# Dice probability when a number is disallowed in the first round

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?

• "This must lower the overall probability of a 7 being rolled for the rest of the game..." Do you mean you want to compute the probability of rolling a 7 for subsequent rolls? This probability is unaltered. How could the dice remember that a 7 was not allowed in the first round? Perhaps instead you mean "What is the probability of rolling a 7 during the entire game (including the first round)?". This will indeed be a little lower, but one would need to know the number of rolls in a game to compute it. – Austin Mohr May 17 '13 at 22:09
• your correct, I was wondering the probability of rolling a 7 for the entire game. – greatwitenorth May 18 '13 at 0:33

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.

• Great answer! Thanks. – greatwitenorth May 18 '13 at 0:30
• how would I go about calculating the probability for the entire game and not just the first round? – greatwitenorth May 18 '13 at 2:19
• @greatwitenorth I assume you mean the average probability of a seven for the whole game. On average, if there are $N$ rounds, the first will have a $0$ percent chance and the other $(N - 1)$ will have a $\frac16$ chance, so the average probability will be $\frac{1 \cdot 0 + (N-1)\frac16}{N} = \frac{N-1}{6N}$. As a mildly educated guess, let's say since you have to get 10 victory points to win, and you already have 2, the game will take 2 rounds for each additional victory point, in total 16 rounds. Then we have that the average probability is about $\frac{15}{6 \cdot 16} = \frac{5}{32}$. – 6005 May 18 '13 at 5:43
• This is approximately $.156$. So about 15.6 percent of all dice rolls in the game are sevens. If there were no special rules for the first round, about 16.7 percent (1/6) of all rolls would be sevens. – 6005 May 18 '13 at 5:45
• Let me know if this is what you wanted or if there is anything else you wanted to ask for. – 6005 May 18 '13 at 5:57