Probability number comes up and then comes up again before another number I was trying to follow the logic in a similar question (Probability number comes up before another), but I can't seem to get it to work out.
Some craps games have a Repeater bet.  You can bet on rolling aces twice before rolling a 7, rolling 3 three times before 7, etc.  The patent for this game (https://patents.google.com/patent/US20140138911) says the odds for aces twice before 7 is 48:1.  The wizard of odds (https://wizardofodds.com/games/craps/appendix/5/) says the probability is 0.020408 (which is 1/49).
I tried calculating this by multiplying the odds of the two events 1/36 for rolling aces and (1/36)/((1/36)+(1/6)) for rolling aces before 7.  I got (1/36)*((1/36)/((1/36)+(1/6))) = 0.003968253968253969 which is like 1/252.
I'm obviously missing something, but can't see what.
Edit:...sorry...after typing this up i figured it out.  The bet has to be made and then aces has to roll before 7 twice.  So if 7 rolls before the first aces the bet loses, so I was wrong by using the 1/36 for the first aces.
((1/36)/((1/36)+(1/6)))*((1/36)/((1/36)+(1/6)))
0.020408163265306128
I still don't understand why one says 48:1 when its 1/49

Let $E_1$ denote the event that you roll snake eyes before a 7.
Let $E_2$ denote the event that you roll snake eyes before a 7, given that event E_1 has already occurred.
In fact, the chance of $E_2$ is the same as the chance of $E_1$. 
I simply separated the events for clarity.
The key formula here is that if
$A$ and $B$ are mutually exclusive events, and you are trying to compute the chance of $A$ happening before $B$, the probability is
$\frac{p(A)}{p(A) + p(B)}$, 
where $p(A)$ refers to the chance of event A occurring and 
$p(B)$ refers to the chance of event B occurring.
Here, there are 6 ways out of 36 to roll a 7 and 1 way out of 36 to roll snake eyes.  Therefore, it is immediate that 
$p(E_1) = \frac{1/36}{(1/36) + (6/36)} = \frac{1}{7}.$
Once event $E_1$ occurs the chance of event $E_2$ then occurring is similarly $= \frac{1}{7}.$
Thus the chance of events $E_1$ and then $E_2$ both occurring is 
$\frac{1}{7} \times \frac{1}{7}.$
