Craps Repeater Side Bet Odds A craps side bet named Repeater has the following true odds: Repeating "2" two times before rolling 7, odds is 48:1 Repeating "3" three times before rolling 7, odds is 63:1 Repeating "4" four times before rolling 7, odds is 80:1 Repeating "5" five times before rolling 7, odds is ~100:1 Repeating "6" six times before rolling 7, odds is ~112:1 I want to know the logic and mathematical formulas in the establishment of the above odds.

Here's how to do the first one, as an illustration:

Note that the only throws that matter are $2's$ and $7's$, we can disregard the rest. The ordinary probability of getting a $2$ is $\frac 1 {36}$ while the ordinary probability of getting a $7$ is $\frac 6{36}$. Hence, if we only consider those two possibilities the adjusted probability of getting a $2$ is $\frac 17$ while the adjusted probability of getting a $7$ is $\frac 67$.

The problem is now quite simple. The only "winning" path for this bet is the one that starts $2,2$ and the probability of that is $(\frac 17)^2=\frac 1{49}$. that probability corresponds to odds of $48:1$.