Roulette Ideas

@Smith given you math knowledge, I wonder might you shed light on this.,

It is a conditional probability. Not a betting strategy.

Every number gets 2.7 % to come out at every spin.
What is not a conditional probability.

8 numbers * 2.7 % = 21.6 % chance to win each spinning

Long term it will hit at every 4th or 5th spinning.

Short term this win will come later often.
I did not tested it with 8 numbers but I know what happen if you cover 6 numbers only.
6 number covering means (6 * 2.7 %) 16.2 % chance to win (long term every 6th spin comes a win) and 6 number can miss 50-60 times too.
50-60 missing is very rare. I start the betting after 25 missing when I play my double street strategy. And I take bet to next 29 spinning.

At the lucky 9 strategy (9 numbers side by side) if was not hit 25 times then I start betting to these straight numbers. Not earlier because I can not take bet to many spinning.
9 * 2.7 % = 24.3 % the chance to win per spin here by the way. Long term every 4th spin comes a win.
 

Hi Hardman,

The formula is correct, but is only valid for finding the probability of exactly one hit over 16 spins for a probability of 8/37, which isn't what you want. You would have to use the formula twice with the corresponding parameters : p = 8/37 for 12 spins, then p = 6/37 for 4 spins, then multiply the probabilities. But I don't think this is what you're after, assuming your intention is to quit in both these series immediately on a hit?

The Binomial formula will only tell you the chance of one (or more) hit(s) if you play out each series to the end.

The formula you need is one which calculates the prob. that at most n spins are needed until the first win, which is

1 ─ (1 ─ p)ⁿ

for the first series where p = 8/37 and n = 12,

p₁ = 1 ─ (1 ─ 8/37)¹²

and for second where p = 6/37 and n = 4,

p₂ = 1 ─ (1 ─ 6/37)⁴

Multiplying these together (since you require both to be fulfilled) gives a final probability of ~48%

Actually, it turns out that this is very close to the value you would have got using the binomial formula, but that isn't always the case.

OK, thanks; that was introductory example to get us on the same page. Might you now look at the bottom paragraph?

Why I focused on the binomial probability formula is, because this gives me the result on the exac number of hits; since I'd like to apply it then to more blocks such the one in the last paragraph .. given that some games get to require several hits, usually close by, to resolve.

So, if two hits above make on combo (hit), then it would require more combos .. & as well of different types = various payouts adjusted for the exposition,

Can you clarify what you mean in the last paragraph? This is how I understand it but I'm not sure : for the first 12 spins you start betting 8 #s for up to 4 spins, if a win go straight to betting 6 #s for up to 4 spins.

If a loss, increase #s to 12 for up to 4 spins, if a win go straight to betting 6 #s for up to 4 spins.

If a loss, increase #s to 16 for up to 4 spins, if a win go straight to betting 6 #s for up to 4 spins.

And you want the probabilities for each of these possible scenarios?

Yes, that's correct.

I don't have time to calculate the probabilities for all the possible outcomes, but this should get you started. Here are the probabilities you need-

        Hit      No hit
Prob.
8/37    0.623    0.377
12/37   0.792    0.208
16/37   0.896    0.104
6/37    0.507    0.493

For each possibility you just need to multiply them. e.g. the chance of losing the first 3 bets (first 12 spins) is 0.377 × 0.208 × 0.104 = 0.81%

If you win the 8/37 bet and lose the 6/37 bet the chance is 0.623 × 0.493 = 30.71%
If you win the 8/37 bet and win the 6/37 bet the chance is 0.623 × 0.507 = 31.6%

etc...