Math Seems to Work

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November 19th, 2019 at 8:20:45 AM permalink

Any series of 50/50 bets that pay even money will result in an expectation of 0. You can flat bet, martingale, reverse martingale, Fosbury or try the Double Fanucci. If your math says otherwise, you've made a mistake.

protosapien

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November 19th, 2019 at 8:22:53 AM permalink

Would you kindly find the error for me?

This is not my strategy, I found it posted here on a thread hidden somewhere, I'll link it later.

But I did break it down, and I can't seem to find where the error is.

sabre

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November 19th, 2019 at 8:31:53 AM permalink

No, I'm not searching through a post with 50 numbers in it to find a math error. A series of 50/50 bets that pays even money will have an expectation of 0. Period.

protosapien

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November 19th, 2019 at 8:37:16 AM permalink

Quote: sabre
No, I'm not searching through a post with 50 numbers in it to find a math error. A series of 50/50 bets that pays even money will have an expectation of 0. Period.

I don't believe that you can ethically make that statement without analyzing the data, so the only thing incorrect at this point in time is your posts.

Thank you for the input though.

By the way,

If you have a series of 2 bets and the sizes are not the same, the expectation will never be 0.

Mission146

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November 19th, 2019 at 8:40:52 AM permalink

Quote: protosapien
I don't believe that you can ethically make that statement without analyzing the data, so the only thing incorrect at this point in time is your posts.

Thank you for the input though.

Can one ethically state that 2 + 2 = 4 without knowing two of what provided the items are identical?

50/50, even money, long run expectation of zero. Thank you for the input though.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

protosapien

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November 19th, 2019 at 8:41:53 AM permalink

Quote: Mission146
Can one ethically state that 2 + 2 = 4 without knowing two of what provided the items are identical?

50/50, even money, long run expectation of zero. Thank you for the input though.

Please refer to edited post.

Mission146

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November 19th, 2019 at 8:43:09 AM permalink

Quote: protosapien
Please refer to edited post.

If one 50/50 bet paying even money has an expectation of zero, then they all do.

You�re conflating expectation with actual results. Actual results would say that two 50/50 bets where one unit is bet the first time and two units are bet the second time cannot equal zero...but that�s meaningless. The first bet by itself also cannot have an actual result of zero. Doesn�t change expected result.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

protosapien

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November 19th, 2019 at 8:46:39 AM permalink

Any series of 50/50 bets with different bet sizes will never have an expectation of 0.

The series listed in OP caps at 15 bets.

I have checked the math and can't find an error.

I've seen this statement made by a hundred people, so if you want to explain further, or find the error in the math, then please do so.

Otherwise, please refrain from debunking a string of numbers without analyzing it.

I didn't post for people to look at this the same way it has been looked at countless times before.

sabre

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November 19th, 2019 at 8:49:16 AM permalink

Quote: protosapien
Any series of 50/50 bets with different bet sizes can only have an expectation of 0.

I fixed that for you

Mission146

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November 19th, 2019 at 8:51:23 AM permalink

Every 50/50 bet returning even money has an expectation of zero, for God�s sake, it just can�t have an actual result of zero.

Again, you�re talking about actual results. A Martingale on an even money proposition cannot have an actual result of zero, doesn�t change anything. A Martingale with an infinite bankroll also results in only one of two things: Winning one unit or playing for all of eternity.

Anyway, I don�t need to look at it to know the expected result is zero. But, I might look at it later when I�m on a computer.

If you�re so sure of it, quit talking to me and go get rich somewhere.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

protosapien

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November 19th, 2019 at 8:51:42 AM permalink

If the series is limited to 2 bets,

1 bet is 1

1 bet is 2

The outcomes are +3, +1, -1, or -3.

I don't get 0 out of this.

The series in OP is capped at 15 and compares the number of times a streak is expected to happen vs the amounts bet when it does happen.

rsactuary

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November 19th, 2019 at 8:54:39 AM permalink

Quote: protosapien
If the series is limited to 2 bets,

1 bet is 1

1 bet is 2

The outcomes are +3, +1, -1, or -3.

I don't get 0 out of this.

The series in OP is capped at 15 and compares the number of times a streak is expected to happen vs the amounts bet when it does happen.

Expectation = 0.25 * 3 + .5* 1 + .5* -1 + .25* -3 = 0 QED

Mission146

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November 19th, 2019 at 8:57:22 AM permalink

Quote: protosapien
If the series is limited to 2 bets,

1 bet is 1

1 bet is 2

The outcomes are +3, +1, -1, or -3.

