# Tennis Probability Problem

There are 12 friends who meet once a week to play tennis. Each friend only plays one game. The tennis court is hired at $5. How much would the total cost of hire be for each friend to play every other friend twice? How many weeks are needed for all these matches to take place? Can you find the cost of hire and the time needed for larger groups of friends playing under the same conditions? I know that each friend would play 22 games, the cost would therefore be$110 for each and this would take 22 weeks, but how could you find the cost of hire and the time needed for "larger groups of friends playing under the same conditions"?

• Do you know about combinations? – Test123 Feb 11 '17 at 12:00

The number of possible pairs is $\binom {12}{2} =66\,\,$. Since each individual plays two games with each friend, the total number of games is $2 \cdot 66=132\,\,\,\,$. Playing $6$ games a week (as suggested by the fact that each friend only plays one game each week), we need $132/6=22\,\,$ weeks, as correctly noted in the OP. Assuming that the cost of $5$ dollars refers to a single game, the total cost of hire is $132 \cdot 5=660\,\,\,$ dollars, corresponding to a payment of $55\,$ dollars for each player (note that the value of $110\,$ reported in the OP must be halved because each player pays $2.5$ dollars for each game).
For larger groups, simply apply the same method by counting the possible pairs of $n$ elements and considering that, if each friend plays one game each week, there are $n/2\,$ games a week.