I think so...media made it into a story to sell news. A close bout is a far better story, even though Romney was never going to win.
This were at a big betting firms webpage some days ago:
2012 US Presidential Race
odds
Barack Obama 1.22
Mitt Romney 4.33
Disregarding all political aspects (not allowed in this forum), and only focusing on the probabilities, were the election really that decided on forehand?
If I understand it correctly I would have gotten 4.33 times my money if Romney had been elected and I had betted on him?
All media (perhaps in self interest) portrayed the election as "close", and in the end very few votes separated the two - though the difference were larger considered the system with elector votes. But was it really so "not close" that it was a good idea to offer 4.33 the money on Romney?
Is media lying to get my attention (yes?), can only true economic indications like betting odds be trusted?
I think so...media made it into a story to sell news. A close bout is a far better story, even though Romney was never going to win.
If you had bet £25 on Romney at decimal odds of 4.43, your total return in the event that he won would be:
£25 x 4.43 = £110.75
Because your original stake was £25, the above return would give you a bottom-line profit of £85.75
You can calculate your potential profit only from any decimal bet (instead of the total return which includes your stake) by deducting 1 from the odds before doing the above maths i.e, if you wanted to calculate your profit only from the same £25 bet it would be:
4.43 – 1 = 3.43
£25 x 3.43 = £85.75
Either way, you're correct - 4.43 would give you a significantly higher return than 1.22
(just realised I've used 4.43 and not 4.33 in my example, but the theory is the same!)
I believe one bookie paid out on Obama bets at the weekend - before the election!
pauly, pretty sure the original poster understands odds ( though I believe unlike in your explanation, in the UK you get your stake back for a successful bet ) but was intrigued why the odds appeared to so strongly favour Obama in what was seen as a close race.
My own view is that like trackside turf accountants, bookies attempt to hedge as much as possible. So if there was a big influx of money on an Obama victory a few days back, they will raise their odds ( payback ) on a Romney win in order to attract more ( losing, the bookies hope ) bets.
Amusingly, Americans are not allowed to bet on the outcome of the election for President of the USA, Land of the Free.
They can, however, trade options on it
Instructive to view the options history graph, this one from Intrade, which maintained a full market in Obama/Romney options up until the result.
This option is valueless if Obama loses, and worth $10 if he wins.
The options history for Romney :
Again, worth $10 for a Romney win or zero if he loses.
The market favoured Obama for a while.
Paul
Last edited by Tokyo Tokei; 8th November 2012 at 05:02.
The option had a strike price of $10. The graph is in percentage of strike price. In the case of Obama, you can see that at almost all times, the price of a $10 option was higher than Romney, meaning lower odds. In the case of Romney, you can see a late September fall where the option falls to somewhere in the $2 range. If you had bought this option then, and Romney had one, you would have received $10 after the election result. Odds roughly 4 or 5 to 1, similar to what you saw.
Again I would make the distinction between gambling odds, which are influenced by the bookmaker's need to hedge his potential payout, and pure political odds. Where the bookmaker has collected a lot of bets over time on a candidate to win, even if the odds were only marginally in favour, he may offer higher odds on the other candidate ( assuming it's an either/or result, as here ) in order to hedge his position. Political odds have no such baggage.
Paul
Last edited by Tokyo Tokei; 11th November 2012 at 01:06.
Thank you for explaining! Very interesting!
At election day it seemed Bronco Bamma* were
(using the return on investment as probability)
[wrong?
(1/.35 -1)/ (1/.7 -1)
= circa
4.33]
or perhaps
(1/.35)/ (1/.7)
=
2
as likely to win? Any thoughts on [wrong? which (if any) of 4.33 or] 2 might be correct?
Any ideas if arbitrage would be possible?
Still, I feel the media really overplayed how close the election was, at least when you factor in the elector voting.
*
http://www.youtube.com/watch?v=OjrthOPLAKM
Last edited by youveboughtwhat; 12th November 2012 at 05:19.