. We've added counted it ${6\choose4}=15$ times and subtracted it $4{6\choose5}=24$ times, so we need to add back in $10$ times the number of ways to choose $6$ correct pairs.

We still have to deal with the possibility of choosing $8$ correct pairs and to do the arithmetic. (Notice that it is impossible to choose exactly $7$ correct pairs, since the eighth pair will necessarily be correct.)

I think you'll be able to complete the problem with the hints I've given you, but feel free to ask more questions at need.

EDIT

I get $4705$ for the numerator, giving a probability of approximately $0.0023211.$ A simulation with $10$ million trials resulted in $23291$ wins, in close agreement with this result. (Earlier, I posted an incorrect result of $\approx.0022106,$ but the simulation results led me to review my calculations and find the mistake.)

saulspatz
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  • Thank you @saulspatz for your answer, but I have not been able to understand why we added back 10 to the number of ways to choose 6 correctly. I have tried to visualize it as a set. – osita unegbe Mar 14 '19 at 17:09
  • Take a case where $6$ are chosen correctly. We've added in $15$ times and subtracted it $24$ times in the previous two steps. That means we've counted it $-9$ times. We want to count it once, so we add it back in $10$ times, because $-9+10=1.$ – saulspatz Mar 14 '19 at 22:30