formulas to predict win% based off past win/loss?

**evo34** · 10-05-16, 04:16 AM

It might be time to retire this sub-forum.

**Alfa1234** · 10-05-16, 04:26 AM

Yes!

You divide the number of wins by the number of games played and multiply this number by 100.

So, say a team has played 20 games and won 10, you divide 10 by 20 and multiply this by 100 so that would give you 50%.

Elementary school.

**tsty** · 10-05-16, 04:53 AM

sorry it came wrong

english isn't my language

what I mean is calculating a teams chance to beat another team based off their previous performance

just their win/loss

so in a 16 team competition if they played each other all 2 times and a team was playing tomorrow

what would their chance be to win someone relative to how much w/l they had each

**Buckandadime** · 10-05-16, 08:44 AM

See above post..

**Johnny Roger** · 01-25-17, 09:38 AM

I believe you are looking for something like this ... ?

Log5 - Wikipedia

https://en.wikipedia.org/wiki/Log5

**Johnny Roger** · 01-25-17, 07:06 PM

Probability of A defeating B = Win%A-(win%A-win%B)/((win%A+win%B)-2*win%A*win%B)

**gojetsgomoxies** · 01-25-17, 08:11 PM

why are people so mean and stupid on this forum?

i think it's good question as to what is the probability that a team with 70% winning percentage beats a team with 50% winning percentage? i.e. simple power rating based on record.

i think last poster might have helped the person........... of course, anything will assume team's have identical schedule and identical standard deviation of scores/results/etc?

similarly with normal power ratings i've wondered the same... spreads by and large seem smaller than most basic power ratings out there would suggest by simple subtraction.

**gojetsgomoxies** · 01-25-17, 08:12 PM

Originally posted by Alfa1234

Yes!You divide the number of wins by the number of games played and multiply this number by 100.
So, say a team has played 20 games and won 10, you divide 10 by 20 and multiply this by 100 so that would give you 50%.
Elementary school.

not so fast.. what about opponent's winning %?

**gojetsgomoxies** · 01-25-17, 08:19 PM

Originally posted by Johnny Roger

Probability of A defeating B = Win%A-(win%A-win%B)/((win%A+win%B)-2*win%A*win%B)

johnny, thank you very much for this..........

but i put in A% = 70.... B% = 40.... so i'd have 70 - (70-40) / ((70 + 40 - 2 x 70 x 40)).... so 70 -30 /(110-56)= 70 - 30/56 = 70 -54 = 16%

MINOR CORRECTION..... 70 -30/54 ... 70 -56 = 14%.... eyeballed the very final division calculation

**gojetsgomoxies** · 01-25-17, 08:25 PM

i scanned this really fast.... http://sabr.org/research/probabiliti...-team-matchups

i thought this might be the right number:

A(1-B)/(A(1-B)+B(1-A)

so if A is 70 and B = 40........ .7 x .6 /(.7 x .6 + .4 x .3)..... .42 / .54...... eyeball at 81%...... EDIT: oops, i see 79% in that table in the article.....

**Johnny Roger** · 01-25-17, 08:45 PM

I may have gotten carried away with my parentheticals...

Here is the link from where I got the equation ... https://en.wikipedia.org/wiki/Log5

With that equation I have Team A (70%) winning 77.78% and Team B (40%) winning 22.22

**Johnny Roger** · 01-25-17, 08:49 PM

Originally posted by gojetsgomoxies

why are people so mean and stupid on this forum?

Right!? Seems like a pretty legit question to me!

Originally posted by gojetsgomoxies

I think last poster might have helped the person........... of course, anything will assume team's have identical schedule and identical standard deviation of scores/results/etc?

Agree with this. You technically need a "True Winning Percentage"

**Waterstpub87** · 01-25-17, 09:28 PM

Originally posted by Johnny Roger

Right!? Seems like a pretty legit question to me!

Agree with this. You technically need a "True Winning Percentage"

People are mean to you because they don't want you lose money betting stupidly. I would rather people tell me that I am wrong, rather than paying to figure out that I am wrong.

**Johnny Roger** · 01-25-17, 10:07 PM

Originally posted by Waterstpub87

People are mean to you because they don't want you lose money betting stupidly. I would rather people tell me that I am wrong, rather than paying to figure out that I am wrong.

Yeah, but this guy asked a legit question to help him not bet stupidly and he was ridiculed ...

**gojetsgomoxies** · 01-26-17, 12:35 AM

people are mean on here because they have nothing else to do...... negative sarcastic comments have nothing to do with helping people.

Johnny Roger, is your name based on nebraska TB of early 70's?

i think this works out the same as what i posted earlier...

but i found (A - AB) divided by (A + B - 2AB) in that link you provided....

so for 70% and 40%... it would be numerator = 70- 70x40..... = 42... denominator = 70 +40 - 2x40x70 = 110 -56 =54

so 42/54 = approx 78%....... and BTW i think when i eyeballed earlier that this is the figure i should have got to when i look at it again.

if interested it's called log5 developed by Bill James although i have to figure the idea/calc was already out there. finetuning idea and naming i'd think.......... EDIT: ok it looks like that wiki article says it's exactly the same as ELO chess ratings....

**evo34** · 02-07-17, 01:30 AM

Originally posted by Johnny Roger

Yeah, but this guy asked a legit question to help him not bet stupidly and he was ridiculed ...

Asking how to project a future game based only on 16 games of binary (W/L) data is a not legitimate question for the Think Tank. This is not the 1980s. And, yes, the OP is incredibly naive.

**Johnny Roger** · 02-09-17, 09:45 PM

Originally posted by evo34

Asking how to project a future game based only on 16 games of binary (W/L) data is a not legitimate question for the Think Tank. This is not the 1980s. And, yes, the OP is incredibly naive.

It would be 30 games in the scenario he provided (16 teams, each played each other twice). And that's just a hypothetical. It doesnt mean the formula doesnt exist ...