Now let's generalize to the case where we have a large number of wagers, say 100, and we win 55 of them and lose 45 of them. All wagers pay off at even money. If the same fraction F is wagered every time, after 100 wagers our capital will increase (or decrease depending on the value of F) to

(( 1 + F ) ^ 55) * (( 1 - F ) ^ 45)

where ^ means exponentiation. i.e.

( 1 + F) ^ 3 is equivalent to

( 1 + F ) * ( 1 + F )* ( 1 + F )

It can be shown algebraically that the same result can be obtained by calculating

G ^ 100

where

G = (( 1 + F ) ^ .55) * (( 1 - F ) ^ .45)

G is interpreted as the increase (or decrease) in capital per game. If P is the probability of winning the game and Q is the probability of losing the game then

G = (( 1 + F ) ^ P) * (( 1 - F ) ^ Q)

The total probability is 1 so P + Q must equal 1.

What Kelly does is find the value of F that maximizes G subject to assumed values of P and Q. You do this by using differential calculus. This is virtually the only time you need to resort to anything beyond high school algebra when doing sportsbetting calculations.

Without going into details, the answer you get is

F = P - Q

i.e. the optimum value of F when the probability of winning is P, the probability of losing is Q and the payoff is even money is P - Q.

Consider a concrete example. You wager only on games where the payoff is even money. You win 60 percent of these games so your value of P is 0.60 and your value of Q is 0.40. The optimum fraction of your capital to bet is P - Q or .20 - i.e. 20 percent.

To prove it to yourself, assume you wager on 10 games. You have a starting capital of 1000 dollars. Build a table where the first column is the fraction of your capital that you are going to bet. The second column is the value of your capital after 10 games. The value is calculated as

1000 * (( 1 + F ) ^ 6) * (( 1 - F ) ^ 4)

Computers come in very handy at this stage.

F Value

-----------

.15 1207

.16 1213

.17 1217

.18 1220

.19 1222

.20 1223

.21 1222

.22 1220

.23 1217

.24 1212

.25 1206

The results are not exact - they are rounded down to the next lower integer value.

Note that if you have a positive expectation of winning - i.e. P is greater than Q - then you can not lose if you wager less than the optimum value of F. You can lose however if the fraction that you wager exceeds the optimum by too great a margin. If F equals .40 then your capital after 10 games will be 975.

Note too that the optimum fraction to bet is the same as your potential profit per wager. If you bet 1 dollar, win 1 dollar with probability P and lose 1 dollar with probability Q then the average profit per dollar wagered is equal to

P * 1 - Q * 1 = P - Q

More to come but not today...

(( 1 + F ) ^ 55) * (( 1 - F ) ^ 45)

where ^ means exponentiation. i.e.

( 1 + F) ^ 3 is equivalent to

( 1 + F ) * ( 1 + F )* ( 1 + F )

It can be shown algebraically that the same result can be obtained by calculating

G ^ 100

where

G = (( 1 + F ) ^ .55) * (( 1 - F ) ^ .45)

G is interpreted as the increase (or decrease) in capital per game. If P is the probability of winning the game and Q is the probability of losing the game then

G = (( 1 + F ) ^ P) * (( 1 - F ) ^ Q)

The total probability is 1 so P + Q must equal 1.

What Kelly does is find the value of F that maximizes G subject to assumed values of P and Q. You do this by using differential calculus. This is virtually the only time you need to resort to anything beyond high school algebra when doing sportsbetting calculations.

Without going into details, the answer you get is

F = P - Q

i.e. the optimum value of F when the probability of winning is P, the probability of losing is Q and the payoff is even money is P - Q.

Consider a concrete example. You wager only on games where the payoff is even money. You win 60 percent of these games so your value of P is 0.60 and your value of Q is 0.40. The optimum fraction of your capital to bet is P - Q or .20 - i.e. 20 percent.

To prove it to yourself, assume you wager on 10 games. You have a starting capital of 1000 dollars. Build a table where the first column is the fraction of your capital that you are going to bet. The second column is the value of your capital after 10 games. The value is calculated as

1000 * (( 1 + F ) ^ 6) * (( 1 - F ) ^ 4)

Computers come in very handy at this stage.

F Value

-----------

.15 1207

.16 1213

.17 1217

.18 1220

.19 1222

.20 1223

.21 1222

.22 1220

.23 1217

.24 1212

.25 1206

The results are not exact - they are rounded down to the next lower integer value.

Note that if you have a positive expectation of winning - i.e. P is greater than Q - then you can not lose if you wager less than the optimum value of F. You can lose however if the fraction that you wager exceeds the optimum by too great a margin. If F equals .40 then your capital after 10 games will be 975.

Note too that the optimum fraction to bet is the same as your potential profit per wager. If you bet 1 dollar, win 1 dollar with probability P and lose 1 dollar with probability Q then the average profit per dollar wagered is equal to

P * 1 - Q * 1 = P - Q

More to come but not today...

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