Dice throwing probability

Question

In a round of a certain game, you throw a fair dice. If the outcome of the dice is greater or equal to 4, you score the number shown on the dice, and the round ends. If the dice shows a number strictly less than 4, you throw the dice again forgetting the outcome of the first throw and scoring the new number shown on the dice at the second throw. Thus each round consists of two throws at most. You play two rounds of this game. What is the probability that the sum of the outcomes of the two rounds is equal to 10?

Answer 1

Let's start by calculating the probabilities of each outcome (1...6) of a single round. We get $$ P(1)=P(2)=P(3) = \frac{3}{6} \cdot \frac{1}{6} = \frac{1}{12} $$ and $$ P(4)=P(5)=P(6) = \underbrace{\frac{3}{6}\cdot \frac{1}{6}}_{\text{two throws}} + \underbrace{\frac{1}{6}}_{\text{one throw}} = \frac{1}{4} $$ Now, if we have two rounds, we can find all the ways of getting a total of $10$ points: $$ 4+6 : \quad P(4)\cdot P(6) = \frac{1}{8} $$ $$ 5+5 : \quad P(5)\cdot P(5) = \frac{1}{8} $$ $$ 6+4 : \quad P(6)\cdot P(4) = \frac{1}{8} $$ Those are all the cases where we can obtain 10. Therefore, the final probability is $\frac{3}{8}$.

Answer 2

You have 4 'paths' in total (T stands for Throw):

1)1T, 1T: (4,5,6), (4,5,6)

2)2T,1T, (1...6), (4,5,6)

3)1T,2T (4,5,6), (1...6)

4)2T,2T (1...6), (1...6)

$$ P(S) = \sum_{k=1}^{4} P(S|Path_k)P(Path_k) = \frac{3}{9} \times \frac{1}{4} + \frac{3}{18}\times \frac{1}{4} + \frac{3}{18}\times \frac{1}{4} + \frac{3}{36}\times\frac{1}{4} $$

Answer 3

Because in one throw, total number$\leq 4$,thus we have three paths;

(1)T1(456);T2(123)(123456);

(2)T1(123)(123456);T2(123)(123456);

(3)T1(123)(123456);T2(456);

(4)T1(456);T2(456);

while in two throws,total number$ N \in[2,9]$,

$P(N=2)=P(N=9)=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$;

$P(N=3)=P(N=8)=\frac{1}{6}\times\frac{1}{6}\times 2=\frac{1}{18}$;

$P(N=4)=P(N=5)=P(N=6)=P(N=7)=\frac{1}{6}\times\frac{1}{6}\times 3=\frac{1}{12}$;

therefore $P(1)=\frac{1}{6}\times\frac{1}{12}\times 3=\frac{1}{24}$;

$P(2)=\frac{1}{36}\times\frac{1}{18}\times 2+\frac{1}{18}\times\frac{1}{12}\times 2+\frac{1}{12}\times\frac{1}{12}\times 3=\frac{43}{1296}$;

$P(3)=\frac{1}{24}$

$P(4)=\frac{1}{6}\times\frac{1}{6}\times 3=\frac{1}{12}$;

$P=P(1)+P(2)+P(3)+P(4)=\frac{259}{1296}$