I don't get 0 out of this.

The series in OP is capped at 15 and compares the number of times a streak is expected to happen vs the amounts bet when it does happen.

If you can�t differentiate actual from expected result, I�m done here. Again, you don�t need two bets, ONE 50/50 bet at even money CANNOT have an actual result of zero.

(.50 * 1) - (.50 * 1) = 0

(.50 * 1) + (.50 * -1) = 0

Expectation of 0. Always and forever. Even on Tuesdays, sorry.

What do you get when you add:

0+0+0+0+0+0+0+0+0+0+0+0+0+0+0 = x

Let me know if you want a hint.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

protosapien

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November 19th, 2019 at 9:00:07 AM permalink

I'm not analyzing whether someone can win more than 50% of bets.

I'm analyzing the impact of sizing in comparison to the expected occurrence of streaks.

Mission146

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November 19th, 2019 at 9:03:22 AM permalink

Quote: protosapien
I'm not analyzing whether someone can win more than 50% of bets.

I'm analyzing the impact of sizing in comparison to the expected occurrence of streaks.

The impact of sizing changes the distribution of your results, which all have an expectation of zero, thereby resulting in a total expectation of 0. If your math says otherwise, then your math is wrong.

I never said anything about winning more than, less than or precisely 50% of the time.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

Mission146

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November 19th, 2019 at 10:15:56 AM permalink

Okay, what we are going to do is greatly simplify the problem. The first assumption is that we are going to see the system through to conclusion, so no taking a positive result and leaving. We are going to simplify this by looking at the probability of each net outcome and then multiplying it by the result of each net outcome. It appears that the system starts over upon any win, so here we go:

Quote: protosapien

1- 0.5000000 b1 (+1)
2- 0.2500000 b2 = -3 (+1)
3- 0.1250000 b4 =-7 (+1)
4- 0.0625000 b8 =-15 (+1)
5- 0.0312500 b16 = -31 (+1)
6- 0.0156250 b5 512x = -36 (-26)
7- 0.0078125 b12 256x = -48 (-24)
8- 0.0039063 b30 128x = -78 (-18)
9- 0.0019531 b60 64x = -138 (-18)
10- 0.0009766 b70 32x = -208 (-68)
11- 0.0004883 b150 16x = -358 (-58)
12- 0.0002441 b200 8x = -558 (-158)
13- 0.0001221 b300 4x = -888 (-258)
14- 0.0000610 b500 2x = -1388 (-358)
15- 0.0000305 b700 = (-2088)

(.5) * 1 = .5
(.5^2) * 1 = .25
(.5^3) * 1 = .125
(.5^4) * 1 = .0625
(.5^5) * 1 = .03125
(.5^6) * (-26) = -0.40625
(.5^7) * (-24) = -0.1875
(.5^8) * (-18) = -0.0703125
(.5^9) * (-18) = -0.03515625
(.5^10) * (-68) = -0.06640625
(.5^11) * (-58) = -0.0283203125
(.5^12) * (-158) = -0.03857421875
(.5^13) * (-258) = -0.03149414062
(.5^14) * (-358) = -0.02185058593
(.5^15) * (-2058) = -0.06280517578
(.5^15) * (-658) = -.0200805664

(.5+.25+.125+.0625+.03125) - (.40625+.1875+.0703125+.03515625+.06640625+.0283203125+.03857421875+.03149414062+.02185058593+0.06280517578+0.0200805664) = 2.0000002e-11

In other words, 0.000000000020000002 due to rounding errors.

In other words, zero, but I already knew that, didn't I?

Also, improve your sales pitch. A loss on the fifteenth attempt loses 2,058, not 2,088. A win on the fifteenth tier is a net loss of 658, not 1358.

You forgot there are two possible Tier 15 results, as did whoever is peddling this particular brand of snake oil.

Thank you for playing and enjoy the rest of your day.

Last edited by: Mission146 on Nov 19, 2019

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

protosapien

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November 19th, 2019 at 10:20:27 AM permalink

Quote: Mission146

Also, improve your sales pitch. A loss on the fifteenth attempt loses 2,058, not 2,088. A win on the fifteenth tier is a net loss of 688, not 1388.

Thank you for playing and enjoy the rest of your day.

The bracketed number is your net win/loss at the end of the streak. The running negative total during a loss streak is listed left of the bracketed numbers.

