MrGoldenSun
MrGoldenSun
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June 17th, 2016 at 11:05:10 AM permalink
But we didn't just declare that, and then it became so. It's math. We only have so much power. :)

Your example doesn't quite seem applicable to me to prove your case, because we all agree there is no value of n where a series of n roulette trials has positive EV. The question is about a martingale of unlimited trials. We can't "work backward" from infinity because we can't ever be there.

What would you say is the EV of a martingale with no limits on number of trials, bet size, or bankroll?
DiscreteMaths2
DiscreteMaths2
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June 17th, 2016 at 11:19:10 AM permalink
I actually don't know how to calculate EV like that (or even if you can). Martingale with the constraints you present as far as I know cannot be represented by a finite or infinite sum due to the fact that you do logical branching. Maybe you can calculate the EV by doing the integral of a function or something ?
Assume the worst, believe no one, and make your move only when you are certain that you are unbeatable or have, at worst, exceptionally good odds in your favor.
DiscreteMaths2
DiscreteMaths2
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June 17th, 2016 at 11:52:29 AM permalink
Flat betting $1:
Expected value of a single spin: 1*(18/38) - 1(20/38) = -1/19
Expected value of 100 spins: sum 1*(18/38) - 1(20/38) from 1 to 100 = -100/19 which is -1/19 a spin
Expected value of an infinite amount of spins: sum 1*(18/38) - 1(20/38) from 1 to infinity = the sum diverges, the expected value does not exist

Martingale starting at $1:
Expected value of the first spin: 1*(18/38) - 1(20/38) = -1/19

Expected value of the second spin given that you won the first: 1*(18/38) - 1(20/38) = -1/19
Expected value of the second spin given that lost the first: 2*(18/38) - 2(20/38) = -2/19

Expected value of the third spin given that you won the first and second spins: 1*(18/38) - 1(20/38) = -1/19
Expected value of the third spin given that you won the first and lose the second spin: 2*(18/38) - 2(20/38) = -2/19
Expected value of the third spin given that you lost the first and second spins: 4*(18/38) - 4(20/38) = -4/19

and so on. How you write that mathematically I don't know.
Assume the worst, believe no one, and make your move only when you are certain that you are unbeatable or have, at worst, exceptionally good odds in your favor.
PeeMcGee
PeeMcGee
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June 17th, 2016 at 10:18:40 PM permalink
Quote: DiscreteMaths2

Flat betting $1:
Expected value of a single spin: 1*(18/38) - 1(20/38) = -1/19
Expected value of 100 spins: sum 1*(18/38) - 1(20/38) from 1 to 100 = -100/19 which is -1/19 a spin
Expected value of an infinite amount of spins: sum 1*(18/38) - 1(20/38) from 1 to infinity = the sum diverges, the expected value does not exist

Martingale starting at $1:
Expected value of the first spin: 1*(18/38) - 1(20/38) = -1/19

Expected value of the second spin given that you won the first: 1*(18/38) - 1(20/38) = -1/19
Expected value of the second spin given that lost the first: 2*(18/38) - 2(20/38) = -2/19

Expected value of the third spin given that you won the first and second spins: 1*(18/38) - 1(20/38) = -1/19
Expected value of the third spin given that you won the first and lose the second spin: 2*(18/38) - 2(20/38) = -2/19
Expected value of the third spin given that you lost the first and second spins: 4*(18/38) - 4(20/38) = -4/19

and so on. How you write that mathematically I don't know.


There is no question that each spin of roulette has negative EV…which is equal to the size of the bet divided by 19. That’s the only thing your math shows.

You are disregarding a very important facet of the infinite Martingale system…

In each of your calculations, you are leaving open the possibility of ending on a losing spin—which cannot happen with infinite money and time. Because you would simply place another bet. Let me reiterate…you will never end on a losing bet.

And because you will never end on a losing bet, your probability of losing is zero. And since your probability of losing is zero, your probability of winning is one.

A winning spin in a Martingale system will always net 1-unit profit. So you would have a 100% chance to win 1-unit. Your EV is +1.

This is easily calculable with real math, as I have done so a few posts ago.
Ibeatyouraces
Ibeatyouraces
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June 17th, 2016 at 10:39:04 PM permalink
It's impossible to have infinity + 1
DUHHIIIIIIIII HEARD THAT!
TomG
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June 17th, 2016 at 10:42:05 PM permalink
Quote: PeeMcGee


A winning spin in a Martingale system will always net 1-unit profit. So you would have a 100% chance to win 1-unit. Your EV is +1.



With an infinite amount of money, the amount of money you have remains the same after every win. There is zero profit.
billryan
billryan
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June 17th, 2016 at 10:47:32 PM permalink
Gambling with an infinite amount of money would be like an episode of The Twilight Zone.
The older I get, the better I recall things that never happened
OnceDear
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June 18th, 2016 at 12:35:03 AM permalink
Quote: PeeMcGee

You are disregarding a very important facet of the infinite Martingale system…

In each of your calculations, you are leaving open the possibility of ending on a losing spin—which cannot happen with infinite money and time. Because you would simply place another bet. Let me reiterate…you will never end on a losing bet.

And because you will never end on a losing bet, your probability of losing is zero. And since your probability of losing is zero, your probability of winning is one. . . .
This is easily calculable with real math, as I have done so a few posts ago.


