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Sorry this is taking so long. I was worried that some of the numbers I was getting from Visual FoxPro were inaccurate so I took some time off to check them with equivalent programs in Mathematica. In Mathematica you can get results to any desired degee of accuracy but at the price of having to program with a much more difficult language. Anyway, the results I posted earlier are indeed the right answers.
Following are the results for betting 1, 2, 3, 4... 15 games simultaneously when the values of P, Q and B are 0.50, 0.50 and 1.2 respectively. The first column is the number of games played per day, the second is the optimal F value and the third is the G value corresponding to the optimal F.
1 0.08333 1.00416
2 0.08267 1.00414
3 0.08199 1.00412
4 0.08130 1.00411
5 0.08061 1.00409
6 0.07989 1.00407
7 0.07916 1.00405
8 0.07841 1.00404
9 0.07763 1.00402
10 0.07680 1.00400
11 0.07592 1.00398
12 0.07491 1.00396
13 0.07355 1.00393
14 0.07073 1.00390
15 0.06651 1.00385
As before, note that F drops slowly as more and more games are bet. When you exceed the number of games where you would be betting your whole bankroll the F value comes closer and closer to BANKROLL / #games but never quite gets there. For example when you are betting 15 games per day you are betting 99.765 percent of your bankroll. This is relatively safe since the chances of not getting any wins in 15 tries are .00305 percent.
So, what are the actual G formulae for betting 1, 2, 3... games at a time? For 1 game it is the formula used in earlier threads
G = (( 1 + B * F ) ^ P) * (( 1 - F ) ^ Q)
For betting 2 games you first have to calculate the change in capital after winning 2 games, 1 game and 0 games. The change in capital is
G2 = ( 1 + 2 * B * F - 0 * F ) for 2 wins
G1 = ( 1 + 1 * B * F - 1 * F ) for 1 win
G0 = ( 1 + 0 * B * F - 2 * F ) for 0 wins
We can refer to Gn as the change in capital for winning n games.
The probability of winning 2 games is the probability of winning the first game times the probability of winning the second game or P * P. Similarly the probability of losing 2 games is Q * Q. The probability of winning 1 game is P * Q but since this can happen in 2 ways you have to multiply P * Q by 2 to get 2 * P * Q.
So it seems that G should be equal to
G2 ^ ( P * P ) * G1 ^ ( 2 * P * Q ) * G0 ^ ( Q * Q )
but this is not quite correct. Remember that we are betting 2 games per day. To calculate the G per game we have to find the square root of the above expression.
G = ( G2 ^ ( P * P ) * G1 ^ ( 2 * P * Q ) * G0 ^ ( Q * Q ) ) ^ 0.5
where the notation X ^ 0.5 or X ^ ( 1 / 2 ) means the square root of X.
For 3 games we have
G3 = ( 1 + 3 * B * F - 0 * F ) for 3 wins
G2 = ( 1 + 2 * B * F - 1 * F ) for 2 wins
G1 = ( 1 + 1 * B * F - 2 * F ) for 1 win
G0 = ( 1 + 0 * B * F - 3 * F ) for 0 wins
and
G =
( G3 ^ ( P ^ 3 ) *
G2 ^ ( 3 * P * P * Q ) *
G1 ^ ( 3 * P * Q * Q ) *
G0 ^ ( Q ^ 3 ) )
^ ( 1 / 3 )
Anyone with some backgound in algebra can probably see the pattern emerging here. If NG is the number of games bet and "n" is the number of wins then
Gn = ( 1 + n * B * F - ( NG - n ) * F )
The probabilities for the Gn's are the individual terms from the expansion of
( P + Q ) ^ NG
For example
( P + Q ) ^ 2 = P * P + 2 * P * Q + Q * Q
( P + Q ) ^ 3 = P * P * P + 3 * P * P * Q + 3 * P * Q * Q + Q * Q * Q
...
Since P + Q = 1 then ( P + Q ) ^ NG must also equal 1. Therefore the sum of the terms in the expansion of ( P + Q ) ^ NG is also 1.
So is there any penalty for betting more than 1 game at a time? There is but it's not large. In the following table we have the average return after 120 games betting 1, 2, 3... 15 games at a time. We get these numbers by raising the G's in the preceding table to the 120 th power.
