I know there's a pretty simple formula to determine the books edge. What I'm trying to figure out is with any given line, what is the formula used to determine what win % someone would need to make a profit off that line. For example I know that with a line at -110 you would need to win something like 52.4 % to make a profit. What I am looking for is how do you get that number? Any help is appreciate thanks.
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On the bottom of the page, it has break-even winning percentages for different lines.
So, for -110, 110/(110 + 100) = 0.5238 = 52.38 %
-120, 120/(120 + 100) = 0.5454 ~ 54.55 %
-130, 130/(130 + 100) = 0.5652 = 56.52 %
and so onComment -
wow awesome link, thanks man.Comment -
-110 is 110/210, or 52.38%.
It's very easy to do. My own personal preference is to do the following : In your example, you want to know how to figure what % you have to be correct to achieve profits at -110. Well, I like to base it on 1,000 games at $100 a game. Each wager is $100 to win $90.91, so we get back $190.91 for every winning bet. You are making $100,000 in bets, so let's say you want to know how many you have to win to achieve a 5% profit. Obviously, that's $105,000, so divide 105,000 by 190.91, and that will give you 550, or 55% to achieve a 5% profit. (the easier method is just take the market price you bet at 52.38% and multiply that by 1.05)
Let's say you want to know how many -140 bets you need to win to make a 5% profit. Same formula, except this time we'll divide 105,000 by 171.43, and that gives us 612.50, so 61.25%. (or take the -140 percentage of 58.33 and multiply it by 1.05) It really doesn't matter what the price is , all that matters is that you end up winning at a higher % than the average price you bet the games at. You will be equally profitable if you can win -300 bets at 78.75% as you will winning -110 bets at 55%. Of course, this only works if you risk the same amount on every bet.
Actually, if you are betting $1000 a game at -110, it's not a bad idea to bet some -240 type games as long as you're only risking the same $1000, and as long as you think your selection has the same size edge, because higher probability bets cut down on your variance.
Figuring out odds to percentages and house edges are easy. If a bet is +240, just take 100 and divide into 340 = 29.41%, just always divide 100 into the price and add 100. If the favorite is -300, thats 300 into 400, or 75%. In this example, you have a 75% and a 29.41% offered on the head to head game. Simply add the two together and you have 104.91. Take the amount over 100 (4.91) and divide that into the total market (104.91) and that will give you the house edge or hold %, which in this case is 4.68%.
This also makes it easy to make your own entries. For example, lets say you want to bet the Jets and the Ravens to win the AFC. The Jets are +1500 and the Ravens are +600. You want to know what the combined price is, and also how much to bet on each to get the same amount back no matter which team wins. Very simple to do. Take the Jets 100/1600 (6.25) and the Ravens 100/700 (14.29) and add them together., you get 20.54% Take 100 minus 20.54 (79.46) and divide that by 20.54 and you get the entry price of +387.
Now, lets say you want to bet $1000 on the Jets/Ravens entry to win $3870. Just take the 20.54 and determine what % of that is the Ravens and the Jets. The Ravens is 14.29/20.54 or 69.57%, so thats how much of your stake needs to go on the Ravens. In this case, youd bet $695.70 on the Ravens, and of course 304.30 on the Jets, and this way, you get back just about $4870 no matter which one wins, or +387 on your $1000 bet.Comment -
Thanks man!Comment -
Thread title is misleading, what you are asking has nothing to do with edge. You are merely calculating break even points.Comment