PMG.
Where you are going wrong is in treating Infinity as a number which can be equated and manipulated with simple maths equations. Infinity is something that a series can approach but NEVER reach or equal.

An Infinite bankroll cannot exist by definition. It just can't.

Saying that Martingale will have +ev of 1 unit if you have infinite bankroll is like saying that you can stroke a unicorn at the petting zoo if unicorns are real. When you know damned well that unicorns, by definition, are not real.
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
TwoFeathersATL
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June 18th, 2016 at 5:34:46 AM permalink
I beg to differ.
I saw a unicorn standing on the side of the road 'bout 3:00 in the morning.
Just 'cause you didn't see it doesn't mean it wasn't there.
Don't know why they don't have them at the petting zoo ;-)
Youuuuuu MIGHT be a 'rascal' if.......(nevermind ;-)...2F
PeeMcGee
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June 18th, 2016 at 9:38:08 AM permalink
Quote: OnceDear

Where you are going wrong is in treating Infinity as a number which can be equated and manipulated with simple maths equations. Infinity is something that a series can approach but NEVER reach or equal.


I am doing no such thing. The math I am using is completely valid. Ironically, it is the other side of the disagreement whom is treating infinity as a number.

By the way, this series is not approaching infinity. It's a convergent series, and it is equal to 1.



Quote:

Quote: Ibeatyouraces

It's impossible to have infinity + 1


Quote: TomG

With an infinite amount of money, the amount of money you have remains the same after every win. There is zero profit.


The size of the bankroll does not change the fact that the game will pay out.

If I had infinite money, and you gave me $10—that does not change the fact that you gave me $10. Just because my buying power didn’t change doesn’t negate that fact.

And it is probably beyond the scoop of this argument, but there are different sizes of infinity.


Quote: OnceDear

An Infinite bankroll cannot exist by definition. It just can't.


Which I (and MrGoldenSun) have pointed out many times. To quote myself; ‘the question is not is it possible?. The question is what if it were possible?'.

We are starting with the assumption that an infinite bankroll and infinite time does exists. And from that assumption, we are asking is Martingale positive.

Of course, this is purely theoretical—no one is denying that.

By the way, you could kind of simulate an infinite bankroll in the physical world—for example, if you were allowed to verbally declare any money amount and have it be honored. It’s really the infinite time that is the difficult obstacle.


Quote: OnceDear

Saying that Martingale will have +ev of 1 unit if you have infinite bankroll is like saying that you can stroke a unicorn at the petting zoo if unicorns are real. When you know damned well that unicorns, by definition, are not real.


If unicorns are real (and assuming they are stroke-able/tamable), then why wouldn’t we be able to stroke one at a petting zoo?
Ibeatyouraces
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June 18th, 2016 at 10:09:57 AM permalink
If I give you $10 out of my infinity, I now don't have infinity. It's impossible to have infinity - 10.

Also, although highly unlikely, but it is mathematically possible to never win another bet for the rest of eternity.
DUHHIIIIIIIII HEARD THAT!
PeeMcGee
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June 18th, 2016 at 10:46:01 AM permalink
Quote: Ibeatyouraces

If I give you $10 out of my infinity, I now don't have infinity. It's impossible to have infinity - 10.


It’s not impossible. If I take away ten numbers from the set of even numbers, the set will still contain infinitely many elements. And it will be a different set.

You are doing what I was accused of—treating infinity as a number. Which it is not.

Quote: Ibeatyouraces

Also, although highly unlikely, but it is mathematically possible to never win another bet for the rest of eternity.


Mathematically, the probability of that is exactly ZERO.
billryan
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June 18th, 2016 at 10:47:04 AM permalink
If I have an infinite amount of money, how would losing ten dollars or a million dollars change that?
The older I get, the better I recall things that never happened
Ibeatyouraces
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June 18th, 2016 at 11:06:09 AM permalink
Quote: billryan

If I have an infinite amount of money, how would losing ten dollars or a million dollars change that?


What's the answer to this equation?

Infinity - 1,000,000 = ?

Surely it can't be infinity, correct?
DUHHIIIIIIIII HEARD THAT!
DiscreteMaths2
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June 18th, 2016 at 11:06:22 AM permalink
Quote: PeeMcGee

There is no question that each spin of roulette has negative EV…which is equal to the size of the bet divided by 19. That’s the only thing your math shows.

You are disregarding a very important facet of the infinite Martingale system…

In each of your calculations, you are leaving open the possibility of ending on a losing spin—which cannot happen with infinite money and time. Because you would simply place another bet. Let me reiterate…you will never end on a losing bet.

And because you will never end on a losing bet, your probability of losing is zero. And since your probability of losing is zero, your probability of winning is one.

A winning spin in a Martingale system will always net 1-unit profit. So you would have a 100% chance to win 1-unit. Your EV is +1.

This is easily calculable with real math, as I have done so a few posts ago.



Not to sound rude but I think you have some mistakes on how you think EV is calculated. At no point in this thread did you calculate EV for martingale. Yes you posted a series that does indeed converge to 1 but that is not how you calculate EV.
Assume the worst, believe no one, and make your move only when you are certain that you are unbeatable or have, at worst, exceptionally good odds in your favor.
Ibeatyouraces
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June 18th, 2016 at 11:08:08 AM permalink
Quote: PeeMcGee

Mathematically, the probability of that is exactly ZERO.


