Variance is never zero for a non-trivial random variable. In a coin flip game, the variance is 0.5*1^2 + 0.5*(-1)^2 = 1.Originally Posted by Tthree
Expected Growth is the geometric mean of growth (decline) of bankroll. If you are handy with derivatives, you can calculate your Kelly bet from the Expected Growth function for a bet with known outcomes. Below is a simplified example of Blackjack with 3 possible outcomes for a bet of 1 unit:
1: Win 1.5 with probability 0.023669;
2: Win 1 with probability 0.480414;
3: Lose 1 with probability 0.495917.
EV for this bet is 1.5*0.023669 + 1*0.480414 -1*0.495917 = 0.02. Variance is 1.0292.
log(EG) = 0.023669*log((B+1.5)/B) + 0.480414*log((B+1)/B) -0.495917*log((B-1)/B) [B represents bankroll]
Calculate 1st derivative with respect to B. Set derivative = 0, which will solve Kelly bankroll for a 1 unit bet.
The derivative degenerates to a quadratic equation, where:
a=-0.04
b=1.999172
c=3
Plugging the above constants into the quadratic equation, B equals 51.437.
And the Kelly fraction is:
1/51.437 = 0.01944, which is closely approximated by the shorthand formula for Kelly fraction, EV/Variance = 0.02/1.0292 = 0.01943.
And EG is:
1.000194
With a doubling rate of:
log(2)/log(1.000194) = 3,566 bets.
EV/Variance is a good approximation for Kelly fraction when EV is small. When EV gets larger, this will diverge from the actual Kelly fraction. For example, assuming the above game has EV 0.2 (with same likelihood of 1.5x), Kelly equals 0.1955. EV/Variance equals 0.2021.
You seemed to imply that breaking even gambling but spending a lot of money in expenses will somehow not cause your EG to go down. Any reduction in bankroll without a concurrent reduction in bet size(s) will reduce EG, assuming the player was betting at levels to achieve max EG. At the point the player is unable to reduce bet sizes to sustain +EG, the game becomes unplayable and the player must stop. But the player has not gone broke.
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