# Re: Are there really 37 possible outcomes?

Tags: unhit spin spins
Quote from: Colbster on October 07, 2014, 12:49:22 AM
After 15 spins (9 singles and 3 doubles), there are 25 spaces left unhit but only 22 chances to hit them.  The odds of any of the unhit numbers falling are now 22/25. Although we don’t know which 3, there are at least 3 numbers that simply CANNOT fall during the course of 37-spin cycle.

The advantage we have in such a situation can be quantified.  We have 22 chances to lose 12 units (22 * -12 = -264).  We have 12 spaces where we can win 24 units if they hit (12 * 24 = 288).  Net of 288-264, we have an expected return of 24 units which would be spread over the 22 spins, or 1.091 units expected return per spin positive.

I agree that there are not anymore enough spins for all 25 unhit numbers to hit so there will be some left unhit in the 37-spin cycle. BUT I think the problem still remains that in the remaining 22 spins we don't know WHEN will a repeater be or when a new unhit number will occur. aka we don't know in a specific spin what the outcome will be. We do know that in the end  some will stay unhit but we do not know  in a specific hit the outcome and HOW/in which way this sure knowledge of us for the end result will eventually form.

So what I think is that even knowing that 3 will stay unhit for a specific spin the numbers that make us lose are still 25 and not 22. Which would lead to:
We have 25 chances to lose 12 units (25 * -12 = -300). We have 12 spaces where we can win 24 units if they hit (12 * 24 = 288).  Net of 288-300 = -12units. we have an expected return of -12 units which would be spread over the 22 spins, or -0.545 units expected return per spin negative.

I HOPE I'm wrong but what you guys think? :/

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Re: Are there really 37 possible outcomes?

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