Infinitely incorrect. Close to zero, but NOT zero.
DUHHIIIIIIIII HEARD THAT!
billryan
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June 18th, 2016 at 11:19:00 AM permalink
Quote: Ibeatyouraces

What's the answer to this equation?

Infinity - 1,000,000 = ?

Surely it can't be infinity, correct?



Then I never had an infinite amount in the first place.
The older I get, the better I recall things that never happened
PeeMcGee
PeeMcGee
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June 18th, 2016 at 1:27:03 PM permalink
Quote:

Quote: billryan

If I have an infinite amount of money, how would losing ten dollars or a million dollars change that?


Quote: Ibeatyouraces

What's the answer to this equation?

Infinity - 1,000,000 = ?

Surely it can't be infinity, correct?


Quote: billryan

Then I never had an infinite amount in the first place.


Again, infinity is not a number. But if you have a set with infinitely many elements and took away any finite number of elements, the set will still contain infinitely many elements.


Quote: Ibeatyouraces

Infinitely incorrect. Close to zero, but NOT zero.


It can be difficult to grasp…but it is in fact zero. In fact it is…. Which is zero.


Quote: DiscreteMaths2

Quote: PeeMcGee

There is no question that each spin of roulette has negative EV…which is equal to the size of the bet divided by 19. That’s the only thing your math shows.

You are disregarding a very important facet of the infinite Martingale system…

In each of your calculations, you are leaving open the possibility of ending on a losing spin—which cannot happen with infinite money and time. Because you would simply place another bet. Let me reiterate…you will never end on a losing bet.

And because you will never end on a losing bet, your probability of losing is zero. And since your probability of losing is zero, your probability of winning is one.

A winning spin in a Martingale system will always net 1-unit profit. So you would have a 100% chance to win 1-unit. Your EV is +1.

This is easily calculable with real math, as I have done so a few posts ago.



Not to sound rude but I think you have some mistakes on how you think EV is calculated. At no point in this thread did you calculate EV for martingale. Yes you posted a series that does indeed converge to 1 but that is not how you calculate EV.


I did—the first post of mine on page 4. I calculated the EV just fine. If you don’t agree, please tell me where you believe the flaw lies. I can make it a little more formal (which it seems you are seeking).

Assuming infinite bankroll and time, a Martingale system will have expected value as follows.

Let W be a discrete random variable (winnings) taking values w0, w1, … with probabilities p0, p1, … respectively. Then the expected value of this random variable is:
TomG
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June 18th, 2016 at 1:58:50 PM permalink
Quote: PeeMcGee

The size of the bankroll does not change the fact that the game will pay out.

If I had infinite money, and you gave me $10—that does not change the fact that you gave me $10. Just because my buying power didn’t change doesn’t negate that fact.



Your original claim was that it had a positive expected value. That is different than saying the game will pay out after a win: https://en.wikipedia.org/wiki/Expected_value
PeeMcGee
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June 18th, 2016 at 2:50:52 PM permalink
Quote: TomG

Your original claim was that it had a positive expected value. That is different than saying the game will pay out after a win: https://en.wikipedia.org/wiki/Expected_value


The game will pay out an expected amount of 1-unit. The probability-weighted average of all possible values is 1. That is the expected value. That has been my claim the entire time. The bettor’s purchasing-power being unchanged does not negate that fact.

I am not sure why you provided that link. Nothing in that article contradicts this claim.
Ibeatyouraces
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June 18th, 2016 at 2:56:49 PM permalink
Quote: PeeMcGee

...It can be difficult to grasp…but it is in fact zero. In fact it is…. Which is zero...


I'll say this again. You are incorrect. If something has a probability of even 0.0000(insert 5,000,000 quadrillion quadrillion quadrillion more zeros)000001% chance of happening, it's still NOT zero.
DUHHIIIIIIIII HEARD THAT!
OnceDear
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June 18th, 2016 at 3:06:05 PM permalink
Quote: billryan

If I have an infinite amount of money, how would losing ten dollars or a million dollars change that?


How would it change your life? Win or lose?
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
DiscreteMaths2
DiscreteMaths2
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June 18th, 2016 at 3:20:39 PM permalink
Quote: PeeMcGee



Let W be a discrete random variable (winnings) taking values w0, w1, … with probabilities p0, p1, … respectively. Then the expected value of this random variable is:



Because you are leaving out where you lose money. First spin is 1*(18/38) - 1(20/38) not just 1*(18/38).
Assume the worst, believe no one, and make your move only when you are certain that you are unbeatable or have, at worst, exceptionally good odds in your favor.
billryan
billryan
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June 18th, 2016 at 4:06:16 PM permalink
Quote: OnceDear

How would it change your life? Win or lose?



I'd tip more on wins.
The older I get, the better I recall things that never happened
FleaStiff
FleaStiff
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June 18th, 2016 at 4:53:29 PM permalink
>>>>I was in a casino playing roulette (I really love this game), but have yet to find
>>>>a way of increasing my odds for the short term :)

To increase your odds for the short term: CHEAT,

To increase your odds for the longer term: CHEAT LONGER.

