odds of consecutive streets in roulette

Can anyone assist with the following roulette question: What would be the odds of there being ten (10) different streets in succession? That is ten different streets without one of them repeating? Ditto for the entire 12 streets occurring without a repeat? Thanks in advance, any help muchly appreciated. Cheers bellbiz

There are three numbers in a street bet, so the probability of ten consecutive unique streets is:

36/38 * 33/38 * 30/38 * 27/38 * 24/38 * 21/38 * 18/38 * 15/38 * 12/38 * 9/38 = 0.002252597249283921 = 1/0.002252597249283921 = 1 in 443.932

For all twelve:

36/38 * 33/38 * 30/38 * 27/38 * 24/38 * 21/38 * 18/38 * 15/38 * 12/38 * 9/38 * 6/38 * 3/38 = 0.00002807946709633696 = 1/0.00002807946709633696 = 1 in 35,613.211

I also assumed zeroes caused the streak to fail, just replace the '38's' in the formula with '36's' if you want zeroes not to cause it to fail.

Awesome! Thanks heaps.

No problem!