Math Question for the Brains on here

Enfuego said:

You have two teams playing each other.

Team A has a win % of 62%
Team B has a win % of 48%

How do you figure out the % of how wins Team A should have?

If it's 62 and 38 it would be easy and of course they should win 62% of the time but what if it equals something other than 100%?

Typically you would just add up the two winning percentages and then divide the winning percentages by the total. In this example Team A should win 56% of the time over Team B but I don't think that's right. Surely a team with a 62% win percentage should beat a sub-500 team more than 56% right or am I off?

Help?

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PatsFan1283 said:

Also no to the 62% and 38% teams and thus that being what they win.

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Enfuego said:

Huh?

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Murphy8276 said:

The difference between 56% and 44% would make or break a bettor over time.
63%+ / 37% makes more sense as patsfan tried to explain.

-murph

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PatsFan1283 said:

If a 62% win team plays a 38% win team they would win much more than 62% of the time. Their 62% win-rate is theoretically based off of playing teams of a 50% win rate

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Enfuego said:

Yes, I see what you're saying but what about a 48% team. Is it really only 56%?

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PatsFan1283 said:

lol this is wrong too. A 62% win team isn't 12% better than their opponent, they are 24% better on average. If you bet on a 62% win team at even money your long-term ROI would be 24%.

The work he has done makes no sense actually and he knows that, that is why he is asking.

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PatsFan1283 said:

lol this is wrong too. A 62% win team isn't 12% better than their opponent, they are 24% better on average. If you bet on a 62% win team at even money your long-term ROI would be 24%.

The work he has done makes no sense actually and he knows that, that is why he is asking.

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PatsFan1283 said:

Assuming all teams Team A plays win 50% of the time then 48% is slightly below that but very close to the median.

If team A wins 62% of the time vs all teams that win 50% of the time. Teams that win 48% of the time win 2% less than teams that win 50% of the time. Thus the win-rate of a 62% team is going to be 2% better than their avg win rate cause of the inferior opponent which would be 63.124%

I am pretty sure this is right and I'm not an idiot, if I am wrong someone feel free to correct me.

I am very good at stats/probability but I suck at explaining it, I don't know why. Always been that way....

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Murphy8276 said:

Yea I agree. I updated my thought on it quite a few times and your method makes a lot more sense after it was posted. Enfuego not considering that 56% and 44% was a significant advantage is what made me start typing to bring about a point as that it would make or break a bettor in the long run

-murph

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PatsFan1283 said:

The question as presented could really never apply to sports though for a # of reasons. Home/away, teams playing uneven schedules, etc

So what was the point of the question? Just curious

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Enfuego said:

Point of the question was exactly what I was asking about. Take away home/away and starting pitchers etc. How do you figure the advantage for the team with the higher winning percentage when the two combined percentages don't equal 100.

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PatsFan1283 said:

Well the question can really only be asked in a vacuum but yes assuming exactly equal SOS, neutral field, everyone on both teams healthy for X # of games then yes more or less that is the way to do it.

However, the "adding up to 100" is wrong you see why now right?

Team A at 62% would beat team B at 38% a lot more than 62% of the time.

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