There is NO other way to accomplish this except the old and honest way: Buy the Casino.
RS
RS
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June 18th, 2016 at 5:47:40 PM permalink
Still not sure how the EV gets above 0.


# WagersEV
1-0.052631579
2-0.108033241
3-0.16635078
4-0.227737663
5-0.292355435
6-0.360374142
7-0.431972781
8-0.50733977
9-0.586673442
10-0.67018257
11-0.758086916
12-0.850617806
13-0.948018743
14-1.050546046
15-1.158469522
16-1.272073181
17-1.39165598
18-1.51753261
19-1.650034327
20-1.789509818
21-1.936326124
22-2.090869604
23-2.253546951
24-2.424786265
25-2.605038173
26-2.794777025
27-2.994502131
28-3.204739085
29-3.426041143
30-3.658990676
31-3.904200712
32-4.162316539
33-4.434017409
34-4.720018326
35-5.021071922
36-5.337970444
37-5.671547836
38-6.022681932
39-6.392296771
40-6.781365022
41-7.190910549
42-7.622011105
43-8.075801163
44-8.553474908
45-9.056289377
46-9.585567765
47-10.14270291
48-10.72916096
49-11.34648522
50-11.99630023
51-12.68031603
52-13.40033267
53-14.15824491
54-14.95604728
55-15.79583924
56-16.67983078
57-17.61034819
58-18.5898402
59-19.62088442
60-20.70619412
61-21.84862539
62-23.05118462
63-24.31703645
64-25.64951205
65-27.05211795
66-28.52854521
67-30.08267916
68-31.71860965
69-33.44064173
70-35.25330709
71-37.16137588
72-39.16986935
73-41.284073
74-43.50955052
75-45.85215845
76-48.31806152
77-50.91374897
78-53.64605155
79-56.52215953
80-59.54964161
81-62.73646485
82-66.09101563
83-69.62212172
84-73.33907549
85-77.25165841
86-81.37016675
87-85.70543868
88-90.26888282
89-95.07250824
90-100.128956
91-105.4515327
92-111.0542449
93-116.9518368
94-123.1598282
95-129.694556
96-136.5732168
97-143.8139124
98-151.4356973
99-159.4586287
100-167.9038197
101-176.7934944
102-186.1510468
103-196.0011019
104-206.3695809
105-217.2837694
106-228.7723888
107-240.8656724
108-253.5954447
109-266.9952049
110-281.1002157
111-295.9475955
112-311.5764163
113-328.0278066
114-345.3450596
115-363.5737469
116-382.7618389
117-402.9598304
118-424.2208741
119-446.6009201
120-470.1588633
121-494.9566982
122-521.0596823
123-548.5365077
124-577.4594818
125-607.9047177
126-639.9523344
127-673.6866678
128-709.1964924
129-746.5752552
130-785.9213212
131-827.3382329
132-870.934982
133-916.8262968
134-965.132944
135-1015.982046
136-1069.507417
137-1125.849913
138-1185.157803
139-1247.587161
140-1313.302275
141-1382.476079
142-1455.290609
143-1531.937483
144-1612.618403
145-1697.545688
146-1786.942829
147-1881.045084
148-1980.100088
149-2084.368514
150-2194.124751
151-2309.657633
152-2431.271192
153-2559.285466
154-2694.037332
155-2835.881402
156-2985.19095
157-3142.358895
158-3307.798836
159-3481.946144
160-3665.259099
161-3858.220104
162-4061.336951
163-4275.144159
164-4500.204378
165-4737.109872
166-4986.484076
167-5248.983237
168-5525.298145
169-5816.155942
170-6122.322044
171-6444.602152
172-6783.84437
173-7140.941442
174-7516.833097
175-7912.508523
176-8329.008972
177-8767.430497
178-9228.926839
179-9714.712462
180-10226.06575
181-10764.33237
182-11330.92881
183-11927.34611
184-12555.1538
185-13216.004
186-13911.63579
187-14643.87978
188-15414.66293
189-16226.01361
190-17080.06696
191-17979.07048
192-18925.38998
193-19921.51577
194-20970.06923
195-22073.80972
196-23235.64181
197-24458.62295
198-25745.97153
199-27101.0753
200-28527.50031
201-30029.00033
202-31609.52666
203-33273.23859
204-35024.51431
205-36867.96243
206-38808.43413
207-40851.03593
208-43001.14308
209-45264.41377
210-47646.80397
211-50154.58313
212-52794.35066
213-55573.05333
214-58498.0035
215-61576.89842
216-64817.84045
217-68229.35836
218-71820.42986
219-75600.50511
220-79579.5317
221-83767.98073
222-88176.87446
223-92817.81522
224-97703.01602
225-102845.3327
226-108258.2975
227-113956.1553
228-119953.9003
229-126267.3161
230-132913.017
231-139908.4915
232-147272.149
233-155023.3674
234-163182.5446
235-171771.1522
236-180811.7918
237-190328.2545
238-200345.5837
239-210890.1407
240-221989.6745
241-233673.3942
242-245972.0465
243-258917.9963
244-272545.3119
245-286889.8546
246-301989.3733
247-317883.6035
248-334614.3721
249-352225.7074
250-370763.9552
251-390277.9002
252-410818.895
253-432440.9947
254-455201.0997
255-479159.1049
256-504378.0578
257-530924.324
258-558867.7621
259-588281.9075
260-619244.1658
261-651836.0166
262-686143.228
263-722256.0821
264-760269.6128
265-800283.8555
266-842404.1111
267-886741.2222
268-933411.8655
269-982538.8584
270-1034251.483
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Now, perhaps I did the math wrong....but I don't think so.
RS
RS
  • Threads: 62
  • Posts: 8626
Joined: Feb 11, 2014
June 18th, 2016 at 5:54:44 PM permalink
To me, it appears the EV only decreases with the more wagers you make.

