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#1
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expected outcome
I know this is a sports example, but I figure this is a general probability problem. Is this the correct way to figure out the expected outcome of a game? (Assuming the line is accurate.)
Team1 odds: -135 (Must bet $135 to win $100) Team2 odds: 115 (Must bet $100 to win $115) Actual odds that Team1 wins (removing the house's take): -125 (Potential money won)/(Potential money lost) = 100/125 = 0.8 Therefore, Team1 will win 80% of the time, or 4 out of 5 games. |
#2
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Re: expected outcome
If the actual odds for team 1 is 125:100 , then the probability team 1 wins is 125/(125+100) ~ 55.55%
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#3
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Re: expected outcome
american odds (the odds you are using--+135, -250, etc.) are used because they are so transparent. they will tell you the juice and the expected outcome. the plus number will tell you the odds that the dog will win in ratio form, and then you can take the inverse to figure out how often the favorite will win. example:
jets vs giants jets are +200 giants are -240 juice is obviously 40 cents. lets assume that the juice-free line is jets +220, giants -220. this means that the jets are 2.2 to 1. this also means that they are expected to win 1 game out of every 3.2 games in this situation. (thats the same as winning 5 out of 16.) so obviously the giants are 1 to 2.2, and win 2.2 games out of every 3 in this situation against the jets. so you can look at money lines and just look at the plus money to see that +135 means they are 1.35 to 1, or win 1 game out of every 2.35 games +700 means they are 7 to 1, or win 1 game out of every 8. the end. |
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