4 ms·

Interesting problem statement and good analysis. If I am understanding correctly then this is continuous space discrete time. How might the problem change if ti

Interesting problem statement and good analysis. If I am understanding correctly then this is continuous space discrete time. How might the problem change if time was also continuous, meaning card-draw would be integration of some random continuous function, rather than discrete samples?

kqr 4y ago
I can't seem to access the server with the article right now, but generally the easiest way to get there is as the limit of the discrete case as the timestep δ -> 0 and the payoff scaled down so they still add up to the same macro conditions. I suspect it would be the same but less interesting, though. The smaller you make the increments, the closer the optimal stopping point is to 21. At the limit, the stopping point would be 21, and -- I think -- all N players plus dealer would have an 1/(N+1) chance of reaching it. Basically, without jumps, it just becomes a situation with all players waiting for a continuous counter to reach 21, and then stopping precisely as it does. The one who was lucky to have the fastest counter (on average) wins. (With jumps as in discrete Blackjack, the fastest counter not necessarily the best!) What might make it more interesting is if you did introduce discontinuities in the otherwise continuous process. So basically each player is fed with what looks like the absolute values of stock price movements, and has to stand somewhere. That should be easy to simulate and maybe even fun to play!
Someone 4y ago
If there wouldn’t be a delay between observing the current value and stopping (assuming a reasonably well-behaved distribution P), that problem is trivial: stop when you reach the threshold. So, you would have to add some time delay between observation of a value and actual stopping. Intuitively, I can think of two stopping strategies that might be optimal: call ‘stop’ when the expected value at stop is that ≈ 0.5706 or call stop when the probability of overshooting it is 50%. I’m fairly sure that intuition is wrong in general, though.

kqr 4y ago

I can't seem to access the server with the article right now, but generally the easiest way to get there is as the limit of the discrete case as the timestep δ -> 0 and the payoff scaled down so they still add up to the same macro conditions. I suspect it would be the same but less interesting, though. The smaller you make the increments, the closer the optimal stopping point is to 21. At the limit, the stopping point would be 21, and -- I think -- all N players plus dealer would have an 1/(N+1) chance of reaching it. Basically, without jumps, it just becomes a situation with all players waiting for a continuous counter to reach 21, and then stopping precisely as it does. The one who was lucky to have the fastest counter (on average) wins. (With jumps as in discrete Blackjack, the fastest counter not necessarily the best!) What might make it more interesting is if you did introduce discontinuities in the otherwise continuous process. So basically each player is fed with what looks like the absolute values of stock price movements, and has to stand somewhere. That should be easy to simulate and maybe even fun to play!

Someone 4y ago

If there wouldn’t be a delay between observing the current value and stopping (assuming a reasonably well-behaved distribution P), that problem is trivial: stop when you reach the threshold. So, you would have to add some time delay between observation of a value and actual stopping. Intuitively, I can think of two stopping strategies that might be optimal: call ‘stop’ when the expected value at stop is that ≈ 0.5706 or call stop when the probability of overshooting it is 50%. I’m fairly sure that intuition is wrong in general, though.