Quote:
Originally Posted by Ganchrow
If so, then I guess my audience for this type of post is the remaining 2% of the population.
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But anyway, the executive summary is this:
If you place one 1-unit bet with an edge close to zero, at decimal odds d, the standard deviation of that bet would be
σ = sqrt(d-1)
If you place n 1-unit bets with edges close to zero, at decimal odds, d, the standard deviation of across those n bets would be
σ = sqrt( n * (d-1) )
So for example, if over the course of a season you were to place 200 $100 zero-edge bets at odds of , your standard deviation (in dollars) would be:
σ = $100 * sqrt(200 * 10/21) ≈ $975.90
This means that your 95% confidence interval would be approximately ±1.96 * $975.90 ≈ ±$1,912.73.
In plain English, this means that after all 200 $100 bets there's be a 95% probability that you would have won or lost less no more than $1,912.73.