Notnab
Notnab
Joined: Jul 11, 2012
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July 11th, 2012 at 9:37:37 AM permalink
Quote: guido111


I still say a blind squirrel can choose winning Roulette numbers just as well as any human or computer can.


Absolute junk.

Hmmm ... I guess squirrels understand probability more than mathematicians.
mustangsally
mustangsally
Joined: Mar 29, 2011
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July 11th, 2012 at 9:39:39 AM permalink
Quote: Notnab

I'm predicting that it won't happen in YOUR lifetime, while YOU'RE playing roulette.

How does that help you in choosing winning Roulette numbers?

You want to believe that a Xspin distribution will always turn into a X+Y distribution.

How much are you willing to lose to prove you are correct?
Your betting methods can be easily simulated.

I see also that guido111 shows you to be waiting many,
many spins for just the right combos of numbers to show and not show before you even bet one time.

What kind of bankroll do you have an idea to use?
and
How many times do you expect to make these bets before your bankroll is ruined?

Even I would like to know those numbers.

Sally
I Heart Vi Hart
Notnab
Notnab
Joined: Jul 11, 2012
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July 11th, 2012 at 9:58:04 AM permalink
Quote: thecesspit

Sure thing, but it doesn't matter to improve your chances of picking the right numbers on the next spin.


The fact that it's unlikely to happen is a reliable "stat" that can be borne in mind when developing a strategy that's based on past and current reality.

Roulette is like the weather.

And a roulette player should be like a weatherman studying the roulette table conditions.

When you see particular types of clouds, it's more likely to rain or the sun to shine or to snow.

You can't predict a specific number(s), but you can take advantage when certain conditions naturally arise on a regular basis and at random regular intervals.

Sometimes it snows; sometimes it rains; sometimes the sun shines. And just like the real weather, a real roulette wheel provides it's own natural signals.
strictlyAP
strictlyAP
Joined: Jun 20, 2012
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July 11th, 2012 at 9:59:33 AM permalink
oh this drives me nuts, how many times i try and explain to my gf even if a quarter lands heads 2000 times in a row tails is not due to be next ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
The bet will not be paid- not now not ever
thecesspit
thecesspit
Joined: Apr 19, 2010
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July 11th, 2012 at 10:01:05 AM permalink
It's not a reliable stat. It's an emergent phenomena from the fact there's a 1/38 chance of each number. If I see 1-36 in the last 36 spins... the chance of hitting 0 or 00 is still exactly the same as it was.

I've heard what you've said, but never met a weatherman who can describe the nature of the snow or sun.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
Notnab
Notnab
Joined: Jul 11, 2012
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July 11th, 2012 at 10:14:40 AM permalink
Quote: guido111

I can show some values. You can choose what is the best.

You also be doing a lot of waiting. Why?
Where is the value in waiting so long?

4.81% of every 31 spin sets will on average result in 19 numbers not showing
48,112 out of 1 million trials

win prob = 19/37 for 6 spins

x    prob[X=x]   prob[X<x]  prob[X>=x]  prob[X<=x]   prob[X>x]
0 0.0132564 0.0000000 1.0000000 0.0132564 0.9867436
1 0.0839570 0.0132564 0.9867436 0.0972134 0.9027866
2 0.2215533 0.0972134 0.9027866 0.3187667 0.6812333
3 0.3118157 0.3187667 0.6812333 0.6305823 0.3694177
4 0.2468541 0.6305823 0.3694177 0.8774364 0.1225636
5 0.1042273 0.8774364 0.1225636 0.9816637 0.0183363
6 0.0183363 0.9816637 0.0183363 1.0000000 0.0000000

7.97% of every 32 spin sets will on average result in 18 numbers not showing
79,657 out of 1 million trials

win prob = 18/37 for 5 spins
x    prob[X=x]   prob[X<x]  prob[X>=x]  prob[X<=x]   prob[X>x]
0 0.0357075 0.0000000 1.0000000 0.0357075 0.9642925
1 0.1691408 0.0357075 0.9642925 0.2048483 0.7951517
2 0.3204772 0.2048483 0.7951517 0.5253255 0.4746745
3 0.3036100 0.5253255 0.4746745 0.8289355 0.1710645
4 0.1438153 0.8289355 0.1710645 0.9727508 0.0272492
5 0.0272492 0.9727508 0.0272492 1.0000000 0.0000000

