General Discussion
Related: Editorials & Other Articles, Issue Forums, Alliance Forums, Region ForumsPresidential betting odds from Ladbrokes
Overall winner:
Obama 4/11
Romeny 2/1
Electoral College:
Romney wins 270-289 4/1
Obama wins 270-289 5/1
Obama wins 290-309 9/2
http://sports.ladbrokes.com/en-gb/politics/us-presidential-election/2012-us-presidential-race-e212304268
aquart
(69,014 posts)DMacTX
(301 posts)If you think Romney will win, you'd be betting $1 of your money against $2 of the other guys.
e.g.
if you risk (stake) $11 correctly on Romney you'd win $22
if you risk (stake) $11 correctly on Obama you'd win $4
in all winning bets you get your stake back
The Magistrate
(95,247 posts)Your eleven back, plus four more. This is roughly equivalent to a 70% chance of President Obama winning. Now, the house has to make a cut, and so the pay-out is always a bit worse than what the bookie thinks the true odds are, so they probably consider the probability President Obama will win to be two out of three, or hair less than that; in other words, the true pay-out ought to be five for ten ( five and a half dollars on a bet of eleven dollars ), but the house keeps the difference.
If you bet one dollar on Romney, you would get back three if it came in, your dollar back, plus two more. This is a straight 'two for one', a probability of about 33% for Romney. In this case, the house shading will work in the other direction, and they will consider the true odds for Romney more on the order of, say, three in eight, between one third and two fifths. Again, the house will pay out less than the true odds, and keep the remainder.
muriel_volestrangler
(101,320 posts)The apparent chances the odds indicate:
For Obama: 11/(4+11) = 11/15 = 73.3%
For Romney: 1/(2+1) = 1/3 = 33.3%
The true odds they reckon are probably about 70% for Obama, and 30% for Romney, thus adding up to 100. Though I don't know if it would be more accurate to subtract the same 3.3% from both figures, like that, to get them to add up to 100, or whether the amount taken off should be in rough proportion to the amounts, eg take 4.3% off 73.3, and 2.3 off 33.3, to get about 69% v. 31%. But the difference in the result isn't that significant.
WhaTHellsgoingonhere
(5,252 posts)DMacTX
(301 posts)Change has come
(2,372 posts)DMacTX
(301 posts)Change has come
(2,372 posts)Thanks for the response.
RandySF
(58,874 posts)Even I don't know for sure what they mean.
muriel_volestrangler
(101,320 posts)without going into precise mathematics. The best thing to do, with odds when there are, realistically, just 2 outcomes - Obama wins or Romney wins - is to take the number that looks simplest - "Romney 2/1". If the bookie wasn't trying to make a profit, that would mean they think Romney has a 1 in 3 chance of winning - if you bet £1 on him and he wins, they give you the £1 back, plus another £2. But the bookie is always slightly less generous in what they give you if you win, so that they can make a profit. So, in this case, they must think Romney has a slightly less than 1 in 3 chance of winning, but will only give you the 'fair' amount for a 1 in 3 chance. In percentage terms, rather than 33.3% chance, perhaps they think it's a 30% chance.