If ya wanna say something like, "Martingale wins over an infinite time with infinite bankroll etc. as long as there is a winning spin [and you quit after the winning spin]" then go ahead, but that's just as stupid as saying, "My system wins if I win!" or "Ya can't lose if you walk away a winner!"

BTW: EV is additive. All you have to do to figure out your total (net) EV is to add all the EV you've generated (or plan to generate going forward) together. The EV on roulette is -5.26% or -0.0526 of the wager made. Every time you make a wager in roulette, multiply it by -0.0526. Do whatever the hell ya wanna do, increase the bets, decrease the bets, or make no bets at all. No matter what, the EV will NOT be a positive number. The greatest it can be is 0 (ie: you don't make any wagers).

EDIT: For the EV, I'm getting [1 - (20/38)^INF] - [(20/38)^INF * (2^INF -1)]

Which appears to be 1 - INF

EDIT EDIT: https://www.wolframalpha.com/input/?i=lim+x-%3EINF+:+%5B1+-+(20%2F38)%5Ex%5D+-+%5B(20%2F38)%5Ex+*+(2%5Ex+-1)%5D

You gotta copy/paste it.
Last edited by: RS on Jun 18, 2016
MrGoldenSun
MrGoldenSun
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June 18th, 2016 at 7:13:25 PM permalink
Quote: DiscreteMaths2

Quote: PeeMcGee



Let W be a discrete random variable (winnings) taking values w0, w1, … with probabilities p0, p1, … respectively. Then the expected value of this random variable is:



Because you are leaving out where you lose money. First spin is 1*(18/38) - 1(20/38) not just 1*(18/38).



You are misunderstanding this a bit. The calculation is the sum from n=1 to infinity of (winnings for a series terminating on spin n)*(probability of ending on spin n)

Quote: Ibeatyouraces

Quote: PeeMcGee

...It can be difficult to grasp…but it is in fact zero. In fact it is…. Which is zero...


I'll say this again. You are incorrect. If something has a probability of even 0.0000(insert 5,000,000 quadrillion quadrillion quadrillion more zeros)000001% chance of happening, it's still NOT zero.



That expression is equal to zero. If you think it is not zero, what do you think it is?

PeeMcGee posted a claim the EV is 1 in which he defined the random variable and the probabilities and then calculated the expected value of that variable. I think he is correct. For those of you saying he is wrong, where is the flaw in his argument?

My only question is if the probabilities can be defined the way we have done it. I don't know if the probability density function is problematic in some way.
Ibeatyouraces
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June 18th, 2016 at 7:24:41 PM permalink
The flaw is that he believes he's guaranteed a win at some point which is not true.
DUHHIIIIIIIII HEARD THAT!
RS
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June 18th, 2016 at 7:30:17 PM permalink
I don't know what the exact definition of EV is, but I would say it's something like this:

Prob(win) * payout - Prob(loss) * loss_amount = EV


I see no prob(L)*Loss_Amount in the formulas that "show" the EV equals 1.
AxelWolf
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June 18th, 2016 at 7:49:03 PM permalink
So is the reason people want to make the argument that infinite time and money using martingale or whatever will work is so they can somehow claim that it can work with less time and money as well?
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
gamerfreak
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June 18th, 2016 at 8:39:11 PM permalink
Here is a recap of this entire thread to date, for whoever is interested:

https://www.youtube.com/watch?v=Nfc7G4xKsl8
DiscreteMaths2
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June 18th, 2016 at 8:43:37 PM permalink
Quote: AxelWolf

So is the reason people want to make the argument that infinite time and money using martingale or whatever will work is so they can somehow claim that it can work with less time and money as well?



No basically it was said in passing that a betting system would have positive EV if they had infinite bankroll, limits, time, etc. I challenged that and said no, not even under those conditions would it have positive EV. I am pretty sure everyone is in agreement that this no bearing in real world gambling application, we are just discussing what it means if it could theoretically happen.
Assume the worst, believe no one, and make your move only when you are certain that you are unbeatable or have, at worst, exceptionally good odds in your favor.
DiscreteMaths2
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June 18th, 2016 at 8:53:57 PM permalink
Quote: MrGoldenSun



You are misunderstanding this a bit. The calculation is the sum from n=1 to infinity of (winnings for a series terminating on spin n)*(probability of ending on spin n)

PeeMcGee posted a claim the EV is 1 in which he defined the random variable and the probabilities and then calculated the expected value of that variable. I think he is correct. For those of you saying he is wrong, where is the flaw in his argument?

My only question is if the probabilities can be defined the way we have done it. I don't know if the probability density function is problematic in some way.



