College Fan said:
that doesn't make sense...
if those teams have expected win chances of 58% and 32%, respectively...
how would a sub-par team of 32% against an above average team of 58% still end up with a 35.56% chance to win (better than their NORMAL win pct)???
if they're only winning at 32% against all the other teams, they would certainly have a win expectation of well BELOW 32% against a team of above average capabilities...
this is where averaging goes wrong.
the problem here is the interactions, that is, one probability and the other are for completely different events when ocurring separately but when you want to combine them there are hidden variables that are not incorporated in your numbers
the "when they play against each other" is a WHOLE different situation and thats when the hidden variables play an important factor, the %s that is 58 and 32 are from a merely empirical basis (you see how many wins etc and extract a raw number) and extrapolating that to any situation would be a mere guess (But hey ! thats gaming after all)
one example
if you have two RANDOM events and want to know whats the probability of both of them having at the same time then you can calculate it
but a baseball game etc is not random and when you have two non random events and want to get the odds of that.............its impossible
the "when they play against each other" is a WHOLE different situation and thats when the hidden variables play an important factor, the %s that is 58 and 32 are from a merely empirical basis (you see how many wins etc and extract a raw number) and extrapolating that to any situation would be a mere guess (But hey ! thats gaming after all)
one example
if you have two RANDOM events and want to know whats the probability of both of them having at the same time then you can calculate it
but a baseball game etc is not random and when you have two non random events and want to get the odds of that.............its impossible
Majorcapper said:
Cant be done, each is independent of the other, here is an example, The Kansas City Royals are winning 20% of thier games, today they are playing a little league team that is undefeated, winning 100%. Do you really believe that you can create a realistic win % for either team by a calculation based solely on these numbers, this is an extreme case but the theory is the same even for teams on a level playing field.
Doug said:
Ya I know I just wanted to keep the question simple I put more into than just taking they're actual records I took 5 things that I like to look at when capping
Overall record
Home away record
Last 25 games record
Last 10 games record
Starting pitcher (I use the pitchers overall, home away and last 3 starts records and average them)
Overall record 24-20 55%
Home record 14-10 58%
Last 25 15-10 60%
last 10 6-4 60%
Starting Pitcher 12-8 60%
Now average out the pecents which is 58.6% then I do the same thing for the other team and it come out to 32.2% so now I'd like to combine these two numbers and compare it to the odds for the game to see if theres value there so if it were 60% the line would have to be better than -150 for some value.
Heres what I came up with 58.6-50=8.6 and 50-32.2=17.8 take 8.6+17.8=26.4 and add 50 so the number would be 76.4%
I think thats kind of high but if its right and the line is -300 or better there would be some value but that dont seem right.
College Fan said:
Not sure if this is right or not or what I'm looking for some agree and some dont.
Although this might be it, if you take a great team that 65% and just a good team 55% the great team got a 54% chance of winning that seems about right and if you 55% team against a 55% team this way come out to 50% which is right.
I'm no number guru, but I keep thinking of the old coin-flip adage.
If you've flipped heads 10 straight times, the law of averages dictates that there is a high probability that the next flip will be tails....BUT for every individual flip, the probability of heads/tails is exactly 50/50.
Good luck, I hope you find an answer and make a ton.
If you've flipped heads 10 straight times, the law of averages dictates that there is a high probability that the next flip will be tails....BUT for every individual flip, the probability of heads/tails is exactly 50/50.
Good luck, I hope you find an answer and make a ton.
I retract my previous posts after further thought, I believe that 1/2Team A Win % + 1/2Team B Loss%, and 1/2Team A Loss% + 1/2Team B Win% would give you an answer. From the inital post the results would be .5*58+.5*68 for 63% chance of Team A winning, and .5*32+.5*42 for 37% chance of winning. But I still believe this would be a very unreliable tool with out a huge sample of games with identical conditions for both teams.
OK fellas after thinking about this all day I think my first conclusion was actually right, I tried a few of the thoreys here and the numbers dont work.
Heres what I got
Team A 58%
Team B 32%
Take the difference between the two which is 26 and add that to 50 which is 76% team A would win and 24% team B would win.
Heres a couple more examples and the numbers look pretty good, first number is the team win % say against everyone second would be in theory the win % vs just the other team.
Team A 58% - 58%
Team B 50% - 42%
Team A 75% - 65%
Team B 60% - 35%
Team A 44% - 55%
Team B 39% - 45%
So does this look like it could possible be right the numbers look about right
Heres what I got
Team A 58%
Team B 32%
Take the difference between the two which is 26 and add that to 50 which is 76% team A would win and 24% team B would win.
Heres a couple more examples and the numbers look pretty good, first number is the team win % say against everyone second would be in theory the win % vs just the other team.
Team A 58% - 58%
Team B 50% - 42%
Team A 75% - 65%
Team B 60% - 35%
Team A 44% - 55%
Team B 39% - 45%
So does this look like it could possible be right the numbers look about right
