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Old 08-08-2018, 02:36 PM   #1
SiberianPIMP
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Serious of -EV to +EV sum?

Hi guys. Please tell what you think about this. Is it correct? Is it wrong? Why? All comments are much appriciated. Criticism also welcomed.

COIN FLIP GAME

Rules of a game:

- Flipping a coin

- We do not have infinity amount of units to bet

- Our lowest vs highest bet is 20x

- We decide about our next pick:H or T, after last one falls

- If we do make correct prediction for next H/T we get +0,999 unit

- If we do not make correct prediction for next T/H we lose 1 unit

- Infinity number of flips

First some calculations:

Number of trials:4
Number of combinations:2 on 4=16
All possible combinations( H: Heads; T: Tails)
1.HHHH
2.THHH
3. HTHH
4. HHTH
5. HHHT
6. TTTT
7. HTTT
8. THTT
9. TTHT
10. TTTH
11. HHTT
12. HTHT
13. HTTH
14. THTH
15. TTHH
16. THHT

We bet all the combinations that has as a result 2H+2T, HHHH and TTTT. There are 8 combinations like this, mentioned above (combinations with numbers: 1,6,11,12,13,14,15,16).

How do we bet:


We repeat that to infinity. If we lose before 4th bet, we wait until 5th bet and repeat as mentioned.

More calculations:
N=4
u= 4*(0,5)=2; 0,5 because it is coin flip; u is mean
o2=4*(0,5)*(1,0-0,5)=1; 0,5 because it is coin flip
o=square root 1=1

Let us now take 0,5o and 1,5o.
0,5o=0,382924; rounded on 6th decimal; not in favour of this theory
1,5o=0,866386; rounded on 6th decimal; not in favour of this theory
1-0,866386=0,133614

We are interested in u+-0,5o. This in our case means from 1,5 to 2,5 (2 is mean).
We are also interested in +-1,5o. This in our case means from 0,5 to 3,5 (2 is mean).

If we would have odds 2 for every winning prediction we would need to hit our system in at least:
winning combinations/all combinations
8/16=0,5=50 % of the time.
The numbers from SD calculations tells us that we will hit this system in: 38,2924% + 13,3614 %=51,6538 % of time.
51,6538 % is more than 50 %.

Therefor, we made a +EV bet out of only 0EV bets.

Now, let us calculate EV in our case from the example (with odds 1,999).
In our case we have odds 1,999.

That means that every of our 8 combinations has got odds:
1,999 on 4=15,968; rounded on 3 decimal; not in favour of this system
Together our system has got odds: 15.968
We calculated before, that we will win our system in 51.6538 % of time.
15,968 x 0,516538=8,2480; rounded on 4 decimal;not in favour of our system;
8,2480 is more than 8.

In every 4 flips we make profit of 8,2480 - 8= 0,248 units.
That is profit per flip: 0,248/4=0,062 .
Profit per every flip =0,062/8=0,755%.
The more time we play this game the more units we have (if we are betting like I wrote).

Therefor, we made a +EV bet out of only -EV bets.

If above calculation is correct. This means that those 2 sentences are wrong:


- The EV of the sum of random variables is always equal to the sum of the EVs, whether the variables are independent or not (for example in the same game).

- Optimal strategy is allways the best strategy.

Last edited by SiberianPIMP; 08-08-2018 at 02:46 PM.
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Old 08-08-2018, 05:39 PM   #2
NewOldGuy
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Re: Serious of -EV to +EV sum?

SMH
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Old 08-09-2018, 01:14 AM   #3
RustyBrooks
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Re: Serious of -EV to +EV sum?

I don't actually follow how the betting works, not even with looking at your spreadsheet.

I am not sure what significance you're placing on +/-.5 and 1.5 standard deviations, or why you're interested in them. That might make sense if you were betting on something with a continuous distribution, where you win with those given ranges, but you aren't, you're betting on individual runs of 4.

If you can explain the betting scheme somewhat better, then I can tell you what the EV is. For the purpose of simplicity I would use a 1:1 payoff instead 999:1000 since it makes the numbers simpler and if you could show a +ev result from an even payoff game you've proved the same thing as a positive payoff from a minutely -ev game.
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Old 08-09-2018, 03:07 AM   #4
asymbacguy
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Re: Serious of -EV to +EV sum?

Yep.
How could we manage to bet 8 perfect symmetrical spots having the same probability of the counterparts?
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Old 08-09-2018, 06:25 AM   #5
SiberianPIMP
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Re: Serious of -EV to +EV sum?

Quote:
Originally Posted by RustyBrooks View Post
I don't actually follow how the betting works, not even with looking at your spreadsheet.

I am not sure what significance you're placing on +/-.5 and 1.5 standard deviations, or why you're interested in them. That might make sense if you were betting on something with a continuous distribution, where you win with those given ranges, but you aren't, you're betting on individual runs of 4.