Expected value for a discrete random variable is a weighted average of ALL possible outcomes. You don't get to ignore certain outcomes and still call it EV.
Assume the worst, believe no one, and make your move only when you are certain that you are unbeatable or have, at worst, exceptionally good odds in your favor.
MrGoldenSun
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June 18th, 2016 at 9:34:39 PM permalink
Quote: DiscreteMaths2

No basically it was said in passing that a betting system would have positive EV if they had infinite bankroll, limits, time, etc. I challenged that and said no, not even under those conditions would it have positive EV. I am pretty sure everyone is in agreement that this no bearing in real world gambling application, we are just discussing what it means if it could theoretically happen.



This is accurate. No one here thinks the martingale is positive expectation in the real world. It's a pure thought exercise.
MrGoldenSun
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June 18th, 2016 at 9:36:30 PM permalink
Quote: DiscreteMaths2

Expected value for a discrete random variable is a weighted average of ALL possible outcomes. You don't get to ignore certain outcomes and still call it EV.



Okay, what outcome do you think has been ignored?
PeeMcGee
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June 18th, 2016 at 11:30:47 PM permalink
Quote: Ibeatyouraces

Quote: PeeMcGee

...It can be difficult to grasp…but it is in fact zero. In fact it is…. Which is zero...


I'll say this again. You are incorrect. If something has a probability of even 0.0000(insert 5,000,000 quadrillion quadrillion quadrillion more zeros)000001% chance of happening, it's still NOT zero.


We are going to keep going back and forth on this …but that is zero. The crap in the brackets is equal to 0.999 repeating—which is equal to 1 (there are plenty of proofs out there if interested).

Quote: RS

To me, it appears the EV only decreases with the more wagers you make.


Yes, very true. I said that a few pages ago. The infinite Martingale does not work for any finite number of bets. But something interesting happens when you can make infinitely many bets—the probability of going bankrupt becomes exactly 0. You literally cannot lose. Which is why you don’t see it in the expected value calculation.


Quote: RS

If ya wanna say something like, "Martingale wins over an infinite time with infinite bankroll etc. as long as there is a winning spin [and you quit after the winning spin]" then go ahead, but that's just as stupid as saying, "My system wins if I win!" or "Ya can't lose if you walk away a winner!"


So yea, it is kind of like saying that. I would maybe phrase it “The system wins, because it cannot lose”. And it is kind of a silly thing to say—which is kind of how I meant it to be when I first said it. But there are concepts and math behind it that are extremely important. I had no idea it was going to be this controversial of an idea. Now, I’m just mind blown at the unfamiliarity towards these concepts.

Quote:

Quote: DiscreteMaths2

Because you are leaving out where you lose money. First spin is 1*(18/38) - 1(20/38) not just 1*(18/38).


Quote: RS

I don't know what the exact definition of EV is, but I would say it's something like this:

Prob(win) * payout - Prob(loss) * loss_amount = EV


I see no prob(L)*Loss_Amount in the formulas that "show" the EV equals 1.


Quote: DiscreteMaths2

Expected value for a discrete random variable is a weighted average of ALL possible outcomes. You don't get to ignore certain outcomes and still call it EV.


I touched on this, but let me add more clarity maybe. My calculation is based on a single round of Martingale. Normally, a Martingale round terminates in one of two ways: (1)by a win or (2)by bankruptcy of the bettor.

So you would see ev calculated like
(1-P[of going bankrupt])*1 – P[of going bankrupt]*bankroll
And that is your Prob(win)*payout – Prob(loss)*loss_amount that you so badly want to see.

But as mentioned, the probability of going bankrupt is zero with infinite money. I have shown this mathematically and tried to explained it logically. I mean, really, it just makes sense. If you don’t believe that the probability is zero…then please explain how one would go bankrupt with infinite money.

So with infinite money and time, the round only terminates on a winning spin. So what you see in my calculation is the probability that the round will terminate on exactly spin n (this is pn) multiplied by the amount of money won or loss on spin n (this is wn). Then all of the possible products are added together.

Yes, my math accounts for ALL possible outcomes. If you add up all of the pn's you get 1. In other words, if you add up all of the probabilities of terminating on exactly spin n for every single n…you get 1. The probability space is covered. There are no outcomes being ignored.

Quote: AxelWolf

So is the reason people want to make the argument that infinite time and money using martingale or whatever will work is so they can somehow claim that it can work with less time and money as well?


I can promise you that is nowhere near my motive. I will reiterate, this works if and only if one has infinite time and money. And like RS have shown, the closer you get to infinite time and money the more you are going to lose. I visit these forums as a mathematician, not a gambler.
AxelWolf
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June 18th, 2016 at 11:41:51 PM permalink
Quote: PeeMcGee

Quote: Ibeatyouraces

Quote: PeeMcGee

...It can be difficult to grasp…but it is in fact zero. In fact it is…. Which is zero...


I'll say this again. You are incorrect. If something has a probability of even 0.0000(insert 5,000,000 quadrillion quadrillion quadrillion more zeros)000001% chance of happening, it's still NOT zero.


We are going to keep going back and forth on this …but that is zero. The crap in the brackets is equal to 0.999 repeating—which is equal to 1 (there are plenty of proofs out there if interested).


Yes, very true. I said that a few pages ago. The infinite Martingale does not work for any finite number of bets. But something interesting happens when you can make infinitely many bets—the probability of going bankrupt becomes exactly 0. You literally cannot lose. Which is why you don’t see it in the expected value calculation.


