I found a seemingly simple question in a popular book on probability:

A draw for a football cup is to be made. There are $n$ teams ($n$ of course being even). So there will be $n/2$ matches. You can bet on the pairings of the actual matches. You will get $p$ for each correct guess, e.g. if team $A$ plays against team $B$ and you guessed that you will get $p$.

The question is how much would you be prepared to offer for the right to make the $n/2$ guesses?

This doesn't seem to be as easy as I thought it to be. The reason for this is that on the one hand you will have binomial coefficients but there is another layer of complexity because you draw the matches without replacement so you have to deal with conditional probabilities and the last match will of course be fixed without further drawing. But perhaps I am missing something and there is an easy way out?

Is this a well known problem? Is it normally posed in a different form (e.g. cards)? When not answering directly perhaps you'll have some references for me.

P.S.: I slightly generalized the question in the book but I won't mention the title for the moment because the answer given there is not very well explained, so it doesn't add any value.


You don't need conditional probabilities because of the linearity of expectation. For each of the $n/2$ matches, the probability that you predicted the first partner to play against the second partner out of the $n-1$ possible partners is $1/(n-1)$. By linearity of expectation, the expectation value for the number of correct guesses is just $n/2$ times that, i. e. $\frac n{2(n-1)}$.

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  • $\begingroup$ Thank you, joriki. Why don't you need conditional probabilities? When a match is set the participating teams are not available for the next draw and the last match will be fixed without even drawing because only two teams will be left. $\endgroup$ – vonjd Sep 22 '11 at 8:07
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    $\begingroup$ @vonjd: You don't have to worry about all that. That's what's so nice about linearity of expectation :-). Once you see that the expected number of correct guesses is just $n/2$ times the probability for any given guess to be correct, you don't have to go through all that; all you need is the probability for a single guess. This works only for the expectation value; if you wanted the distribution of the number of correct guesses, you'd have to deal with those complications, since they introduce correlations; but linearity of expectation holds independent of correlations. $\endgroup$ – joriki Sep 22 '11 at 8:57
  • $\begingroup$ @vonjd: Take $n=4$ as an example. The way you're trying to solve this, you'd say that the probability to get the first match right is $1/3$, and then the conditional probability for the second guess would be $1$ if you got the first one right and $0$ if you got it wrong, for a total expectation value of $1/3+1/3\cdot1+2/3\cdot0=2/3$. In my approach, you just take the $1/3$ for the first guess and multiply it by $n/2=2$. The result is the same, by linearity of expectation. The marginal probability must be independent of the order in which you choose to go through the matches. $\endgroup$ – joriki Sep 22 '11 at 9:07

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