How much to bet
Tonight's idea
A coin lands heads with probability \(p > \tfrac12\) and pays even money: stake something, get it back doubled on heads, lose it on tails. What fraction \(f\) of your wealth should you stake each time?
Staking everything is fatal: one tail and you are out, and tails come. Staking nothing earns nothing. In between, after \(H\) heads and \(T\) tails your wealth is \(W_0(1+f)^H(1-f)^T\). Wealth multiplies, so take logs and it adds:
\[\ln\frac{W_N}{W_0} = H\ln(1+f) + T\ln(1-f).\]
Over many rounds \(H/N \to p\) and \(T/N \to 1-p\), so the log of your wealth grows at a steady rate per round,
\(g(f) = p\ln(1+f) + (1-p)\ln(1-f).\)
That is the thing to maximise. Differentiate and set to zero:
\[g'(f) = \frac{p}{1+f} - \frac{1-p}{1-f} = 0 \iff p(1-f) = (1-p)(1+f) \iff f = 2p - 1.\]
\(g''(f) = -\dfrac{p}{(1+f)^2} - \dfrac{1-p}{(1-f)^2} \lt 0\), so this is the maximum. \(f^* = 2p-1\) is the Kelly fraction: your edge, \(p - (1-p)\). For \(p = 0.6\), stake 20%. Stake more and \(g\) falls; past a point it turns negative, and a gambler with a real edge goes broke.
Watch one: \(p = 0.6\), one hundred rounds
Expect 60 heads and 40 tails. Multiply the factors. \(f = 0.2\): \(1.2^{60}\times 0.8^{40} = 7.5\). \(f = 0.1\): \(1.1^{60}\times 0.9^{40} = 4.5\). \(f = 0.5\): \(1.5^{60}\times 0.5^{40} = 0.033\).
Kelly turns 1 into 7.5; half Kelly manages 4.5; betting half your stack each time, with a 60% coin, loses 97% of it. The rate \(g(0.2) = 0.6\ln 1.2 + 0.4\ln 0.8 = 0.0201\) per round, and \(e^{100 \times 0.0201} = 7.5\): the formula and the brute-force product agree.
Twenty gamblers, a thousand rounds, and a slider for \(f\): mathdetours.com/demos/kelly.html.
Your turn
- \(p = 0.55\). Find the Kelly fraction and \(g\) at it, to four decimal places. Then compute \(g(0.2)\), double Kelly, and say what happens to that gambler in the long run.
- For \(p = 0.6\) compute \(g(0.1)\), \(g(0.3)\) and \(g(0.4)\). Between which two of \(0.3\), \(0.4\), \(0.5\) does the growth rate cross zero? What do you notice about \(g(0.1)\) and \(g(0.3)\)?
- Now the bet pays odds \(b\): win \(b\) per unit staked, or lose the stake. Write down \(g(f)\), show the best fraction is \(f^* = p - \dfrac{1-p}{b}\), check it gives \(2p-1\) when \(b = 1\), and evaluate it for a fair coin paying 3 to 1.
- Expected wealth after one round is \(W_0\big(p(1+f) + (1-p)(1-f)\big)\). (a) Simplify, and show it is largest at \(f = 1\). (b) For \(p = 0.6\) and \(N = 10\) rounds at \(f = 1\), find the expected wealth factor and the probability of not being broke. (c) Reconcile (b) with \(g(1) = -\infty\): what is the mean measuring that a typical gambler never sees?
Stretch: half Kelly
Use \(\ln(1+u) \approx u - u^2/2\) to show \(g(f) \approx (2p-1)f - f^2/2\). Find the maximum value, then show that betting half the Kelly fraction keeps three quarters of the growth rate. Check against the exact \(g(0.1)/g(0.2)\) for \(p = 0.6\).
Answers
- \(f^* = 0.1\); \(g(0.1) = 0.55\ln 1.1 + 0.45\ln 0.9 = 0.0050\) per round. \(g(0.2) = -0.0001\): slightly negative, so the double-Kelly gambler drifts to ruin despite the edge. For a small edge, double Kelly is roughly break-even.
- \(g(0.1) = 0.0150\), \(g(0.3) = 0.0147\), \(g(0.4) = -0.0024\). Zero lies between \(0.3\) and \(0.4\) (at \(f \approx 0.39\)). \(g(0.1)\) and \(g(0.3)\) are nearly equal: the curve is close to symmetric about \(0.2\), a parabola near its top.
- \(g(f) = p\ln(1+bf) + (1-p)\ln(1-f)\). \(g' = \dfrac{bp}{1+bf} - \dfrac{1-p}{1-f} = 0\) gives \(bp(1-f) = (1-p)(1+bf)\), so \(f(bp + b(1-p)) = bp - (1-p)\) and \(f^* = p - \dfrac{1-p}{b}\). With \(b = 1\): \(2p - 1\). Fair coin, \(b = 3\): \(f^* = \tfrac12 - \tfrac{1}{6} = \tfrac13\).
- (a) \(W_0\big(1 + (2p-1)f\big)\), increasing in \(f\) when \(p > \tfrac12\), so largest at \(f = 1\). (b) \(1.2^{10} = 6.19\); survive with probability \(0.6^{10} = 0.006\). (c) The mean is 6.19 because one path in 165 reaches \(2^{10} = 1024\) and the rest are at zero. The mean is pulled up by a path almost nobody is on. \(g\) measures the growth of the typical path, which is what one gambler, playing once through, experiences.
Stretch. \(g(f) \approx p\big(f - \tfrac{f^2}{2}\big) + (1-p)\big(-f - \tfrac{f^2}{2}\big) = (2p-1)f - \tfrac{f^2}{2}\). Maximum at \(f^* = 2p-1\), value \(\tfrac{(2p-1)^2}{2}\). At \(f^*/2\): \(\tfrac{(2p-1)^2}{2} - \tfrac{(2p-1)^2}{8} = \tfrac34 \cdot \tfrac{(2p-1)^2}{2}\). Half the stake, three quarters of the growth, and far smaller swings. Exact, \(p = 0.6\): \(0.0150/0.0201 = 0.747\).