Quote: RS

If ya wanna say something like, "Martingale wins over an infinite time with infinite bankroll etc. as long as there is a winning spin [and you quit after the winning spin]" then go ahead, but that's just as stupid as saying, "My system wins if I win!" or "Ya can't lose if you walk away a winner!"


So yea, it is kind of like saying that. I would maybe phrase it “The system wins, because it cannot lose”. And it is kind of a silly thing to say—which is kind of how I meant it to be when I first said it. But there are concepts and math behind it that are extremely important. I had no idea it was going to be this controversial of an idea. Now, I’m just mind blown at the ignorance towards these concepts.



I touched on this, but let me add more clarity maybe. My calculation is based on a single round of Martingale. Normally, a Martingale round terminates in one of two ways: (1)by a win or (2)by bankruptcy of the bettor.

So you would see ev calculated like
(1-P[of going bankrupt])*1 – P[of going bankrupt]*bankroll
And that is your Prob(win)*payout – Prob(loss)*loss_amount that you so badly want to see.

But as mentioned, the probability of going bankrupt is zero with infinite money. I have shown this mathematically and tried to explained it logically. I mean, really, it just makes sense. If you don’t believe that the probability is zero…then please explain how one would go bankrupt with infinite money.

So with infinite money and time, the round only terminates on a winning spin. So what you see in my calculation is the probability that the round will terminate on exactly spin n (this is pn) multiplied by the amount of money won or loss on spin n (this is wn). Then all of the possible products are added together.

Yes, my math accounts for ALL possible outcomes. If you add up all of the pn's you get 1. In other words, if you add up all of the probabilities of terminating on exactly spin n for every single n…you get 1. There are no outcomes being ignored.


I can promise you that is nowhere near my motive. I will reiterate, this works if and only if one has infinite time and money. And like RS have shown, the closer you get to infinite time and money the more you are going to lose. I visit these forums as a mathematician, not a gambler.

I was not referring to you specifically, hell I don't even know who brought it up this time. Others(good movie BTW) have mentioned this in the past and I suspect they were using it as a pro argument for betting systems.
-------------------------------------------------------------------------------------
I think it would be interesting to follow around and document a martingaler or system bettor for a few weeks.
Last edited by: AxelWolf on Jun 19, 2016
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
RS
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June 19th, 2016 at 1:50:30 AM permalink
I disagree about using P(L) = 0 for this equation. P(L) = (20/38)^N for N rounds in a trial.
TwoFeathersATL
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June 19th, 2016 at 3:40:40 AM permalink
Quote: DiscreteMaths2

. I am pretty sure everyone is in agreement that this no bearing in real world gambling application, we are just discussing what it means if it could theoretically happen.

People are strange.
Youuuuuu MIGHT be a 'rascal' if.......(nevermind ;-)...2F
TwoFeathersATL
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June 19th, 2016 at 3:43:13 AM permalink
Quote: RS

I disagree about using P(L) = 0 for this equation. P(L) = (20/38)^N for N rounds in a trial.

Some people are strange AND disagreable ;-)
Youuuuuu MIGHT be a 'rascal' if.......(nevermind ;-)...2F
PeeMcGee
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June 19th, 2016 at 7:58:19 AM permalink
Quote: RS

I disagree about using P(L) = 0 for this equation. P(L) = (20/38)^N for N rounds in a trial.


Please explain how it is possible to deplete an infinite bankroll?
OnceDear
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June 19th, 2016 at 8:36:19 AM permalink
Quote: PeeMcGee

Please explain how it is possible to deplete an infinite bankroll?


Please explain how it is possible to HAVE an infinite bankroll?
or maybe
Please explain how it is possible to increase an infinite bankroll by 1?


An infinite bankroll cannot exist, so this whole thread has been led down a blind alley.
Also PMG.... When you set off with your infinite bankroll intent on Martying until you win, HOW do you know you will win. Your infinite series seems to suggest that you KNOW you will win, but you might be infinitely unlucky and not win even after all eternity has passed. What's ev then?
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
PeeMcGee
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June 19th, 2016 at 9:46:14 AM permalink
Quote: OnceDear

Please explain how it is possible to HAVE an infinite bankroll?


I have said it so many times now—infinite bankroll and time is not possible. But that’s not the question we are asking. This is a hypothetical exercise.

Quote: OnceDear

or maybe
Please explain how it is possible to increase an infinite bankroll by 1?


I have addressed this multiple times as well. The purchasing power of the bankroll has no influence on the expected value of the game.

Quote: OnceDear

An infinite bankroll cannot exist, so this whole thread has been led down a blind alley.
Also PMG.... When you set off with your infinite bankroll intent on Martying until you win, HOW do you know you will win. Your infinite series seems to suggest that you KNOW you will win, but you might be infinitely unlucky and not win even after all eternity has passed. What's ev then?


Well first, assuming infinite time means eternity will not ‘has passed’.
Second, that would simply mean the bet is still in progress/not resolved.
When calculating the expected value of a normal roulette spin, do you include the time when the wheel is spinning?
RS
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June 19th, 2016 at 3:21:57 PM permalink
Quote: PeeMcGee

Please explain how it is possible to deplete an infinite bankroll?