If you can explain the betting scheme somewhat better, then I can tell you what the EV is. For the purpose of simplicity I would use a 1:1 payoff instead 999:1000 since it makes the numbers simpler and if you could show a +ev result from an even payoff game you've proved the same thing as a positive payoff from a minutely -ev game.
Hi, RustyBrooks

It is "some sort of continuous distribution". I am betting 4 individual, but I made them as a whole of 4(with mentioned betting system).

I explained here most of the things that you are also interested in:
https://forumserver.twoplustwo.com/4...to-ev-1719674/

If it is something still unclear, please feel free to ask me on this link.

Last edited by SiberianPIMP; 08-09-2018 at 06:33 AM.
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Old 08-09-2018, 06:30 AM   #6
SiberianPIMP
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Re: Serious of -EV to +EV sum?

Quote:
Originally Posted by asymbacguy View Post
Yep.
How could we manage to bet 8 perfect symmetrical spots having the same probability of the counterparts?
I wrote a "smaller paper" about my theory. In the case I am correct, It basically explains so many things(or put some things closer to solutions at math in general):
-some of the things you mentioned in this thread,
-Parrondo´s paradox,
-st. petersburg paradox,
,...
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Old 08-09-2018, 09:21 AM   #7
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Re: Serious of -EV to +EV sum?

Quote:
Originally Posted by SiberianPIMP View Post
It is "some sort of continuous distribution". I am betting 4 individual, but I made them as a whole of 4(with mentioned betting system).
I'm not 100% sure I've understood what you are doing... It looks like you have correctly calculated the SD for the (discrete) binomial distribution, but then used a normal approximation thereafter:

plot PDF[BinomialDistribution[4, 0.5], x] and PDF[NormalDistribution[4*0.5, 4*0.5*0.5], x]

This explains the 51,6538% / 50% discrepancy.

Juk
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Old 08-09-2018, 09:28 AM   #8
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Re: Serious of -EV to +EV sum?

I also suggest that you learn a little bit of programming so that you can test stuff like this via simulation in the future (just make sure you use a decent pseudo-RNG!!!).

If you can back up your claims with a simulation then people will be a lot more likely to take you seriously...

Juk
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Old 08-09-2018, 10:31 AM   #9
statmanhal
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Re: Serious of -EV to +EV sum?

Quote:
Originally Posted by jukofyork View Post
I'm not 100% sure I've understood what you are doing... It looks like you have correctly calculated the SD for the (discrete) binomial distribution, but then used a normal approximation thereafter:

plot PDF[BinomialDistribution[4, 0.5], x] and PDF[NormalDistribution[4*0.5, 4*0.5*0.5], x]

This explains the 51,6538% / 50% discrepancy.

Juk
This nails it. NickTheGeek in the SMP forum has a similar explanation.
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Old 08-09-2018, 10:37 AM   #10
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Re: Serious of -EV to +EV sum?

Quote:
Originally Posted by SiberianPIMP View Post
It is "some sort of continuous distribution". I am betting 4 individual, but I made them as a whole of 4(with mentioned betting system).
That's not continuous, it's discrete. There are 16 known outcomes - you enumerated them above. You don't need to use standard deviations to calculate the chance of a win - you already know what the chance of a win is, because you can see for yourself that your winning sequences happen 1/2 of the time.
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Old 08-09-2018, 10:58 AM   #11
NewOldGuy
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Re: Serious of -EV to +EV sum?

Yeah, he tried to use a Normal distribution instead of a Binomial distribution.

SiberianPIMP - you can't flip half a head or half a tail, etc. A normal distribution requires that all values be possible (continuous). But since only two values are possible you use the binomial distribution.

If you change that in your calculations, you will arrive at 50%. But as already pointed out here and in the other thread too, there are much easier ways to prove the 50% probability.

How you even contemplated that the real chance to flip a heads was 51% is mind blowing. Did you actually think everyone has been wrong about this for 3000 years?

Last edited by NewOldGuy; 08-09-2018 at 11:05 AM.
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Old 08-09-2018, 02:50 PM   #12
SiberianPIMP
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Re: Serious of -EV to +EV sum?

@jukofyork

TNX for reply.

Yeah, but my betting system is not exactly classic binominal stuff(I think if we assume this betting system is binominal stuff we are wrong)... I wrote some replies why.. in under mentioned link(at the bottom of this post is link;SMP forums). Yeah, it would be great if I could run simulation for this, but I do not know how to.

@RustyBrooks and others that wrote in SMP

TNX.. will wrote constructive comments in SMP forum, but at the moment I am in a hurry...
__________________________________________________
Because I did not get any replies in this thread(at the beggining: when I opened it) I opened another same thread in SMP, therefor I hope whosnext(or somebody else) can close this thread. Eitherway, if this can not be done, I do not see a reason to have same discussions in two threads about same things... I am not discussing anything in this thread anymore. I am answering and having constructive discussions about this matter only in the thread in SMP forums (pointless to have same discussions in two threads).
Here is a link:
https://forumserver.twoplustwo.com/4...to-ev-1719674/

Hope you people understand this. Everybody is welcome to contribute something to the mentioned thread. I will be very glad for every smart contribution.
Thanks...

Last edited by SiberianPIMP; 08-09-2018 at 02:58 PM.
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