protosapien
protosapien
Joined: Nov 19, 2019
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November 19th, 2019 at 6:30:24 AM permalink
Came across this strategy that sparked my interest and when I break it down, the math seems to work, but figured I'd ask the wizards. This is assuming a true 50/50.

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)

You win 32,767 base bets of 1 by the time
15x losses statistically happens

you will lose 13,312 due to 6th tier wins
you will lose 6,144 due to 7th tier wins
you will lose 2,304 due to 8th tier wins
you will lose 1,152 due to 9th tier wins
you will lose 2,176 due to 10th tier wins
you will lose 928 due to 11th tier wins
you will lose 1,264 due to 12th tier wins
you will lose 1,032 due to 13th tier wins
you will lose 716 due to 14th tier wins
you will lose 2,088 due to 15th tier loss and 50% of the time you will lose 1,388 if you have a 15th tier win

32,767-31,116=1,651

LEGEND
0.00 = odds
bXX = bet amount
1x = amount of times this occur before 15x
-XX= total loss of streak
(=/-)= sum or difference at end of streak


The numbers here are rough and don't account for decimal places.

Also, do you have to account for the 1,023 times that you wont win your base unit of 1? If so, it brings your expected value to 628 units every 32,767 bets.

If it does work, my small brain imagines that some of the (b) sizes can be tweaked to attain maximum value.
sabre
sabre
Joined: Aug 16, 2010
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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
protosapien
Joined: Nov 19, 2019
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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
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
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
Mission146
Joined: May 15, 2012
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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.
Vultures can't be choosers.
protosapien
protosapien
Joined: Nov 19, 2019
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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
Mission146
Joined: May 15, 2012
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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.

Youíre 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 thatís meaningless. The first bet by itself also cannot have an actual result of zero. Doesnít change expected result.
Vultures can't be choosers.
protosapien
protosapien
Joined: Nov 19, 2019
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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
sabre
Joined: Aug 16, 2010
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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

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