Calucating Dice Probabilities?

[12AX7]

Yes, I finally found some info that verifies this.

One writer described at the probability of

(1) (P of not a 7) * (P of not a 7)... (P of not a 7)^N

He also wrote that getting a 7 at exactly roll N was

(2)(P of not a 7)^N-1 * (P of a 7)

A related question I'm struggling with is, "How many rolls should I expect it to take to hit a 7?"

Intuitively the mean appears to be 6 since a 7 comes up, on average once in 6 rolls.

But what is the probability of that? Or say 5 or 7 rolls?

I found that when I ran equation (2) through a spread sheet for up to 40 rolls, that the probability just gradually declined. It did not peak at 6 rolls.

I've read the average rolls per shooter is 8.5? And average rolls for a pass line decision is about 3? But that doesn't really tell me "average rolls between 7's".

What I'm trying to do if devise a betting strategy that would make use of the fact that 6 rolls is most likely, to stretch my -EV dollars as far as possible LOL!

So taking a page from blackjack theory, if I'm most likely to hit on the 6th roll, shouldn't I bet more at that time to back fill losses? And possible at the 5th and 7th rolls, while tapering off on either side.

For example, at 8 or more rolls, you could start just betting enough to break even on your current losses, to avoid hitting the table limit any sooner than neccesary, and to stretch your session stakes.

Granted "Any 7" is worst bet on the table. But the math is easiest for it, for me. So it makes it easier for me to see the concepts.

I realize that ultimately, given this example, you are gathering up a 16% statistical deficit on each dollar wagered. However, seems to me you might could tailor the betting progression to put it off as long as possible. [img]/images/graemlins/laugh.gif[/img]