Help with Maths Probability

[Gzgzbee]

Hello,

I was wondering if any of the Math Guys would help!?

The odds of 12 numbers out of 37 not showing in 10 spins is: (If I am right)

37/25 x10 (x10 = To the Power of 10) = 50.42

But what are the odds of this happening twice in a row BUT with the same number?

I.e say my 12 "special" numbers are 1,2,3,4,5,6,7,8,9,10,11,12

If number "1" comes out I bet 10 spins that include the above 12 "special" number and hope one of the number hit within 10 spins.

If none of them numbers come out the odds should be as above i.e 50/1 change.

What i would like to know is what are the odds of next time when number "1" lands that none of the above "special" will show again over 10 spins?

Thanks 

[Gzgzbee]

Anyone!!?

[nottophammer]

Anyone!!?

maybe Turner

[Blue_Angel]

Sorry but didn't understand your question, do you want to bet those 12 numbers or bet against them for 10 consecutive spins??

I have a similar question, a dozen or column can be missing for up to 46 consecutive spins (10 mil trials)

So if we were to bet specifically against a return of a delayed dozen/column we would have to bet the other 2 dozens/columns, thus we would need something more than the double of wins than our losses, correct so far??

A dozen/column has a probability to hit 1 out of 3 times, therefore within this range cannot be categorized as delayed or cold, if we would bet that any cold dozen which sleeps for 3 consecutive times would NOT hit on the 4th consecutive spin we would win  all the times that dozen/column is sleeping for 5 consecutive spins or more and lose when  awakes on specifically the 4th spin.

Would all of the rest times be more than double of those which awakes on the 4th spin??

[Gzgzbee]

What are the Odds of 12 numbers out of 37 on roulette not showing for 10spins?Interested in the odds of this happenin twice in a row with the same number?
Edit

So for example if my 12 numbers I use are 1,2,3,4,5,6,7,8,9,10,11,12

Number 2 comes out so i bet on all of my 12 number above over the next 10 spins and hope i get a hit!

If none of the above hit then its loses!

What are the odds the next time number 2 comes out that again my 12 number above will not show in the next 10 spins?

[Blue_Angel]

What are the Odds of 12 numbers out of 37 on roulette not showing for 10spins?Interested in the odds of this happenin twice in a row with the same number?
Edit

So for example if my 12 numbers I use are 1,2,3,4,5,6,7,8,9,10,11,12

Number 2 comes out so i bet on all of my 12 number above over the next 10 spins and hope i get a hit!

If none of the above hit then its loses!

What are the odds the next time number 2 comes out that again my 12 number above will not show in the next 10 spins?

Now I understood your question, what it has confused me is that if you are looking for the probability of 20 consecutive misses for 12 numbers OR if you are looking for 10 consecutive misses for 12 numbers.

It would be better described as: wait for ANY 12 numbers to miss 10 (or more) consecutive times, then wait for 1 of those 12 numbers to hit and bet those 12 numbers for up to 10 spins or till a hit.

Is this what you mean??

[Gzgzbee]

Thanks for the reply Blue_Angel.

It would be 10 consecutive misses for 12 numbers? But this happening twice in a row with the same number??

Can you help?

I know how to work out the odds of the 12 number not landing in 10 consecutive spins.

37/25 = 1.48 x10 (x10 = To the power of)

= 50.4216617

so i 1/50 change! If thats correct?