Seriously? You're twisting all this around. Probability of a loss is (20/38)^INF. This is NOT 0, but approaches 0. Use this in the formula I gave before. Overall, the EV is -INF (big surprise, matches with theory, of -5.26%). You're taking numbers that approach 0 or 1 and then saying they are 0 or 1 when there's more math to be done.
TomG
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June 19th, 2016 at 4:49:31 PM permalink
Quote: PeeMcGee

Yes, very true. I said that a few pages ago. The infinite Martingale does not work for any finite number of bets. But something interesting happens when you can make infinitely many bets—the probability of going bankrupt becomes exactly 0.



With an infinite bankroll and a finite number of bets, the probability of bankruptcy is also zero. So it really isn't all that interesting.

What is interesting is that everyone agrees you can't have an infinite bankroll, but still allow themselves to pretend or imagine one; yet we can't also imagine losing an infinite amount of times in a row? If you play for in infinite amount of time you will have an infinite amount of losses (along with an infinite amount of wins), why can't they all be consecutive at some point?

What's really interesting is that EV of each bet is -0.027% but EV of a finite series is 1.00? (Ending after a win is finite, not infinite). Even google doesn't give much help in answering these paradoxes.
PeeMcGee
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June 20th, 2016 at 5:59:13 AM permalink
Quote: RS

Seriously? You're twisting all this around. Probability of a loss is (20/38)^INF. This is NOT 0, but approaches 0. Use this in the formula I gave before. Overall, the EV is -INF (big surprise, matches with theory, of -5.26%). You're taking numbers that approach 0 or 1 and then saying they are 0 or 1 when there's more math to be done.


The limit of the total expected value, as the size of the bankroll increases, diverges to negative infinity. You show this with your 1000 row table. No disagreement there.

Your mistake is thinking 'infinite bankroll' simply means 'really big bankroll'. You are still using infinity as a number—while it is more of a conceptual idea. You, and many others, are completely ignoring what it means to have an infinite bankroll.

Infinite bankroll simply means you cannot run out money.

When you have such a bankroll, there is a discontinuity in the expected value as it is equal to +1. Limits do not have to equal the value of the function.

So again, I ask, please explain how it is possible to deplete an infinite bankroll?


Quote: TomG

With an infinite bankroll and a finite number of bets, the probability of bankruptcy is also zero. So it really isn't all that interesting.

What is interesting is that everyone agrees you can't have an infinite bankroll, but still allow themselves to pretend or imagine one; yet we can't also imagine losing an infinite amount of times in a row? If you play for in infinite amount of time you will have an infinite amount of losses (along with an infinite amount of wins), why can't they all be consecutive at some point?

What's really interesting is that EV of each bet is -0.027% but EV of a finite series is 1.00? (Ending after a win is finite, not infinite). Even google doesn't give much help in answering these paradoxes.


I have no qualms with saying that the wheel can infinitely spin one way. That would not deplete the bankroll though. The bet will still be in progress/forever ongoing…meaning no outcome to the bet. Hence, mathematically, this has exactly zero probability.
Last edited by: PeeMcGee on Jun 20, 2016
RS
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June 21st, 2016 at 8:11:59 PM permalink
Quote: PeeMcGee

The limit of the total expected value, as the size of the bankroll increases, diverges to negative infinity. You show this with your 1000 row table. No disagreement there.

Your mistake is thinking 'infinite bankroll' simply means 'really big bankroll'. You are still using infinity as a number—while it is more of a conceptual idea. You, and many others, are completely ignoring what it means to have an infinite bankroll.

Infinite bankroll simply means you cannot run out money.

When you have such a bankroll, there is a discontinuity in the expected value as it is equal to +1. Limits do not have to equal the value of the function.

So again, I ask, please explain how it is possible to deplete an infinite bankroll?



I have no qualms with saying that the wheel can infinitely spin one way. That would not deplete the bankroll though. The bet will still be in progress/forever ongoing…meaning no outcome to the bet. Hence, mathematically, this has exactly zero probability.



I never said or meant to say the BR could be depleted. I'm saying the EV is negative. I could play a -EV game (9/6 JOB straight up) with an infinite bankroll, and my bankroll would not be depleted. But, the game is still -EV.
PeeMcGee
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June 21st, 2016 at 9:28:34 PM permalink
Quote: RS

Quote: PeeMcGee

The limit of the total expected value, as the size of the bankroll increases, diverges to negative infinity. You show this with your 1000 row table. No disagreement there.

Your mistake is thinking 'infinite bankroll' simply means 'really big bankroll'. You are still using infinity as a number—while it is more of a conceptual idea. You, and many others, are completely ignoring what it means to have an infinite bankroll.

Infinite bankroll simply means you cannot run out money.

When you have such a bankroll, there is a discontinuity in the expected value as it is equal to +1. Limits do not have to equal the value of the function.

So again, I ask, please explain how it is possible to deplete an infinite bankroll?



I have no qualms with saying that the wheel can infinitely spin one way. That would not deplete the bankroll though. The bet will still be in progress/forever ongoing…meaning no outcome to the bet. Hence, mathematically, this has exactly zero probability.



I never said or meant to say the BR could be depleted. I'm saying the EV is negative. I could play a -EV game (9/6 JOB straight up) with an infinite bankroll, and my bankroll would not be depleted. But, the game is still -EV.


The difference is, such a game will not guarantee a profit when you win.
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