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It is possible to bet on the Brexit e.g. on this page:

https://sports.ladbrokes.com/en-gb/betting/politics/british/eu-referendum/uk-european-referendum/220800266/

The quotes are 8/15 for remain, and 8/4 for leave.

Can someone derive the implied probabilities for remain/leave?

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The general formula for conversion of "a to b" odds to a probability is $p=\frac{b}{a+b}$

http://www.calculatorsoup.com/calculators/games/odds.php

So 8/15 remain implies remain with probability 0.652

8/4 for leave implies leave with probability 0.333

The amount 1-0.652-0.333 = 0.0145 represents the bid-ask spread or loss that you suffer (and the other side collects) for making both bets.

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It's 6/4 to leave right now. Hence, the implied probabilities are 6/10 = 0.6 to leave and 8/23 to stay. So it's about 2/3 to leave, 1/3 to stay. You can't do much better, since you don't know number of bets and the profit margin for the venue.

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  • $\begingroup$ You can assume that the profit margin is 0 and that the number of bets are equal. But I don't think it matters, and I don't understand your derivation of probabilities. $\endgroup$ – emcor Jun 16 '16 at 15:58
  • $\begingroup$ Then for $m/n$ odds, the probability is $m/(m+n).$ $\endgroup$ – LazyCat Jun 16 '16 at 16:47
  • $\begingroup$ This would imply that odds of 100 to 1 imply a probability of 100/101. I wish this were true because then I could bet on any horse running at 100 to 1 and be sure that I'd win. $\endgroup$ – Captain Nemo Jun 18 '16 at 10:41
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    $\begingroup$ That's rather a pathetic way to say that I've mixed up $m,$ $n.$ $\endgroup$ – LazyCat Jun 18 '16 at 17:11

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