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ThatDonGuy
ThatDonGuy
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June 10th, 2015 at 4:56:23 PM permalink
Quote: DeMango

How about betting banker with a better than 50% win rate? Also at a casino that only charges 4% comission.


I'm not 100% sure about this, but I think you multiply the above number by (q - 0.95 p) / (q - p) for a 5% commission on the win, or by (q - 0.96 p) / (q - p) for a 4% commission on the win.

Note that the initially calculated EV is negative if p > q, but when you divide the number by q - p, that's also negative, so if q > 0.96p, the result will be positive (and if q < 0.96p, then it's an advantage play).
DeMango
DeMango
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June 11th, 2015 at 4:33:19 AM permalink
D'Alembert is a great tool for craps too! Combinations of bets can be used, such as combined 4/10, extreme numbers at crapless, and inside numbers combined. Discounting vig, the odds are 1 to 1, 1 to 1, and 3 to 1. You would have to bet the 11/3 at twice the 2/12. Inside #'s one unit each and three hits or the seven determine whether the bets go up one unit or down. 4/10 also up one unit or down.
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
TwoFeathersATL
TwoFeathersATL
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June 11th, 2015 at 7:52:42 AM permalink
Shoot got nuttin to say except I like the way this started from the original OP, wanted it to stay in the loop.
More later....
Youuuuuu MIGHT be a 'rascal' if.......(nevermind ;-)...2F
JyBrd0403
JyBrd0403
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May 20th, 2016 at 11:32:56 PM permalink
I was thinking about this topic recently, I was wondering why your simulator seems to be breaking The Law of Large Numbers. Your hypothesis backed up by your simulator results are stating that the longer you play the further away from the expected value you will get. The more trials the further away from 0 you will go, thereby ruining the D'Alembert progression. The law of large numbers defined here by wikipedia, suggest the exact opposite, the more trials the closer to the expected value you will get. So, infinity shouldn't be a problem.

- According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed.- Wiki

The law states clearly that the longer the trials go the closer you get to the expected value. So, in our case, let's say after 1 million trials, you are at 49.9% wins, or -1000 units from 0. That in the next million trials you should improve on this number. So, at 2 million trials, instead of 49.9% you would be at 49.99% or -200 from 0. And should tend to get closer and closer until finally reaching the goal of 0. Especially, when we are spacing these at 1 million trial intervals. Each million trials should be closer and closer to the expected value, not further and further away.

Your simulator results, however, goes from -1000 to -2000 to -3000 to -4000, and continues to march away from 0, ignoring the law of large numbers, and thus, thwarting the mathematician D'Alemberts betting system. And, your response is that this is suppose to happen. That the more trials the further you will get from the expected value.

The law seems to support all of my theories about mathematician D'alembert's betting system. So, I tend to agree with it. What am I missing here? How will Infinite trials cause the numbers to stray further and further away from 0?
JyBrd0403
JyBrd0403
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May 20th, 2016 at 11:37:45 PM permalink
Quote: ThatDonGuy

Why would I say that? In fact, didn't I (and a few others) say that Martingale also always wins on a 50/50 game if you don't have to worry about things like time and bankroll? The same applies to D'Alembert.

I think I have been able to calculate the EV for an infinite D'Alembert under the given stop conditions.
It appears that there are (2n)! / (n! (n+1)!) ways to have n wins and n losses such that the first result is a loss and losses > wins until the (2n)th bet.
If p is the probability of winning a particular bet, then the stop conditions are:
(a) the first bet is a win (probability p, result +1)
(b) the first two results are LW (probability (1-p) p, result +2)
(c) the first two results are LL, then n Ls and Ws that do not result in the number of Ws = the number of Ls (including the first 2) at any point, then WW (probability (2n)! / (n! (n+1)!) * pn+2 * (1-p)n+2, result (n+2))

For a 50/50 game, this appears to diverge
For double-zero roulette, it appears to converge at around 5.210526
For a 25% game, it appears to converge at 0.625

Remember the key word - infinite. A finite D'Alembert has expected value zero for 50/50 and negative for a less than 50/50 game.



This was meant to go with the above post.
JyBrd0403
JyBrd0403
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May 24th, 2016 at 12:55:16 AM permalink
Just to finish the thought here. The LLN states that the average of the results (or win percentage) will become closer to the expected value the more trials performed. So, I did a little math experiment, starting at 10,000 trials with a win percentage of 49.1%, or -90 units from 0. I added a .001 percent increase to the win percentage every 10,000 trials. So, a modest move closer to 50% every 10,000 trials. Here's what I got.