11.91% of every 33 spin sets will on average result in 17 numbers not showing
119,080 out of 1 million trials

win prob = 17/37
x    prob[X=x]   prob[X<x]  prob[X>=x]  prob[X<=x]   prob[X>x]
0 0.0853715 0.0000000 1.0000000 0.0853715 0.9146285
1 0.2902632 0.0853715 0.9146285 0.3756348 0.6243652
2 0.3700856 0.3756348 0.6243652 0.7457204 0.2542796
3 0.2097152 0.7457204 0.2542796 0.9554355 0.0445645
4 0.0445645 0.9554355 0.0445645 1.0000000 0.0000000


Thanks in advance for your mathematical analysis.

- Notnab




It should never affect anyone's perspective.

The past only shows what has happened in a sequence of events
and
does no better than a blind squirrel picking the next numbers to win.

Any computer simulation will prove this to be absolutely true using RNG numbers or internet listed actual spins.

Where are the results of your computer simulations?
If you have none, time for you to learn to do them.
There will be power in your findings that none of us here can ever give you.

BTW, your %s are really rounded

no showunique #sprobability
20170.051300%
19180.282190%
18191.174500%
17203.605390%
16218.384150%
152214.835930%
142319.936310%
132420.475660%
122515.925890%
11269.406580%
10274.159970%
9281.364840%
8290.328120%
7300.054960%


Thanks.

Helpful stats. Might prove useful in making a long-term decision.

My main thought has been:

Is it better to bet on 17 numbers with profit (£36 - £17 = £19); with probability 17/37 = 0.4594594, or
Is it better to bet on 18 numbers with profit (£36 - £18 = £18); with probability 18/37 = 0.4864864, or
Is it better to bet on 19 numbers with profit (£36 - £19 = £17); with probability 19/37 = 0.5135135
Keyser
Keyser
Joined: Apr 16, 2010
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July 11th, 2012 at 10:43:10 AM permalink
Notnab,

You remind me of Mauisunset. Are you two friends?
mustangsally
mustangsally
Joined: Mar 29, 2011
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July 11th, 2012 at 10:52:36 AM permalink
Quote: Notnab

Roulette is like the weather.

And a roulette player should be like a weatherman studying the roulette table conditions.

No it is not.

weather predicting is about dependent events.
The probabilities are always changing and can never be exactly calculated.

Roulette spins, either one or a sequence of many, are independent events.
The probabilities are known and are always the same.

The past results, as a pointer to what numbers to bet in the next upcoming results,
in no way affects the next spin or a number of next spins.

If you are choosing to bet 17 numbers that did not show in the last 33 spins
and I choose any 17 numbers at random or by a coin toss.
We both will win and lose the next sequences at the same rate, percentage or average.

You need to understand the difference between dependent events and independent events.
And to make you better than the rest, know the difference between independent events and mutually exclusive events.
These are basic statistical facts.
Then people will listen to you.
a good starting point is here
http://stattrek.com/probability/probability.aspx?Tutorial=Stat
They will not try to sell you anything :)

A computer simulation will easily show the differences between all of these.
I Heart Vi Hart
mustangsally
mustangsally
Joined: Mar 29, 2011
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July 11th, 2012 at 10:56:23 AM permalink
Quote: Notnab

My main thought has been:

Is it better to bet on 17 numbers with profit (£36 - £17 = £19); with probability 17/37 = 0.4594594, or
Is it better to bet on 18 numbers with profit (£36 - £18 = £18); with probability 18/37 = 0.4864864, or
Is it better to bet on 19 numbers with profit (£36 - £19 = £17); with probability 19/37 = 0.5135135



from guido111 binomial distribution tables
to bet on 17 numbers you have a 0.0853715 chance of NOT winning at least 1 time
to bet on 18 numbers you have a 0.0357075 chance of NOT winning at least 1 time
to bet on 19 numbers you have a 0.0132564 chance of NOT winning at least 1 time

Looks right to me.
I Heart Vi Hart
Mission146
Mission146
Joined: May 15, 2012
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July 11th, 2012 at 11:10:27 AM permalink
Remembering always that the wheel doesn't care what numbers hit before:

I'll tell you what I am going to do for fun on WOO free Roulette game later, or if I had a lot of money that I didn't really want. I'd wait for seven different consecutive numbers, which could take awhile, and then run the following system:

Eighth Spin: Seven Numbers at $1.00/each, straight bet getting 36:1.

