6 ms·

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 media

by 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)
  - OscarCunningham 8y ago
    Okay so I ran my program and I didn't manage to find the exact optimal median but I narrowed it down a lot. I'm assuming each bet is a whole number of cents. I found that there was a strategy that achieved $6.84 with probability 1/2 + 1/2^10. So the median is at least $6.84 (which is 3.8 times the median achieved by Kelly criterion!). I found that there was no strategy that achieved $6.87 with probability 1/2 or greater, so the median is less than $6.87. I found that $6.85 and $6.86 could both be achieved with probability 1/2 but no greater. So the median is somewhere in the range [$6.84, $6.855]. The optimal starting bets of these strategies show some interesting behaviour. For example if you want to have a greater than 1/2 chance of achieving $6.84 then your starting bet can be $0.22, $0.26, $0.31, $0.36, $0.40 or $0.50, but no other value. Make of that what you will.

by szemet 8y ago

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)
- OscarCunningham 8y ago
  Okay so I ran my program and I didn't manage to find the exact optimal median but I narrowed it down a lot. I'm assuming each bet is a whole number of cents. I found that there was a strategy that achieved $6.84 with probability 1/2 + 1/2^10. So the median is at least $6.84 (which is 3.8 times the median achieved by Kelly criterion!). I found that there was no strategy that achieved $6.87 with probability 1/2 or greater, so the median is less than $6.87. I found that $6.85 and $6.86 could both be achieved with probability 1/2 but no greater. So the median is somewhere in the range [$6.84, $6.855]. The optimal starting bets of these strategies show some interesting behaviour. For example if you want to have a greater than 1/2 chance of achieving $6.84 then your starting bet can be $0.22, $0.26, $0.31, $0.36, $0.40 or $0.50, but no other value. Make of that what you will.