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Using a little math, here’s Kelly’s answer for how much to bet in the case of the coin. Let n, W and L be the number of tosses, wins and losses, respectively. The win and loss probabilities are p and q = 1 – p. We start with bankroll X0 and bet a fixed fraction f at each trial. After n tosses we have Xn=X0 (1+f )W(1-f )L. With a little rearranging we have Xn/X0=exp{n log([Xn/X0]1/n) = exp{nGn(f)}}where Gn(f)=log ([Xn/X0]1/n) = (W/n) log (1+f) + (L/n) log (1-f) measures the exponential rate of growth per trial. The expected value of Gn(f) is g(f) = p log(1+f) + q log(1-f) which has a maximum at f* = p – q. In this instance f* happens to be the same as the bettor’s edge, namely the expected value of one unit bet.
If g(f) > 0 then limn→∞ Xn = ∞, almost surely, i.e. one’s fortune tends to infinity with probability one: the zone of positive growth. If g(f) < 0 then limn→∞ Xn= 0, almost surely, i.e. one’s fortune tends to zero with probability one: the zone of ruinous over betting. If g(f) = 0 then Xnoscillates wildly from 0 to ∞. With the Kelly strategy Φ*verus any “essentially different” strategy Φ, the ratio Xn(Φ*)/Xn(Φ) tends to infinity with probability one. The expected time to reach any specified goal tends to be least with Kelly.
