View Full Version : Math Question

12-13-2004, 08:45 PM
If you have some various numbers that you want to combine to form an overall score, but each number is based on a different scale, what is the best way to go about this?

For example:

You want to compare two guys that play some sports. Let's say bowling and golf.

During the competition:

Bill bowled a 215 game, and shot a golf score of 94.

Don bowled a 160 and had a round of 79.

Is it possible to give them each an overall score based on some uniform scale, and declare a winner?

My first take on this is to convert them both to 0-100 scales and go from there. Bowling scores can range from 0 - 300, and golf scores from about 60 - 150.

Bill: 215/300 = 71.6
Don: 160/300 = 53.3

Bill: (94-60)/(150-60) = 37
Don: (79-60)/(150-60) = 21

Here's where I'm kind of stuck. The tricky thing is that lower is better in golf, and higher scores are better in bowling, so it doesn't make sense to add each players numbers together, does it?

Bill = 71.6 + 37 = 108.6
Don = 53.3 + 21 = 74.3

Would show bill as the winner, but he shot a worse golf score, and when we add his scores together we are no longer on a 0-100 scale.

There has to be a way to do this, logrithmic scale or something?

Any suggestions would be appreciated, thanks.

12-13-2004, 09:23 PM
Just invert the golf percentage after you calculate it. Subtract your normalized golf score from 100 to get them into the same range.

Look at the worst cases. If you shoot a 60 in golf, that's the 'best' score in your range. That should map to 100%, but using your equations it maps to 0%. Similarly, if you look at the 'worst' score, 150, that calculates to 100%. Both of those are obviously wrong.

Also, just looking at the numbers you computed for your golf scores, Don shot a better game but got a lower score. That just doesn't seem right.

Use your equations for the golf scores, then subtract from 100.

If you want to map the combined scores to a range of 0-100, then an average of the two computed values is probably what you want. Just add the bowling and golf scores and divide by 2. Just adding them won't be in the range of 0-100 like you implied you wanted... think of the case where one player shoots/bowls a perfect game in both games. They'd have a normalized score of 100 in both games... which adds to 200 in your math.

12-13-2004, 09:53 PM
Standardization in statistics means taking a score, subtracting the expected (read: average) value from that score, then dividing by the standardization. It is similar to what you were doing in spirit, but uses standard deviation instead of absolute range (what you were doing).

Maybe you want to consider 180 an average score in bowling (depends on who we're talking about here!), with a standard deviation of 40. Then a score of 240 gets standardized to a score of (240-180)/40 = +1.50. A score below 180 would get a negative value.

The hard part is picking the average and standard deviation ahead of time, in a way that is reasonable. (This is as subjective as you picking your golf ranges.)


12-14-2004, 02:31 PM
This makes sense too, but it also assumes that bowling scores and golf scores follow a normal distribution, right? Which may or may not be the case.

You are right, the hard part is that I will not have an average or standard deviation before doing this calculation. So I'm leaning towards using the 100 scale and the idea suggested above by niin.

Nick M
12-14-2004, 09:36 PM
i think you are in trouble here. See i think the whole problem with this is there is a definitive worst and best for bowling 0-300. That's it. But Golf range is actually 18 and infinite. Obviously 18 is not possible and nobody sucks so much they could never finish a 18 hole course. I think this is a contest between the 2 people, not the 2 people on a scale against the rest of the world and how they could also do in the sport. Do it by percentage. For instance.

A 85
B 97

"A" did 14.1 percent better than "B" did.
(remember in this case the 85 should be equal to 100 not the 97)


A 156
B 195

"B" did 20 percent better than "A" for a difference of 5.9 percent. "B" is the winner.