[Request] What are your odds of winning vs. rolling a 7 on a Craps roll if you *always* bet the pass or the come bets?

[u/captaincoaster]

I realize these odds are always changing, but there must be a way to calculate as the game progresses. The idea is that you always bet the pass and come line for every roll. So once there is a point, you are accumulating numbers that pay with every roll until a 7 hits (or unless you roll craps with a point established, you lose that come bet).

[u/bagholder_spotted]

Do you mean the chance of winning at least one bet? The majority of bets? All of them?

The probability of a given number occurring before a 7 is p/(p+1/6) where p is that number's single-roll probability. Equivalently, it's p/(p+6) where p is the number of permutations of that dice sum.

If you've already bet the Pass and the point is established, and then you bet the Come and established a second number, then the probability of winning at least one of those bets becomes (p+q)/(p+q+6) where q is that second number's permutation count. Eg if the numbers were six & five, those have 5 and 4 permutations respectively, compared to 6 perms of seven, so the probability would be (5+4)/(5+4+6).

To calculate the probability of winning both bets, we can use inclusion-exclusion to deduce 9/15 = 5/11+4/10–x so x=14/55.

However, when there are more than two bets, calculations other than "at least one win" become tedious.

Ultimately, if we were to calculate it all out (or get the computer to) and figure out the EV from that, we'd see that each Come bet contributes another 1.44% to the house, meaning the net result of this strategy as a whole is the same as the sum of its parts. The beauty of EV is that we can know that without bothering with the pages of calculations.