Consider the following game.
A game involves rolling two six-sided dice, followed by rolling a
third six-sided die. To win the game, the number rolled on the third
die must lie between the two numbers rolled previously. For example,
if the first two dice show $1$ and $4$, the game can only be won by
rolling a $2$ or $3$ with the third die.
I am trying to find the probability that a player has no chance of winning before rolling the third die.
My thought process was that this only occurs when the first two rolls are the same number or consecutive numbers. Both of these have probability $1/6$ and so the probability of losing before the third roll is $1/6+1/6=1/3$.
However, the solution states the probability is $4/9$. I do not understand how my logic is incorrect.
Best Answer
Notice the probability of having consecutive numbers is actually $\dfrac{4}{9}$. In this case, the first two dice rolls are distinct, so we have $5 \cdot 2$ ways for the two dice out of $36$ total was. We get $5 \cdot 2$ because we have $(1, 2), (2, 3), \dots, (5, 6)$ and the same thing with the ordered pairs reversed. From $\dfrac{5}{18} + \dfrac{1}{6}$, the answer is $\dfrac{4}{9}$ as your solution states
Hope this helps your understanding!