## Problem no. 30, 2007

### The problem

Your local club awards three prizes every year: Best average for the first half of the year; Best average for the second half of the year and Best average for the whole year.
BigBob is feeling extremely well, because he is going to be the first to win all three prices. He has been extremely successful, playing thirty pairs games, winning them all. His only competitor for the awards, LittleBob, has also played thirty pair games, but he finished second in them all.
During both the first and the second half of the year, BigBob's average was one percentage point higher than LittleBob's. So, it sure looked like BigBob could pocket all three awards!
As expected, BigBob won both half-year prices, but when it was time for the one-year price, BigBob couldn't belive his ears. The club's tournament director stated that the price for the highest average for the whole year should go to ... LittleBob. Outraged, BigBob accused the club for both cheating and incompetence. But when he finally clamed down and could take a look at the calculations, he had to agree that they were correct. The price should go to LittleBob, since he was the player with the highest average for the year.

### Solution

If BigBob and LittleBob have played, say, 15 tournaments each season, BigBob would win. And the same is true if the average is the same for both halves of the year. But if the average is higher for one half, and LittleBob has played more tournaments during that half, he may get a higher overall average than BigBob.
Let's say that BigBob has played ten tournaments during the first half of the year with an average of 65%, and 20 tournaments during the second half of the year with an average of 61%. Then, his average for the whole year is (10 x 65) + (20 x 61) / 30 = 62,33%. LittleBob has played 20 tournaments during the first half of the year and ten during the second half. His overall average will be (20 x 64) + (10 x 60) / 30 = 62,67%.
As some solvers has shown, there is a general formula to solve this problem. LT is the number of tournaments LittleBob has played during the half that he has played more than BigBob. BT is the number of tournaments BigBob has played during the same half. A1 is the average LittleBob had for the half he played most, and A2 is the average LittleBob had during the half he played less tournaments than BigBob. T is the number of tournaments each competitor has participated in. LittleBob gets a higher average than BigBob if:

(LT-ST) x (A1-A2) > T

If we take the figures from above, we get (20-10) x (64-60) = 40. Since 40 > 30, LittleBob gets a higher average than BigBob.

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