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Sep
29
comment Expected travel of random walk in arbitrary game with multiple payouts
@ColmBhandal: Okay done!
Sep
29
revised Expected travel of random walk in arbitrary game with multiple payouts
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Sep
25
comment Expected travel of random walk in arbitrary game with multiple payouts
Sure. Here you go: pastebin.com/985eDTFh - Don't worry about the size - three quarters of it is code for a fast random class I pinched from Stackoverflow. You can use the default Random class if you prefer. Also, if you're not using C#, don't worry too much about converting the convertCSVtoPayoutTable() function as you can hard code the probability and payout arrays yourself if you prefer.
Sep
25
comment Expected travel of random walk in arbitrary game with multiple payouts
Aha, yes I get roughly around 9.38-9.4 after simulating 10M 'universes' (which you call experiments). Let me know if you'd like the code (it's in C# but is easily transferable to other languages).
Sep
24
comment Expected travel of random walk in arbitrary game with multiple payouts
Thanks, great presentation, and probably bounty worthy. The 'tails-not-cancelling' concept made sense too. I'm having a little trouble with interpretation - can you show explicitly plugging in the numbers to the final formula so I can see it more clearly to see how 268.8 is reached (or perhaps rather 18.8 as that's the difference between that and 250). Here they are again: INPUT: outcomes<probability:profit> = {0.75:-1, 0.1:0.5, 0.15:4.5}, mean=-0.025, runs=10000. OUTPUT = 268.8. Excuse the more programmer-like description there. Hopefully it's clear.
Sep
19
comment Expected travel of random walk in arbitrary game with multiple payouts
@ColmBhandal: Thanks, any chance for a closed-form version and/or for a more accurate version that solves the tiny error thing? Also the latter two limit equations use n for the variable towards infinity, but it isn't used.
Sep
19
comment Expected travel of random walk in arbitrary game with multiple payouts
Okay, I tried and it passed. My new zero-mean test was 0.75:-1, 0.1:0.75, 0.15:4.5 = 156.233 apx travel in the simulation which closely matches the formula $\sqrt{2*3.84375*10000/pi}=156.4$. Also tried 0.5:-1, 0.25:1, 0.25:1 which comes out the same as the coin-toss paytable 0.5:-1, 0.5:1 (= 78.8 approx) as expected. I'm pretty confident the simulation is accurate.
Sep
17
comment Expected travel of random walk in arbitrary game with multiple payouts
I plugged in the numbers $-0.025*10000 + \sqrt{2*3.811875*10000/\pi}$ but got apx 94 (or -405 if I tried using negative instead of positive between the terms). I expected 268.8 (or -268.8) as I detailed in my question. Where I have gone wrong? I presume $\sigma^2$ = 3.811875.
Sep
17
comment Expected travel of random walk in arbitrary game with multiple payouts
Just to add I now understand that in the limit where N tends to infinity, $μN$ does indeed equal what I'm after. But not for a finite number such as 10000 runs, so I would like the solution for an arbitrary number of runs. Things are beginning to make a lot more sense now.
Sep
16
comment Expected travel of random walk in arbitrary game with multiple payouts
@mjqxxxx: Thanks that's clear, but $μN$ would then simply equal -0.025*10000 = -250. But my simulation returns around -268. Do you think my simulation is mistaken?
Sep
16
revised Expected travel of random walk in arbitrary game with multiple payouts
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Sep
16
comment Expected travel of random walk in arbitrary game with multiple payouts
@Did: Thanks. Regardless, that still leaves a lot to the imagination with my current level of mathematical knowledge, but I'll open a bounty.
Sep
16
revised Expected travel of random walk in arbitrary game with multiple payouts
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Sep
16
revised Expected travel of random walk in arbitrary game with multiple payouts
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Sep
16
comment Expected travel of random walk in arbitrary game with multiple payouts
@Did: Had time to revisit this. Assuming N=10000, I'm not sure how to resolve $(0.75*-1 + 0.1*0.5 + 0.15*4.5) + o(10000)$ to equal -268.8. The first part of that sum equals -0.025. The second part (o(N)) from what I understand is the set of functions which have a lower rate of growth than N, so it could be almost anything. Somehow, I'm supposed to ascertain around -269 from that? Perhaps I could open a bounty if the expansion is more complex than I anticipated.
Aug
17
comment Expected travel of random walk in arbitrary game with multiple payouts
@Did: Can you expand the formula for clarity? I think by a 'mean increment', you're referring to the expected payout (which is $prob_1*pay_1 + prob_2*pay_2 + prob_3*pay_3$), but I'm not sure how to interpret o(N) from your comment...
Aug
17
comment Expected travel of random walk in arbitrary game with multiple payouts
@Did: Ah, in that case, can you give me the new revised formula which takes into account non-centered paytables?
Aug
17
comment Expected travel of random walk in arbitrary game with multiple payouts
@Did: Thanks for the feedback. Your formula works for the coin and dice rolls, but when I tried for three entries or more, it differs from the simulation. An example is: probabilities = [0.75, 0.1, 0.15] and payout = [-1, 0.5, 4.5]. That's a variance of 3.811875, but the average 'travel' according to your formula is 155.779 (to 3dp), whilst I get approx 268.8. Could you check your formula?
Aug
17
revised Expected travel of random walk in arbitrary game with multiple payouts
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