Also the formula you used only accounts for the expected win/loss value, and doesn't evaluate the amount of times that your bet sizing will occur before a 15x loss streak is reached.

Lastly, I did state that the 15th round would win 50% and lose 50%, as was described in my breakdown directly underneath the set of numbers.

Mission146

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November 19th, 2019 at 10:34:50 AM permalink

Quote: protosapien
The bracketed number is your net win/loss at the end of the streak. The running negative total during a loss streak is listed left of the bracketed numbers.

Also the formula you used only accounts for the expected win/loss value, and doesn't evaluate the amount of times that your bet sizing will occur before a 15x loss streak is reached.

Lastly, I did state that the 15th round would win 50% and lose 50%, as was described in my breakdown directly underneath the set of numbers.

Can you not subtract? -1358 +-700 = -2058, not negative 2,088. The total bets do not add up to 2,088; they add up to 2058. Using 2088 as a potential loss actually makes your system look negative (as opposed to zero) which is why this took me so long. Originally, I just copied the numbers. I assumed the basic addition and subtraction in the opening post was correct...upon further evaluation, it was not. My apologies for making the assumption.

1+2+4+8+16+5+12+30+60+70+150+200+300+500+700 = 2,058

Not 2,088.

Should I show why it is not negative 2,088 in fifteen more different ways, or will this suffice?

The formula I used takes into account the probability of every single net result assuming the system starts over upon a win. Because of this, the number of times blah, blah, blah happens before blah, blah, blah, is already factored in. You would be better to come up with a system that is not fully self-contained, at least Martingale advocates can cling to their argument that the Martingale can never truly lose with an unlimited bankroll and unlimited time...it can just continue forever.

Lastly, you did do that! You just forgot that convenient little part where you list all of those possibilities in your list of possible results. Step 15 is terminal one way or the other, both terminations have a negative net result, but the initial table only lists one possibility of the two...both of which are equally likely.

I'll tell you what you can do instead: Just don't list any of the results that have a negative net effect. System will look really good then.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

protosapien

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November 19th, 2019 at 10:41:04 AM permalink

You're right, I did make a mistake. It should be 2058.

Why do I come up with a number that will cover the losses of a 15x loss streak by the time it is expected to happen?

Mission146

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November 19th, 2019 at 10:42:29 AM permalink

Quote: protosapien
You're right, I did make a mistake. It should be 2058.

Why do I come up with a number that will cover the losses of a 15x loss streak by the time it is expected to happen?

I don't know or care. I've analyzed the entire system, which is self-contained, and the expectation is zero. The number of times it succeeds before failing is irrelevant. A fifteen-step Martingale would succeed (profit v. loss) more frequently than your system does. The good news is, on a 50/50 even money bet, your system is no less effective. Of course, no system is any less effective. Or any more effective.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

protosapien

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November 19th, 2019 at 10:44:11 AM permalink

What the Martingale does not do is account for a finite bankroll.

Mission146

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November 19th, 2019 at 10:45:26 AM permalink

Ssshhh...don't tell them that, you'll hurt their feelings.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

unJon

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November 19th, 2019 at 10:51:37 AM permalink

Mission your patience here is admirable.

Last edited by: unJon on Nov 19, 2019

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

gordonm888
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November 19th, 2019 at 12:35:38 PM permalink

Let me first restate your table and correct a minor math error on the net loss on tiers 13-15 (which does not affect your argument.) I have also written the tier 15 case as two lines -one where you win the 700 unit bet (after 14 prior losses) and one where you lose the 700 unit bet. Your analysis ignored the scenario where you win that Tier 15 bet (which is still a net loss of 658 units) -that is one of your your principal errors.

1- 0.5000000 b1 16384x = -1 (+1)
2- 0.2500000 b2 8192x = -3 (+1)
3- 0.1250000 b4 4096x =-7 (+1)
4- 0.0625000 b8 2048x =-15 (+1)
5- 0.0312500 b16 1024x = -31 (+1)
6- 0.0156250 b5 512x = -36 (-26)
7- 0.0078125 b12 256x = -48 (-24)
8- 0.0039063 b30 128x = -78 (-18)
9- 0.0019531 b60 64x = -138 (-18)
10- 0.0009766 b70 32x = -208 (-68)
11- 0.0004883 b150 16x = -358 (-58)
12- 0.0002441 b200 8x = -558 (-158)
13- 0.0001221 b300 4x = -858 (-258)
14- 0.0000610 b500 2x = -1358 (-358)
15L- 0.0000305 b700 = (-2058)
15W- 0.0000305 b700 = (-658)

Now I will show this information in a different format and with an additional column: "Return "which is defined as the probability of each Tier scenario multiplied by the payoff from from teh given scenario.