10,000 trial intervals.

10k x .491 = 4,910 wins -90 from 0
20k x .492 = 9,840 wins -160 from 0
30k x .493 = 14,790 wins -210 from 0
40k x .494 = 19,760 wins -240 from 0
50k x .495 = 24,750 wins -250 from 0
60k x .496 = 29,760 wins -240 from 0
70k x .497 = 34,790 wins -210 from 0
80k x .498 = 39,840 wins -160 from 0
90k x .499 = 44,910 wins -90 from 0

As you can see, the longer you play the closer the win percentage gets to 50%, and the closer to 50% you get, the closer you get to 0.

Interesting fact, this D'Alembert guy, who gets ridiculed so often, apparently did some work with one-dimensional waves. Looking at the above it kinda looks like a wave doesn't it, starts at -90, comes to an apex at -250, and ends at -90.

Just for fun. 100 trial intervals. 1% increase.

100 x .3 = 30 wins -20 from 0
200 x .31 = 62 wins -38 from 0
300 x .32 = 96 wins -54 from 0
400 x .33 = 132 wins -68 from 0
500 x .34 = 170 wins -80 from 0
600 x .35 = 210 wins - 90 from 0
700 x .36 = 252 wins -98 from 0
800 x .37 = 296 wins -104 from 0
900 x .38 = 342 wins -108 from 0
1000 x .39 = 390 wins -110 from 0 (Apex)
1100 x .40 = 440 wins -110 from 0 (Apex)
1200 x .41 = 492 wins -108 from 0
1300 x .42 = 546 wins -104 from 0
1400 x .43 = 602 wins -98 from 0
1500 x .44 = 660 wins -90 from 0
1600 x .45 = 720 wins -80 from 0

After 1600 trials, you are at 45% on your win rate (pretty terrible run), but your already heading back to 0. See the wave forming, with the apex at a measly -110 units from 0. Funny thing is, the above all makes sense.
ThatDonGuy
ThatDonGuy
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May 24th, 2016 at 8:32:19 AM permalink
Quote: JyBrd0403

Just to finish the thought here. The LLN states that the average of the results (or win percentage) will become closer to the expected value the more trials performed. So, I did a little math experiment, starting at 10,000 trials with a win percentage of 49.1%, or -90 units from 0. I added a .001 percent increase to the win percentage every 10,000 trials. So, a modest move closer to 50% every 10,000 trials. Here's what I got.

10,000 trial intervals.

10k x .491 = 4,910 wins -90 from 0
20k x .492 = 9,840 wins -160 from 0
30k x .493 = 14,790 wins -210 from 0
40k x .494 = 19,760 wins -240 from 0
50k x .495 = 24,750 wins -250 from 0
60k x .496 = 29,760 wins -240 from 0
70k x .497 = 34,790 wins -210 from 0
80k x .498 = 39,840 wins -160 from 0
90k x .499 = 44,910 wins -90 from 0

As you can see, the longer you play the closer the win percentage gets to 50%, and the closer to 50% you get, the closer you get to 0.


Of course the total expected return gets closer to zero if you keep increasing the probability of winning. The trouble is, nobody I know runs any game where this really happens.

What makes you think the overall success rate would jump from 0.498 to 0.499 after the ninth set of 10,000 trials? It can be 0.4981, which is still consistent with the Law of Large Numbers, and yet 90,000 x 0.4981 = 44,829 wins = -171 from 0, which is farther away from 0 than the -160 after 80,000 trials.

However, I did another simulation with a 50/50 bet and the added condition that the streak stops when you make a profit rather than when you win a bet with your original bet amount; so far the largest run needed to turn a profit is 213,928,179 bets, with a bankroll of at least 370,966,364 (and even then, the house has to accept a bet of 29,693); even if you could make 1 bet a second, that would require about 6 3/4 years of 24-hour-a-day nonstop play - and if you stop at any point before then, then D'Alembert failed.
UPDATE: I just had a run of 39,368,476,661 bets needed, with a required bankroll of 55,487,447,683 and a high bet of 375,973; even at 10 bets per second, that's just short of 125 years, and even if the initial bet is one penny, you will need over 550 million dollars to prevent having to stop for lack of funds.


