Quote:
Originally Posted by RustyBrooks
To solve this you don't change the percentages, you change the EV formula.
Let's say you're flipping a coin against someone. If it's heads you gain $1 otherwise you lose $1
EV = 0.5*1 + 0.5*-1 = 0
If instead you lose $1.1 when you lose, because the guy takes juice, then
EV = 0.5*1 + 0.5*-1.1 = -0.05
If it's not a fair flip and instead you have X chance to win, and therefore (1-X) chance to lose, then
EV = X*1 + (1-X)*-1.1
EV = X - 1.1 + 1.1X
EV = 2.1X - 1.1
Thanks for answering. But I don't understand the X*1 + (1-X)*-1.1 and what all this means unless its said in words.
Basically i do understand simple probabilities, but why i am asking this question is from a horse racing point of view. Most horse bettors frame betting markets to 100% so each horse takes a % of probability from that 100% to win the race, so all horses combined equal a 100% in total.
But some bettors prefer to frame their markets to 90% to turn the margin into their favour, as opposed to the bookmaker who bets overound to 110% so its in their favour.
They do this so whatever price they have assigned the horse say $2 decimal in the 90% market its easier for them to see value instantly instead of framing the market to 100% and then say adding 10% to that price $2 and needing $2.20 to back the horse.
So my question is is there some kind of probability error in framing to 90% vs 100%? If there isn't any error is framing the market to 90% then someone can actually frame a market to 10%.
And the horse whos wining probability in a 100% market is 10% would be 1% in a 10% total market value.
Another example is with flipping a coin again since heads and tails are each 50% totaling 100%, but what if instead we used heads 1% heads and tails 1% totaling 2% total.
Does it make a difference if you frame a betting market to 100% 90% or 10% as in is there any probability error in maths etc?