Caldazar said:

It does seem puzzling. To interpret, I belive you are saying that: On a given don’t pass hand with a $10 bet, let’s say **the point is a 10.** Right now the odds are 2 to 1 in your favor of winning your bet (66%). If you put down an additional $10, that additional $10 odds has (50%). So why would you want to average your winning bet with the 50% bet, you’re just bring your overall average down, eh?

Which is sound reasoning as long as you stick to the percentages. However, you have also increased your bet to $20 instead of $10.

Before odds:

1 out of 3 times you lose $10

2 out of 3 times you win $10

Expected value =

(1/3)*(-10) + (2/3)*(+10) = +$3.33

With single odds:

1 out of 3 times you lose $20

2 out of 3 times you win $15

Expected value =

(1/3)*(-20) + (2/3)*(15) = -6.66 + 10 = +$3.33

Therefore, we deduce there ain’t a damn bit of difference in the probability side of things, the only difference is the amount of action you got going. As you said, the *expectation percentage* has changed, but the actual *expected value* has not. The percentage has changed in exact proportion to the amount of extra money you’ve put out behind the line.

This particular example works if the point is 4 or 10, I feel pretty dang confident the math will check for the other points also.

Hope this helps to clarify.