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nottophammer

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Okay on todays J247.com sheet you can see at the 37th spin theres 22 1x's.
So we bet the 22,1X's,like below, use +1/-1   +£58
  • 38-22,w
    39-21,w
    40-20,L
    41-40,w
    42-19,w
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nottophammer

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I have a question that seems obvious on the face of it but I don't know how to approach what should be an easy answer.

We all know that during the course of 37 spins, there will be a given number of repeats (I have heard minimum of 4, but the actual number is irrelevant for this question unless the minimum is 0).  At the end of the 37 spins, we will have 4 or 6 or 10 spaces that haven't shown.  Looking back at what is now history, we could argue that those spaces were never possible options.  We didn't know what the numbers were that wouldn't hit, but we don't have to if we are playing for repeats.

If the minimum is actually 4, are not our odds of getting a hit on say 9 spaces covered 9/33 rather than 9/37?  It doesn't matter which 4 don't hit, they simply won't hit.  It becomes a discussion of misses being any of the 24 losing spots instead of the 28 "possible" spots, as we know they won't all hit.  The implications of this are fairly drastic over just a few spins.  In 1 spin, it is 27.27% vs. 24.32%.  However, over the course of 3 spins, it moves to 62.71% instead of 56.66%.  5 spins, our odds of hitting move to 88.52% against 75.18%.

If our "real" odds are not 9/37 and 28/37 respectively, but rather 9/33 and 24/33 in the worst case scenario of only 4 repeats, am I incorrect that there would be a positive expectation?
I give colbster the thumbs up here.
Mogul
the line in red, works the same for non-hit,the more spins you wait for a particular non-hit past your known average to hit, the better the chance you'll win, of course it might hit by its known average to hit.
Now if you was to collect the non-hits where you play in the good old USA, after collecting 100 games of at least 60 spins, you would have an average to start working with.
I'd bet your dollar that you'd see for up to the 19th non-hit they would average to hit in just 2 spins, there max to hit ? but the more games you collect the max will appear.
See if you can work with this, its not been updated for months, but these averages are the same as others i have.
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nottophammer

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There will come a time (if this observation proves true) where the expected return turns negative if we shift the ratios.  For instance, if after 15 spins, we have 3 repeats and 9 singles, our bet is 12 units for a possible +24 (+35 - 11 losers).  If we shift the number of potential numbers left to hit down by the number of repeats, there are really only 34 possible outcomes.  We take 12/34 * 24 for our expected wins and get 8.471.  We have 22/34 * -12 for our expected losses and get -7.765, netting us 0.71.  We have better than even odds that we will be positive at the end of all this. This would be our cue to start betting on all 12 of the units for the remainder of the 37-spin series.[/color]

Prior to this spin (which I am claiming as a repeat for the sake of this argument), we had 2 repeats and 10 singles.  With only 2 repeats, there are the potential for 35 unique numbers.  12/35 * 24 gives us 8.229 and 23/35 * -12 gives us -7.886 for a net +0.34.  We could also have played beginning with this spin and would have had a win as indicated above.

If the spin before this was also a repeat and we had 1 repeat and 11 singles to this point, our math changes to 12/36 * 24 and 24/36 * -12, or 0.  At this point, we were neutral and there was no justification for a bet.  Anything before this would have been neutral or negative prior to the repeat.   I think that with some basic tracking, we can identify the exact spin where our expectations move from negative to neutral to positive.  I'm working a spreadsheet now to test by hand, but I bet there is a simple mathematical equation that could solve the dilemma for us with less hoopla to just indicate if we are positive or negative on expectation.
The above in maroon, is when you will win, with flat bet, the only part is, what winnings you are going to take? if using £1 units, i'd stop as soon as +£24 is reached.
Don't believe, i'd suggest you test and soon as 12non-hit are left start flat betting.
Remember Nottophammer told you this  :thumbsup:
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nottophammer

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Just played on MPR 25th came

35-R
34-26th
27-27th
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nottophammer

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Now this is betting non-hit, but with 30 seconds would not get on,but when live table is busy it can be done, so theres your answer, why they like to spin fast, so you cant spread the large non-hits at the start.

But like Colbster says whether inadvertently bet the last 12 non-hit.
From me flat bet works, you just have to keep your faith.
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nottophammer

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This should be in KTF but look at the trot, 7,+2, 12,+2, the old average so next 10 spins could see 3 0x's, countback showing how fast and look at spin 37, Luck of Irish shows 24 at 37. countback saying could see 21, so you know its fast.
After 25th, flat bet the remaining 12,0x's, made the +24 so stop.

WTF, do you reckon Celtic would have reset after 29,29, looking further, i think he'd have reset at spin 10
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BellagioOwner

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After 15 spins (9 singles and 3 doubles), there are 25 spaces left unhit but only 22 chances to hit them.  The odds of any of the unhit numbers falling are now 22/25. Although we don’t know which 3, there are at least 3 numbers that simply CANNOT fall during the course of 37-spin cycle.

The advantage we have in such a situation can be quantified.  We have 22 chances to lose 12 units (22 * -12 = -264).  We have 12 spaces where we can win 24 units if they hit (12 * 24 = 288).  Net of 288-264, we have an expected return of 24 units which would be spread over the 22 spins, or 1.091 units expected return per spin positive.

I agree that there are not anymore enough spins for all 25 unhit numbers to hit so there will be some left unhit in the 37-spin cycle. BUT I think the problem still remains that in the remaining 22 spins we don't know WHEN will a repeater be or when a new unhit number will occur. aka we don't know in a specific spin what the outcome will be. We do know that in the end  some will stay unhit but we do not know  in a specific hit the outcome and HOW/in which way this sure knowledge of us for the end result will eventually form.

So what I think is that even knowing that 3 will stay unhit for a specific spin the numbers that make us lose are still 25 and not 22. Which would lead to:
We have 25 chances to lose 12 units (25 * -12 = -300). We have 12 spaces where we can win 24 units if they hit (12 * 24 = 288).  Net of 288-300 = -12units. we have an expected return of -12 units which would be spread over the 22 spins, or -0.545 units expected return per spin negative.

I HOPE I'm wrong but what you guys think? :/
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nottophammer

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Don't believe, i'd suggest you test and soon as 12non-hit are left start flat betting.
Remember Nottophammer told you this 
So from further testing we see it should be when only 9 non-hit remain of the starting 37.


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BellagioOwner

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What you think guys on my reply#201? Can there really be any positive expectation per spin the way suggested?  I have good results so far but it could be just luck.

PS: Wishing you a very good and productive year as well in every field of your life.  And with health to you and your loved ones :)
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MrG

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What you think guys on my reply#201? Can there really be any positive expectation per spin the way suggested?  I have good results so far but it could be just luck.

PS: Wishing you a very good and productive year as well in every field of your life.  And with health to you and your loved ones :)

I would like to know who is right, but I'm not a math boy. But I tested it some time ago and posted the results somewhere in this thread.

 

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