Dice probability when a number is disallowed in the first round

Question

In the game Settlers of Catan a player starts each turn rolling 2 six sided dice. There's a variation of the game where if a 7 is rolled in the first round of a game (a 'round' is when each player has taken a turn) it doesn't count an needs to be re-rolled. This must lower the overall probability of a 7 being rolled for the rest of the game (and I'd imagine increase the odds for all the other numbers). How would this probability be calculated?

Answer 1

There are exactly 6 ways to roll a seven with two dice:

\begin{array}{|c|c|} \hline \text{First die} & \text{Second die} \\ \hline 1 & 6 \\ \hline 2 & 5 \\ \hline 3 & 4 \\ \hline 4 & 3 \\ \hline 5 & 2 \\ \hline 6 & 1 \\ \hline \end{array}

The total number of ways of rolling two dice is $6 \times 6 = 36$, since there are $6$ ways of rolling each die individually. Therefore, if you don't reroll sevens, you will get a seven with probability $\frac{6}{36} = \frac16$.

If you do reroll sevens, during the time you reroll them you will have $0$ probability of getting a seven. You will keep rerolling until you get some other case, and each other case is still equally likely, so now there are only $36 - 7 = 29$ cases left, instead of the usual $36$.

The effect of the first round of the game has no bearing on the probability of a seven being rolled for the rest of the game. This is the nature of probability; the two parts of the game are entirely independent. However, one other thing we can compute that maybe you wanted to know is how much rerolling sevens increases the probability of rolling other things. Well, each individual case (such as 1, 1 or 3, 1) now has probability $\frac1{29}$ of being rolled instead of the original $\frac{1}{36}$, so the probability has increased by $$ \frac{1/29}{1/36} = \frac{39}{29} \approx 1.24137 $$ when sevens are rerolled. So each number (2,3,4,5,6,8,9,10,11,12) appears about $24$ percent more often than normal during the first round.