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Betting everything maximizes your mean wealth. Using the Kelly Criterion maximizes your median wealth. Let's say there are 1024 (2^10) people with $1 each, and

Betting everything maximizes your mean wealth. Using the Kelly Criterion maximizes your median wealth. Let's say there are 1024 (2^10) people with $1 each, and a bet which is 50 / 50 of returning either 3x or 0x (i.e. you bet $1 and either get back $3 or $0). Let's further specify that there will be 10 rounds of betting. Using the "bet everything" strategy, you end up with the following distribution of outcomes: 10 wins, 0 losses -- 1 person: $59,049 1 or more losses -- 1023 people: $0 The average ending balance is $57.67, while the median ending balance is $0. Using the Kelly Criterion, one will bet 25% of the bankroll each time. You will end up with the following distribution of outcomes. 0 wins, 10 losses -- 1 person: $57.66 1 wins, 9 losses -- 10 people: $28.83 2 wins, 8 losses -- 45 people: $14.41 3 wins, 7 losses -- 120 people: $7.20 4 wins, 6 losses -- 210 people: $3.60 5 wins, 5 losses -- 252 people: $1.80 6 wins, 4 losses -- 210 people: $0.90 7 wins, 3 losses -- 120 people: $0.45 8 wins, 2 losses -- 45 people: $0.22 9 wins, 1 losses -- 10 people: $0.11 10 wins, 0 losses -- 1 person: $0.05 The average ending balance is $3.25, while the median ending balance is $1.80. Incidentally, the choice of maximizing the median is somewhat arbitrary. You can also use a similar approach to maximize, say, the 25th percentile outcome, at the expense of average and median outcomes (You would bet 10% of the stake each time, yielding a 25th percentile of $1.08, a median of $1.25, and a mean return of $1.28).

szemet 8y ago
"the median is somewhat arbitrary" I don't know if it matters, but: in the bet everything scenario you end up with 1023 people who will be poorer than they would be in the Kelly scenario. In the 25th percentile maximizing case, you end up with 852 people who will be poorer than in the Kelly case. Imagine you are running a bank, then in both of your alternative cases a large majority of people would be upset: "If you used Kelly Criterion we would have been richer"
OscarCunningham 8y ago
If you're allowed to bet a different proportion of your money in each turn then you can push the median even higher than the Kelly criterion. For example consider the strategy that bets nothing if it has more than $2.98, whatever is needed to reach $2.98 on the next bet if it has less than $2.98, or bets everything if it can't reach $2.98 on the next bet. This strategy has a greater than 50% chance of ending up with $2.98 since it has a 50% chance of success on the first bet and a nonzero probability of success after that. So its median outcome is $2.98. I don't know which strategy actually maximises the median, but I suspect it involves being conservative after you've been lucky and agressive after you've been unlucky.
- szemet 8y ago
  I guess you are right. A minor issue is that you have to beat the 1.8 value of OP in (or in less than) ten rounds because if you have infinite rounds then median can grow infinitely (am I right?). It is not clearly visible in your solution, that it really happens in less then 11 rounds. But it is easy to extend the strategy with this: First get half of the people above 1.8 by betting 0.400000000...1. Now half of the people have 1.80...1. They are finished. Half of the people have .5999. They can do all-in twice, then 25% of them will be above 1.8. That is 3 rounds in total. edit: the maximum of this exact 3 round strategy is betting 8/11$=0.72 in first round, then in round 3 we will have 62.5% of people at 2.45$ edit 2: got the numbers wrong in previous calculation, fixed (Now someone should chime in, and present the actual optimal solution. ;)
  - OscarCunningham 8y ago
    The reason I chose $2.98 was that if you follow my strategy then that will lead to you betting $0.99 in the first round. If you win then you end up with exactly $2.98 and stop betting. If you lose then you end up with $0.01. If you were to bet all of it for the next 9 rounds then there is a 1/2^9 chance of you ending up with $0.01×3^9 = $196.83. Since this is more than $2.98 my strategy must have at least a 1/512 chance of achieving $2.98 even if it loses the first round. Since the chance of winning the first round is 1/2 the overall chance of ending up with $2.98 is greater than 50% and hence $2.98 is the median. I think I can run a computer search to find the actual optimal strategy. I'll report back with the results.
    - szemet 8y ago
      Ok. An additional thing is, if you find the optimal strategy in the form of: "in n round the best achievable median is x betting this and this way", you should check the strategy when n approaches infinity - I will not have been surprised if it turns out to be the Kelly Criterion after all - and OP will be right in this sense (though presented the fact somewhat badly by limiting the number of rounds)
      - 8y ago

szemet 8y ago

"the median is somewhat arbitrary" I don't know if it matters, but: in the bet everything scenario you end up with 1023 people who will be poorer than they would be in the Kelly scenario. In the 25th percentile maximizing case, you end up with 852 people who will be poorer than in the Kelly case. Imagine you are running a bank, then in both of your alternative cases a large majority of people would be upset: "If you used Kelly Criterion we would have been richer"

OscarCunningham 8y ago

If you're allowed to bet a different proportion of your money in each turn then you can push the median even higher than the Kelly criterion. For example consider the strategy that bets nothing if it has more than $2.98, whatever is needed to reach $2.98 on the next bet if it has less than $2.98, or bets everything if it can't reach $2.98 on the next bet. This strategy has a greater than 50% chance of ending up with $2.98 since it has a 50% chance of success on the first bet and a nonzero probability of success after that. So its median outcome is $2.98. I don't know which strategy actually maximises the median, but I suspect it involves being conservative after you've been lucky and agressive after you've been unlucky.

szemet 8y ago
I guess you are right. A minor issue is that you have to beat the 1.8 value of OP in (or in less than) ten rounds because if you have infinite rounds then median can grow infinitely (am I right?). It is not clearly visible in your solution, that it really happens in less then 11 rounds. But it is easy to extend the strategy with this: First get half of the people above 1.8 by betting 0.400000000...1. Now half of the people have 1.80...1. They are finished. Half of the people have .5999. They can do all-in twice, then 25% of them will be above 1.8. That is 3 rounds in total. edit: the maximum of this exact 3 round strategy is betting 8/11$=0.72 in first round, then in round 3 we will have 62.5% of people at 2.45$ edit 2: got the numbers wrong in previous calculation, fixed (Now someone should chime in, and present the actual optimal solution. ;)
- OscarCunningham 8y ago
  The reason I chose $2.98 was that if you follow my strategy then that will lead to you betting $0.99 in the first round. If you win then you end up with exactly $2.98 and stop betting. If you lose then you end up with $0.01. If you were to bet all of it for the next 9 rounds then there is a 1/2^9 chance of you ending up with $0.01×3^9 = $196.83. Since this is more than $2.98 my strategy must have at least a 1/512 chance of achieving $2.98 even if it loses the first round. Since the chance of winning the first round is 1/2 the overall chance of ending up with $2.98 is greater than 50% and hence $2.98 is the median. I think I can run a computer search to find the actual optimal strategy. I'll report back with the results.
  - szemet 8y ago
    Ok. An additional thing is, if you find the optimal strategy in the form of: "in n round the best achievable median is x betting this and this way", you should check the strategy when n approaches infinity - I will not have been surprised if it turns out to be the Kelly Criterion after all - and OP will be right in this sense (though presented the fact somewhat badly by limiting the number of rounds)
    - 8y ago