If win: $36 - $7 = $29

If loss: -$7 AND

Eight numbers at $1.00 each

If win: $36 -$8 -$7 = $21

If loss: -$15 AND

Nine Numbers at $1.00 each

If win: $36 -$9 -$8 -$7 = $12

If loss: -$24 AND:

Ten Numbers at $1.00 each

If win: $36 -$10 -$9 -$8 -$7 = $2

If loss -$34 AND:

Eleven Numbers at $2.00 each

If win: $72 -$22, -$10, -$9, -$8 -$7 = $16

If loss: -$56 AND

Twelve Numbers at $3.00 each

If win: $108 -$36 -$22 -$10 -$9 -$8 -$7 = $16

If loss: -$92 AND:

Thirteen Numbers at $5.00 each

If win: $180 -$65 -$92 = $23

If loss: -$157 AND

Fourteen Numbers at $8.00 each

If win $288 -$112 -$157 = $19

If loss: -$269 AND

Fifteen Numbers at $13 each

If win: $468 -$195 -$269 =$4

If Loss: -$464 AND

Sixteen Numbers at $24 each

If win: $864 - $384 -$464 = $16

If loss: -$848 AND

---In my experience, the system would end in failure at this point because most $5.00 Min. tables I have seen have a $500 Maximum. If the maximum were $1,000, however:

Seventeen Numbers at $45 each:

If win: $1,620 -$848 -$765 = $7

If Loss: -$1,613

PROBABILITY OF FAILURE

30/37 * 29/37 * 28/37 * 27/37 * 26/37 * 25/37 * 24/37 * 23/37 * 22/37 * 21/37 ($500 Limit) * 20/37 ($1,000 Limit)

$500 Limit: .022673482 or 2.2673482%

EV on Failure = .022673482 * 848 = -$19.2271126 ($19.23)

$1000 Limit = .012255936 or 1.2255936%

EV on Failure = .012255936 * 1613 = $19.76882497 ($19.77)

Expected Returns

29 * (7/37) = $5.49

21 * (7/37) * (8/37) = $0.86

12 * (7/37) * (8/37) * (9/37) = $0.12

2 * (7/37) * (8/37) * (9/37) * (10/37) = $0.01 (Rounding Up by .0463)

16 * (7/37) * (8/37) * (9/37) * (10/37) * (11/37) = $0.01 (Rounding down .2791886)

16 * Above () and (12/37) = $0.01 (Rounding up impermissibly by .00585128)

23 * Above () and (13/37) = $0.00 (Not even close) In fact, we're going to say the rest of the results total and MAYBE round up to $0.01

Expected Return (Any Success): $6.51

Total Expected Values

$500 Limit

-$19.23 + $ 6.51 = -$12.72

$1,000 Limit

-$19.77 + $6.51 = -$13.26

Hold Percentage: 12.72/848 = 0.015 or 1.5% of total bets for $500 Limit

Hold Percentage: 13.26/1613 = .00822 or .822% of total bets for $1,000 Limit

CAN SOMEONE PLEASE VERIFY MY HOLD PERCENTAGES?

In the event that my Math is correct at the end (I'm good on everything before that) this system actually, somehow, reduces the HE. The problem is that no win after the second play exceeds the total amount that you have out in bets, so system failure is obviously devastating. In fact, the third play only gives you a profit of HALF as much as you have out in current/previous bets at that point.

Coincidentally, it is also the third bet where you are more than 50% to succeed (or have already succeeded) by then:

30/37 * 29/37 * 28/37 = .480919195 or 48.1% chance of three consecutive failures, inversely, 51.9% chance of success.