UPDATE #2: After some serious number crunching, I have managed to calculate the EV of a "true D'Alembert" (where you keep betting until you win with a minimum bet) for win probabilities from 0.4 to 0.499:


[
pEV
0.499105.83620371
0.49862.70465528
0.49742.16124641
0.49631.74399952
0.49525.4925
0.49421.32433333
0.49318.34664286
0.49216.113
0.49114.37538889
0.49012.985
0.48911.84713636
0.48810.89866667
0.48710.09588462
0.4869.40757143
0.4858.81083333
0.4848.2885
0.4837.82744118
0.4827.41744444
0.4817.05044737
0.4806.72
0.4796.42088095
0.4786.14881818
0.4775.90028261
0.4765.67233333
0.4755.4625
0.4745.26869231
0.4735.08912963
0.4724.92228571
0.4714.76684483
0.4704.62166667
0.4694.48575806
0.4684.35825
0.4674.23837879
0.4664.12547059
0.4654.01892857
0.4643.91822222
0.4633.82287838
0.4623.73247368
0.4613.64662821
0.4603.565
0.4593.48728049
0.4583.41319048
0.4573.34247674
0.4563.27490909
0.4553.21027778
0.4543.1483913
0.4533.08907447
0.4523.03216667
0.4512.97752041
0.4502.925
0.4492.87448039
0.4482.82584615
0.4472.77899057
0.4462.73381481
0.4452.69022727
0.4442.64814286
0.4432.60748246
0.4422.56817241
0.4412.53014407
0.4402.49333333
0.4392.45768033
0.4382.42312903
0.4372.38962698
0.4362.357125
0.4352.32557692
0.4342.29493939
0.4332.26517164
0.4322.23623529
0.4312.2080942
0.4302.18071429
0.4292.15406338
0.4282.12811111
0.4272.10282877
0.4262.07818919
0.4252.05416667
0.4242.03073684
0.4232.00787662
0.4221.9855641
0.4211.96377848
0.4201.9425
0.4191.92170988
0.4181.90139024
0.4171.8815241
0.4161.86209524
0.4151.84308824
0.4141.82448837
0.4131.80628161
0.4121.78845455
0.4111.77099438
0.4101.75388889
0.4091.73712637
0.4081.72069565
0.4071.70458602
0.4061.68878723
0.4051.67328947
0.4041.65808333
0.4031.64315979
0.4021.6285102
0.4011.61412626
0.4001.6



However, these numbers imply infinite bankroll and infinite time - I still need to see if I can determine any statistics (e.g. mean, median) concerning the number of bets necessary.
Last edited by: ThatDonGuy on May 24, 2016
ThatDonGuy
ThatDonGuy
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May 24th, 2016 at 9:17:14 PM permalink
Quote: ThatDonGuy

UPDATE #2: After some serious number crunching, I have managed to calculate the EV of a "true D'Alembert" (where you keep betting until you win with a minimum bet) for win probabilities from 0.4 to 0.499


Slight correction: a true D'Alembert ends when the bet returns to the minimum, not when you win a bet with the minimum.

Also, I have figured out the "per bet" EV for infinite time and bankroll (I can't stress that enough - this does not apply if there is any sort of limit where you could walk away a loser) if the per-bet win probability is p:
You win your first bet with probability p: the "run" ends with you being +1
You lose your first bet with probability 1-p: the "run" ends with N wins and N losses, and you being +N
The total EV per bet is (p x 1) + ((1 - p) x N / 2N) = p + (1 - p) / 2 = (1 - p) / 2
Yes, the EV is higher if the win probability is lower; this is because you will need more bets to make it back to the starting point, and your resulting profit when you get there will be higher.

Note that if there is any sort of time or bankroll limit, then the EV = the player's edge (i.e. -1 x the HE) of the bet x the expected total amount of money bet, which is always negative if the HE is positive. This is not as easy to calculate; for example, if a run is LLLWWW, the bets are 1, 2, 3, 4, 3, 2, which is 15, but if the run is LLWLWW, the bets are 1, 2, 3, 2, 3, 2, which is 13.
JyBrd0403
JyBrd0403
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May 24th, 2016 at 11:57:58 PM permalink
Probability of winning is 50/50. Technically,the win rate wouldn't be used according to the LLN, but instead the average results every 10,000 trials. So, if the average results obtained means that you have to take the average wins every 10,000 trials. Then the average results would increase every 10,000 trials. Which....... kinda looks exactly the same.

