Craps Place Number Question

The average roll of a shooter in craps is 8.5 rolls before they seven out.

How many of those rolls will be place numbers?

Place numbers are 4, 5, 6, 8, 9 and 10.

4.67? Not much time to make a point or hit a single place bet. I do a 5-count on the DC when other people shoot unless they've been hot shooters before. Place a bet on the 4th roll since a point is established (which counts as roll #1) then add a 2nd bet at roll 9, and a 3rd bet roll 14; but I reset the count if a point is made. These 5-counts bring me past the 8.5 roll marker so a bet can be resolved. The problem is that shooters will roll a 7-11 when I just put my bet down on the DC and it loses. I should bet larger with $30 multiples of odds.

Are you saying that 4.67 rolls will be place numbers out of the 8.5 rolls?

Let me 2nd guess that
2/3 x 7.5 = 5 rolls
Subtract one roll because place bets are off during the come-out roll to establish a point assuming no other come-out rolls.

A 3rd guess would be to subtract two rolls because one roll is the come-out roll and the other is the 7-out.
2/3 x 6.5 = 4.33 rolls

If I wanted to count more than one come-out roll, somebody with more patience for math could step in and get a lower number than mine.
Some people roll 6 come-out rolls in a row, then they point 7-out.

The more you cover different numbers, the more you lose when they don't hit.

So, the shooter should roll 5 place numbers out of 8 rolls?

So, it should be possible to get 3 place numbers per shooter.

Would you say that is a correct mathematical statement?