How would I calculate the odds of either of two outcomes happening?

[yisman]

Say a soccer/futbol book has the odds of Team A winning at -110, the odds of Team B winning at +300, and the odds of a draw happening at +240.

What would be the odds of betting against Team A winning? IOW, the latter two outcomes combined.

I would guess it would be at somewhere between -110 and -130, but I'm mostly curious about how to apply this to other situations.

cheers

[reno cool]

1/2.1= .476

1/4=.25

1/3.4=.294 total youd have to bet to return 100 for all3 is 102 ... 1/1.02= 98.03 ret

the chance of the other 2 happening is .25+.294= 54.4%

consider what you'd have to bet to return the same amount regardless which bet wins.
29.40 at +240 returns 99.96
25.00 at +300 returns 100

so you would have to bet 54.40 to win 45.60 looks like -119

[Ganchrow]

foo

Say a soccer/futbol book has the odds of Team A winning at -110, the odds of Team B winning at +300, and the odds of a draw happening at +240.

What would be the odds of betting against Team A winning? IOW, the latter two outcomes combined.

If you're looking for the effective wagering odds of betting against A, then recalling that +300 and +240 are expressed as 4 and 3.4 in decimal odds, respectively:

Odds(not A) = -100*(4 + 3.4)/(4*3.4 - 4 - 3.4) ≈ -119.35

If, on the other hand, you're looking for the fair odds of betting against A then:

Overround = 110/(110+100) + 100/(300+100) + 100/(240+100) ≈ 106.793%

Pr(A) = 110/(110+100)/106.793% ≈ 49.049%
Pr(not A) ≈ 1-49.049% ≈ 50.951%

Odds(not A) ≈ -100*50.951%/(1-50.951%) ≈ -103.88

See http://forum.sbrforum.com/handicappe...rcentages.html and http://forum.sbrforum.com/handicappe...ical-hold.html.

[yisman]

thanks.

I'm looking for effective wagering.

Ganchrow, I'll use your method for something else and see if I got it right.

Odds are +225 and +240

Odds(not A) = -100*(3.25 + 3.4)/(3.4*3.25 - 3.4 - 3.25) ≈ x

-100 * 6.65 / 11.05 -6.65

-665 / 4.4 ≈ -151.14 = x


Did I do that correctly?

[Ganchrow]

thanks.

I'm looking for effective wagering.

Ganchrow, I'll use your method for something else and see if I got it right.

Odds are +225 and +240

Odds(not A) = -100*(3.25 + 3.4)/(3.4*3.25 - 3.4 - 3.25) ≈ x

-100 * 6.65 / 11.05 -6.65

-665 / 4.4 ≈ -151.14 = x


Did I do that correctly?

Yes, that is correct.

A potentially more intuitive way of looking at this would be in terms of implied probabilities:

Event X: +225 ⇒ Prob. = 100/(100+225) ≈ 30.769%
Event Y: +240 ⇒ Prob. = 100/(100+240) ≈ 29.412%

Hence:

Prob(Not A) = Prob(X or Y) ≈ 30.769% + 29.412% ≈ 60.181%

Prob. of 60.181% ⇒ decimal odds = 1/60.181% ≈ 1.6617

1.6617 converted to US-style odds = -100/(1.6617-1) ≈ -151.14

[yisman]

Seems the books are pretty precise on this (or at least Bodog is).

England/Spain today.

Bodog had Spain at -110, England at +330, and a draw at +210.

I used the formula in your last post to eight decimal places. Answer was exactly -125.