Tier (scenario)	Prob, P	Payout	Return, P x Payout
1	0.5	+1	0.5
2	0.25	+1	0.25
3	0.125	+1	0.125
4	0.0625	+1	0.0625
5	0.03125	+1	0.03125
6	0.015625	-26	-0.40625
7	0.0078125	-24	-0.1875
8	0.00390625	-18	-0.0703125
9	0.001953125	-18	-0.03515625
10	0.000976563	-68	-0.06640625
11	0.000488281	-58	-0.028320313
12	0.000244141	-158	-0.038574219
13	0.00012207	-258	-0.031494141
14	6.10352E-05	-358	-0.021850586
15L	3.05176E-05	-2058	-0.062805176
15W	3.05176E-05	-658	-0.02008
Total	1		0

Note that the total return, the summation of all the numbers in the return column, is zero. So, summing over all scenarios, this betting system neither wins or loses money; this was expected because every wager is a 50/50 proposition.

Edit: Just realized that Prof. Mission has already posted an analysis which appears to be identical to mine. My post may have some formatting that makes it easier to follow.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

OnceDear

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November 19th, 2019 at 3:09:59 PM permalink

Quote: protosapien
If the series is limited to 2 bets,

1 bet is 1

1 bet is 2

The outcomes are +3, +1, -1, or -3.

I don't get 0 out of this.

Average(+3, 1, -1, -3) = (3 + 1 - 1 - 3) / 4 = 0 / 4 =0
QED.

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.

RS

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November 20th, 2019 at 2:46:41 AM permalink

What even is this thread? If I'm understanding this correctly, he's using a progressive system on a 50/50 wager that pays even money. And he's saying the EV is something other than 0. Is that correct? If so, why even bother responding (beyond sabre's initial response)? If not, can someone clarify what he's saying?

Mission146

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November 20th, 2019 at 3:10:45 AM permalink

Debunker�s gonna debunk.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

DeMango

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November 20th, 2019 at 9:54:24 AM permalink

Quote: Mission146
Debunker�s gonna debunk.

Only reason to follow this thread is for the snarky comments!

When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.

michael99000

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November 20th, 2019 at 1:23:07 PM permalink

It�s a classic WOV battle.

The martingaler vs the debunkers.

Martingaling reality smacks you in the face when you�re shoving that $2088 bet into the circle, and you realize the fact that you�ve lost the previous 14 bets makes this one no more likely to be a winner.

You just cannot replicate that empty feeling beforehand with a nicely organized chart.

lilredrooster

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November 21st, 2019 at 10:16:32 AM permalink

if you do a reverse martingale on a coin flip - betting heads or tails - no house edge - and you continue until you lose - you will always lose money

mathematics says it's zero sum - you have an expectation of winning nothing and losing nothing

but in reality you will always lose........................

i.e...............:

$100.... win....$200......win........$400......win.......$800.......win..........$1600................win............$3200...............lose................................

net loss of ($100)

the foolish sayings of a rich man often pass for words of wisdom by the fools around him

DeMango

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November 21st, 2019 at 12:52:06 PM permalink

Well the standard answer to combat Martingale is table limits. Why would this not apply here? Win $3200, table limit $3000. Gosh, gee, gotta take the money and run!

When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.

RS

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November 21st, 2019 at 1:12:55 PM permalink

Quote: lilredrooster
if you do a reverse martingale on a coin flip - betting heads or tails - no house edge - and you continue until you lose - you will always lose money

mathematics says it's zero sum - you have an expectation of winning nothing and losing nothing

but in reality you will always lose........................

i.e...............:

$100.... win....$200......win........$400......win.......$800.......win..........$1600................win............$3200...............lose................................

net loss of ($100)

Maybe my math is wrong, but isn't the following true?


lim x->infinity [ (1/2)^x ] * [2^x - 1] - [ 1 - (1/2)^x ] = 0

sabre

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November 21st, 2019 at 3:21:11 PM permalink

I love it when people come in with a stupid system they think is going to break the bank. They expect adulation and praise. They get cold facts that their system is useless. They demand that you go through their badly formatted garbage post and find their math error. Then when someone inevitably does, they go radio silent. Not even an "oops, my bad".

Mission146