1 1.64531
2 1.64199
3 1.63866
4 1.63529
5 1.63189
6 1.62845
7 1.62497
8 1.62143
9 1.61783
10 1.61414
11 1.61033
12 1.60633
13 1.60194
14 1.59605
15 1.58626
Betting 1 game at a time, a capital of 1 dollar becomes a capital of 1.64531 dollars after 120 wagers. Equivalently a 10000 dollar stake would become 16453.1 dollars. Betting 4 games at a time we would end up with 16352.9 dollars.
Actually "average" here is a misnomer. The average return from Kelly wagering is actually much higher than these numbers would indicate. The numbers in the above table are not really averages - they are "expected" values. An expected value is the value that you most often expect to get. Kelly averages are higher because when you do well at Kelly you do REALLY well. When you don't do well you DON'T do well. The good results over time however swamp the poor results. This is what scares a lot of people away from Kelly - the fluctuations in your bankroll can be staggering.
To give you some idea of the fluctuations consider the following table. In it I have run a Kelly simulation using P = 0.5, Q = 0.5, B = 1.2 and 4 wagers per day. I have used a random number generator to generate wins and losses for a "season" of 120 wagers. I have done this 10000 times. How often does Kelly win, how often does it lose and how big are the wins and losses?
-6 2
-5 2
-4 42
-3 271
-2 864
-1 1918
0 2667
1 2497
2 1238
3 401
4 90
5 8
The rows in the table are read as follows. The first column is an exponent, specifically the exponent of the number 2. If we call the value in the first column "i" then the value in the second column is the number of simulations in which the finishing capital is greater than 2 ^ i times the starting capital but less than 2 ^ ( i + 1 ) times the starting capital. So the row with a zero in column 1 means that 2667 times out of 10000 the finishing capital was greater than 2 ^ 0 times the starting capital and less than 2 ^ 1 times the starting capital. Since 2 ^ 0 is 1 and 2 ^ 1 is 2 then 2667 times out of 10000 we finished with more than our starting capital but less than double our starting capital. The next row means that 2497 times out of 10000 we ended with more than twice our starting capital but less than 4 times our starting capital.
We made money or broke even 6901 times out of 10000. We ended up with at least 8 times our starting capital 499 times. We lost money 3099 times.
More to come.
Following are the results for betting 1, 2, 3, 4... 15 games simultaneously when the values of P, Q and B are 0.50, 0.50 and 1.2 respectively. The first column is the number of games played per day, the second is the optimal F value and the third is the G value corresponding to the optimal F.
1 0.08333 1.00416
2 0.08267 1.00414
3 0.08199 1.00412
4 0.08130 1.00411
5 0.08061 1.00409
6 0.07989 1.00407
7 0.07916 1.00405
8 0.07841 1.00404
9 0.07763 1.00402
10 0.07680 1.00400
11 0.07592 1.00398
12 0.07491 1.00396
13 0.07355 1.00393
14 0.07073 1.00390
15 0.06651 1.00385
As before, note that F drops slowly as more and more games are bet. When you exceed the number of games where you would be betting your whole bankroll the F value comes closer and closer to BANKROLL / #games but never quite gets there. For example when you are betting 15 games per day you are betting 99.765 percent of your bankroll. This is relatively safe since the chances of not getting any wins in 15 tries are .00305 percent.
So, what are the actual G formulae for betting 1, 2, 3... games at a time? For 1 game it is the formula used in earlier threads
G = (( 1 + B * F ) ^ P) * (( 1 - F ) ^ Q)
For betting 2 games you first have to calculate the change in capital after winning 2 games, 1 game and 0 games. The change in capital is
G2 = ( 1 + 2 * B * F - 0 * F ) for 2 wins
G1 = ( 1 + 1 * B * F - 1 * F ) for 1 win
G0 = ( 1 + 0 * B * F - 2 * F ) for 0 wins
We can refer to Gn as the change in capital for winning n games.
The probability of winning 2 games is the probability of winning the first game times the probability of winning the second game or P * P. Similarly the probability of losing 2 games is Q * Q. The probability of winning 1 game is P * Q but since this can happen in 2 ways you have to multiply P * Q by 2 to get 2 * P * Q.
