# Qestion on Craps

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.