1) 4,910 wins
2) 4,930 wins (avg. of 1,2) 4,920 wins .492
3) 4,950 wins (avg. of 1,2,3) 4,930 wins .493
4) 4,970 wins (avg. of 1,2,3,4) 4,940 wins .494
5) 4,990 wins (avg. of 1,2,3,4,5) 4,950 wins .495
6) 5,010 wins (avg. of 1,2,3,4,5,6) 4,960 wins .496
7) 5,030 wins (avg. of 1,2,3,4,5,6,7) 4,970 wins .497
8) 5,050 wins (avg. of 1,2,3,4,5,6,7,8) 4,980 wins .498
9) 5,070 wins (avg. of 1,2,3,4,5,6,7,8,9) 4,990 wins .499

The wave disappears doing it this way, but you can still do it the other way and see the wave :) But, it's pretty much the same thing. Which means that the wins move closer to the EV the more trials you run. The simulator results are still showing the exact opposite, the more trials the further from 0 you go. I'm sure if your simulator shows the avg. results it would have to be breaking the law of large numbers. I can't see any way the wins could keep straying away from 0 without breaking the law.

I don't understand your Table, are you saying for infinite bankroll and infinite time the EV is positive for the D'Alembert? Even at a .40 win probability?
ThatDonGuy
ThatDonGuy
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May 25th, 2016 at 7:11:55 AM permalink
Quote: JyBrd0403

Probability of winning is 50/50. Technically,the win rate wouldn't be used according to the LLN, but instead the average results every 10,000 trials. So, if the average results obtained means that you have to take the average wins every 10,000 trials. Then the average results would increase every 10,000 trials. Which....... kinda looks exactly the same.

1) 4,910 wins
2) 4,930 wins (avg. of 1,2) 4,920 wins .492
3) 4,950 wins (avg. of 1,2,3) 4,930 wins .493
4) 4,970 wins (avg. of 1,2,3,4) 4,940 wins .494
5) 4,990 wins (avg. of 1,2,3,4,5) 4,950 wins .495
6) 5,010 wins (avg. of 1,2,3,4,5,6) 4,960 wins .496
7) 5,030 wins (avg. of 1,2,3,4,5,6,7) 4,970 wins .497
8) 5,050 wins (avg. of 1,2,3,4,5,6,7,8) 4,980 wins .498
9) 5,070 wins (avg. of 1,2,3,4,5,6,7,8,9) 4,990 wins .499

The wave disappears doing it this way, but you can still do it the other way and see the wave :) But, it's pretty much the same thing. Which means that the wins move closer to the EV the more trials you run. The simulator results are still showing the exact opposite, the more trials the further from 0 you go. I'm sure if your simulator shows the avg. results it would have to be breaking the law of large numbers. I can't see any way the wins could keep straying away from 0 without breaking the law.


The fraction of wins approaches 0.5, but that does not necessarily mean that the wins minus losses value approaches zero - I have already shown a counterexample to that statement.
Here is the definition of the Law of Large Numbers from Wolfram Mathworld (emphasis mine):

A "law of large numbers" is one of several theorems expressing the idea that as the number of trials of a random process increases, the percentage difference between the expected and actual values goes to zero.

Quote: JyBrd0403

I don't understand your Table, are you saying for infinite bankroll and infinite time the EV is positive for the D'Alembert? Even at a .40 win probability?


Yes - similar to the principle of "infinite monkeys and infinite typewriters," you "should" "eventually" catch up for any positive probability bet. It might take quadrillions of years, but it "should" happen "eventually." Of course, keep in mind that, under the same conditions, Martingale has positive EV as well.

I ran a 50/50 bet simulation using D'Alembert with the stop condition that you had to stop if you wiped out a bankroll of one billion (with a B) initial bets. At first, every sequence won, and the "per run" profit got as high as 6000 - then, about one billion runs into the simulation, it hit a bet that kept trying to crawl its way back to success but eventually hit the bankroll limit, and the per run profit (including all of the ones that led to the 6000 per run rate up to that point) was now a 6000 per run average loss. Note that the overall bet success was 49.99893%. This is in line with a general principle of pretty much every system - the fraction of wins may be high, but the ratio of average loss amount to average win amount makes up for it.

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