So it seems that G should be equal to
G2 ^ ( P * P ) * G1 ^ ( 2 * P * Q ) * G0 ^ ( Q * Q )
but this is not quite correct. Remember that we are betting 2 games per day. To calculate the G per game we have to find the square root of the above expression.
G = ( G2 ^ ( P * P ) * G1 ^ ( 2 * P * Q ) * G0 ^ ( Q * Q ) ) ^ 0.5
where the notation X ^ 0.5 or X ^ ( 1 / 2 ) means the square root of X.
For 3 games we have
G3 = ( 1 + 3 * B * F - 0 * F ) for 3 wins
G2 = ( 1 + 2 * B * F - 1 * F ) for 2 wins
G1 = ( 1 + 1 * B * F - 2 * F ) for 1 win
G0 = ( 1 + 0 * B * F - 3 * F ) for 0 wins
and
G =
( G3 ^ ( P ^ 3 ) *
G2 ^ ( 3 * P * P * Q ) *
G1 ^ ( 3 * P * Q * Q ) *
G0 ^ ( Q ^ 3 ) )
^ ( 1 / 3 )
Anyone with some backgound in algebra can probably see the pattern emerging here. If NG is the number of games bet and "n" is the number of wins then
Gn = ( 1 + n * B * F - ( NG - n ) * F )
The probabilities for the Gn's are the individual terms from the expansion of
( P + Q ) ^ NG
For example
( P + Q ) ^ 2 = P * P + 2 * P * Q + Q * Q
( P + Q ) ^ 3 = P * P * P + 3 * P * P * Q + 3 * P * Q * Q + Q * Q * Q
...
Since P + Q = 1 then ( P + Q ) ^ NG must also equal 1. Therefore the sum of the terms in the expansion of ( P + Q ) ^ NG is also 1.
So is there any penalty for betting more than 1 game at a time? There is but it's not large. In the following table we have the average return after 120 games betting 1, 2, 3... 15 games at a time. We get these numbers by raising the G's in the preceding table to the 120 th power.
1 1.64531
2 1.64199
3 1.63866
4 1.63529
5 1.63189
6 1.62845
7 1.62497
8 1.62143
9 1.61783
10 1.61414
11 1.61033
12 1.60633
13 1.60194
14 1.59605
15 1.58626
Betting 1 game at a time, a capital of 1 dollar becomes a capital of 1.64531 dollars after 120 wagers. Equivalently a 10000 dollar stake would become 16453.1 dollars. Betting 4 games at a time we would end up with 16352.9 dollars.
Actually "average" here is a misnomer. The average return from Kelly wagering is actually much higher than these numbers would indicate. The numbers in the above table are not really averages - they are "expected" values. An expected value is the value that you most often expect to get. Kelly averages are higher because when you do well at Kelly you do REALLY well. When you don't do well you DON'T do well. The good results over time however swamp the poor results. This is what scares a lot of people away from Kelly - the fluctuations in your bankroll can be staggering.
To give you some idea of the fluctuations consider the following table. In it I have run a Kelly simulation using P = 0.5, Q = 0.5, B = 1.2 and 4 wagers per day. I have used a random number generator to generate wins and losses for a "season" of 120 wagers. I have done this 10000 times. How often does Kelly win, how often does it lose and how big are the wins and losses?
-6 2
-5 2
-4 42
-3 271
-2 864
-1 1918
0 2667
1 2497
2 1238
3 401
4 90
5 8
The rows in the table are read as follows. The first column is an exponent, specifically the exponent of the number 2. If we call the value in the first column "i" then the value in the second column is the number of simulations in which the finishing capital is greater than 2 ^ i times the starting capital but less than 2 ^ ( i + 1 ) times the starting capital. So the row with a zero in column 1 means that 2667 times out of 10000 the finishing capital was greater than 2 ^ 0 times the starting capital and less than 2 ^ 1 times the starting capital. Since 2 ^ 0 is 1 and 2 ^ 1 is 2 then 2667 times out of 10000 we finished with more than our starting capital but less than double our starting capital. The next row means that 2497 times out of 10000 we ended with more than twice our starting capital but less than 4 times our starting capital.
We made money or broke even 6901 times out of 10000. We ended up with at least 8 times our starting capital 499 times. We lost money 3099 times.
More to come.
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