By the time you reach the point that probability says you should succeed, the total risk is no longer worth the reward.

By comparison, laying the 4 for $24 in Craps would yield a profit of $12, if successful, but you have a 66.666-% chance of that happening!

You don't even have a fixed profit total for this system! It will be anywhere between $2-$29 depending upon what stage of the system at which you win.

With the $500 Limit, you fail 1:44.1 attempts.

With the $1000 Limit, you fail 1:81.59 attempts.

Martingaling the Red starting at $5, however, gives you seven shots at a $500 Limit starting at $5 and eight shots at a $1,000 Limit:

The probability of Martingale failure is:

(19/37) * (19/37) * (19/37) * (19/37) *(19/37) * (19/37) * (19/37)-$500 or * (19/37) -$1,000

$500 = .009415928 or .94%

$1000 = .004835206 OR .48%

$500 Limit Loss EV ($635 Lost) 635 * .009415928 = 5.97911428 -$5.98

$1000 Limit Loss EV ($1275 Lost) 1275 * .004835206 = 6.16488765 or -$6.16

$5.00 Win (Any Point) = 5 * .990584072 = 4.95292036 or $4.95 on $500 Limit

or

$5.00 Win (Any Point) = 5 * .995164794 = 4.97582397 or $4.98 on $1,000 Limit

EV on $500 Limit -$1.03

EV on $1000 Limit -$1.18

1.03/635 = .001622 or .1622% House Hold on $635 Bet

1.18/1275 = .00092549 or .0925% House Hold on $1275 Bet

COMPARISON

THE GOOD:

1.) My goof-off system up top has a better return percentage in relation to the amount bet (assuming a win) than the Martingale with exception only to the Ten Number Pick vs. the Fourth Martingale Spin:

The fourth Martingale spin has you winning $5 against $75 bet or 6.666% of total bets

The Goof-Off system has you winning $2 on a bet of $34 or 5.8824% of total bets

Remember, this is a return only in relation to the bets. It does not reflect the EV of actually winning at this time, nor does it reflect the EV of winning any individual trial. Comparing the two systems, the Martingale is ALWAYS more likely to win any individual trial. If we assume the $500 Limit, the goof-off system is 16/37 to succeed on the final bet before failure and the Martingale is always 18/37 to succeed on any individual turn.

2.) You will usuallly have less same spin money riding (incuding current and previous-lost-bets) with the goof-off system. The only exceptions are the 1st Spin $7 v. $5 and the second spin, which is $15 in both cases.

THE BAD:

Everything Else, specifically:

1.) The goof-off system has a higher chance of failure. It fails slightly less than 2.5 times more than the Martingale at a $500 limit and slightly more than 2.5 times the Martingale at the $1,000 limit. 2.56, to be specific.

The goof-off system also has a higher risk of failure on any individual spin.

The average spin at the $500 limit has a 11.5/37 single spin chance of winning, or 31.08%

The average spin at the $1000 imit has a 12/37 single spin chance of winning, or 33.33333%

The Martingale has a constant 18/37 single spin chance of winning, or 48.6%

2.) In addition to having a higher risk of failure, the goof-off system also loses more actual money on a failure. It loses $213 more for a $500 Table Max failure and $338 more for a $1,000 Table Max failure.

The Point

There's no point in really trying the goof-off system, except for fun. The wheel and the ball have no concern nor control over what happened before or what will later happen in relation to previous events. Because this is the case, you could just start it off with seven individual random numbers as opposed to waiting for six consecutive non-recurring numbers, at least then you don't run the risk of being bored waiting for non-recurring numbers.

The best way for this system to succeed would require a ridiculously high limit and a crapload of money. I may do the math on it at a later time. However, what you would do is basically just start picking one number (because that is the cheap end, anyway) and then increasing your bets and picking additional numbers in a system simillar to the one described above. The fail point would be a thirty-five number pick bet losing 35/37 and that's 94.6% to win by itself with nothing else considered. It would be almost impossible to lose, but the results would be catastrophic if you did.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

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