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I'm reading an article about expected value in betting, I got stuck at the very beginning:

Let’s use a coin toss as an example of calculating expected value. Assuming the coin and the toss are fair, each outcome (heads or tails) has an equal probability of 50% - therefore the odds offered on a fair market would be 2.0.

This would result in an EV of 0 for either a Head or Tail - because the probability of the two outcomes is the same, so if you tossed a coin infinitely it would theoretically end up all square.

I think odds = p / (1 - p) so in this case, where the probability of landing head or tail is 0.5 so the odds should be 0.5 / (1 - 0.5) = 1? And the next point,

If however you were offered odds of 2.15 for the coin to land on heads, this is a value bet.

If you placed £10 on the coin landing on heads at 2.15, the EV is calculated likewise:

(11.50 X 0.5) – (10 X 0.5) = 0.75 This shows an EV of 0.75. Therefore you would expect to make an average profit of 75p for each £10 bet, because the odds received are better than the implied odds of the coin toss.

How is the expected value actually calculated? Where is that number "11.5" from?

The article link: https://help.smarkets.com/hc/en-gb/articles/214554985-How-to-calculate-expected-value-in-betting

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2 Answers 2

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I think I get it now. The odds mentioned in the article is the decimal odds (not the "standard" odds ratio that I was thinking). Having understood the decimal odds, I can calculate the expected value.

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If the odds would be 1 than each player who wins the coin toss just get back his bet instead of winning anything from the other player. The expected value is calculated by the probability of winning x the potential outcome in the case of winning - the probability that you lose x your loss.

The 11.50 can be calculated by 10 (your bet) x 1.15 (your expected winning ratio).

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