roulette math question

Could anyone tell me the probability of having a 1, 2 and 3 all spin before a 4, 5, 6, 7, 8 or 9 spins? And could those probabilities be broken down further, to include if the 1 would spin last of the 3 ones? Thanks in advance.

Quote: allinriverking

Could anyone tell me the probability of having a 1, 2 and 3 all spin before a 4, 5, 6, 7, 8 or 9 spins? And could those probabilities be broken down further, to include if the 1 would spin last of the 3 ones? Thanks in advance.

Yes they could. There are a lot of smart math guys who could do that here.

I expect that exactly zero of them want to, and one of them (at least) will, and after they do, nobody will be any better off.

There is no math that you can learn that will reduce the cost of playing roulette without reducing the amount that you bet on it.

There is one exception to that rule, and that would be if you happen to like to play ( 0, 00, 1, 2, 3 ) In this case, it's better to make a lower percentage cost bet. All the others are the same.

Roulette is one of the simplest games in the casino and if you don't understand the math of roulette, you have no business doing math trying to improve your chances to win at any game in the casino.

Some people consider all gambling a regressive tax on the mentally inept and/or math-challenged.

I don't. But for roulette, I have to make an exception.

Do yourself a favor and quit gambling and find a less expensive and more satisfying hobby. That's my advice.

Quote: allinriverking

Could anyone tell me the probability of having a 1, 2 and 3 all spin before a 4, 5, 6, 7, 8 or 9 spins? And could those probabilities be broken down further, to include if the 1 would spin last of the 3 ones? Thanks in advance.

Your second part of the question is easy. It is 3 times whatever you figure out for the answer to the main question.
The probability of having a 1, 2 or 3 spin before the six other numbers is 3 in 9. Once that has happened the probability of having one of the two remaining numbers spin before your 6 bad numbers is 2 in 8. Once that has happened the probability of the last number spinning before your 6 bad numbers is 1 in 7.
So.... 3/9 x 2/8 x 1/7 = 1/84, or approximately 1.2%

If you require '1' to be the last number, it is then 1/252, or approximately 0.4%

I hope one of the real math guys can tell me if I did this right!.

Thanks Soopoo,
I thought that might be the way, I also thought at first
3/6×2/6×1/6=6/216=1/36=2.78%

Quote: allinriverking

...the probability of having a 1, 2 and 3 all spin before a 4, 5, 6, 7, 8 or 9 spins?....the 1 would spin last of the 3 ones?...

I'm reading Q1 as all of 1 2 and 3 have to spin before any 4 thru 9 (but repeats of 1 2 or 3 are allowed). The chances are 3/9 * 2/8 * 1/7 = 1 / 84. The chances of the 1 being last are, by symmetry, 1 in 3; hence Q2 = 1 / 252. So yes I agree with the answer above.

My interpretation of the question is what is the probability that the numbers 1, 2, and 3 will all be spun within x spins. Here is my answer:

Spins	single-zero	double-zero
3	0.000118	0.000109
4	0.000455	0.000420
5	0.001091	0.001009
6	0.002094	0.001939
7	0.003518	0.003261
8	0.005404	0.005016
9	0.007785	0.007234
10	0.010684	0.009937
15	0.033231	0.031066
20	0.068639	0.064476
25	0.114718	0.108254
30	0.168563	0.159750
35	0.227272	0.216265
40	0.288292	0.275379
45	0.349548	0.335089
50	0.409453	0.393835
55	0.466865	0.450467
60	0.521017	0.504191
65	0.571445	0.554501
70	0.617922	0.601122
75	0.660393	0.643951
80	0.698930	0.683016
85	0.733693	0.718435
90	0.764897	0.750386
95	0.792791	0.779086
100	0.817638	0.804773

The general formula for this kind of question is:

Pr(A and B and C) = Pr(A) + Pr(B) + Pr(C) - Pr(A and B) - Pr(A and C) - Pr(B and C) + Pr(A and B and C)

Quote: Wizard

Pr(A and B and C) = Pr(A) + Pr(B) + Pr(C) - Pr(A and B) - Pr(A and C) - Pr(B and C) + Pr(A and B and C)

You mean P(A or B or C) on the left hand side.

In this case it's easiest to take 1 minus the probability of not getting all 3, so that would be

P(A and B and C) = 1 - P(A' or B' or C')

where the ' means not.

= 1 - [P(A') + P(B') + P(C') - P(A' and B') - P(A' and C') - P(B' and C' ) + P(A' and B' and C')

Since A,B and C have the same probability here, this is

1 - [3*P(A') - 3*P(A' and B') + P(A' and B' and C')]

With N the total numbers (37 for European and 38 for American), and s= number of spins, this is

1 - [3*((N-1)/N)^s - 3*((N-2)/N)^s + ((N-3)/N)^s].

And picking a random ones, like N = 38 and s = 10, it looks like that agrees with what you computed.