# Probability of Losing a Dice Game

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.

• The probability of the first two rolls being consecutive numbers isn't $1/6$. You could roll $12$, $23$, $34$, $45$, or $56$, in either order, for a total of $10$ possibilities, and a probability of $10/36=5/18$. (Otherwise, your reasoning is correct.) Jul 27, 2023 at 3:17
• Silly question, but if we can roll $12, 23, 34, 45$ or $56$ in any order, why do we not have $12$ possibilities of rolling the same number (i.e. while $11$ is the same as $11$, can we not distinguish these)?
– Bell
Jul 27, 2023 at 3:38
• As in your comment, you can roll 12, 23, 34, 45, 56 or their reversed orders, which are just $10$ possibilities out of $36$. Jul 27, 2023 at 3:53
• Right... 11 is the same as 11, we can't distinguish those (because they're identical). Another way to see this: after the first roll, whatever it is, there's only one possibility for the second roll that makes the two rolls identical (probability is $1/6$). There's either one possibility ($1/3$ of the time) or two ($2/3$ of the time) for the second roll that makes the two rolls consecutive (probability is $1/18 + 4/18 = 5/18$). Jul 27, 2023 at 4:55

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
Probability of having the same number is $$\frac16$$ however to have consecutive numbers the probability is $$\frac5{18}$$ which sums to $$\frac49$$
• Why is the probability $5/18$ and not $5/36$? Have you assumed that the roll $1$ and $2$ is different from the roll $2$ and $1$?
Simple, you’re being asked the probability of the second dice being equal in value to the first having having a difference in value of one. If the first die is 2,3,4 or 5, the the probability is clearly $$3/6=1/2.$$ However if the first result is 1 or 6, then the probability is $$2/6=1/3$$. And finally $$2/3*1/2+1/3*1/3=4/9$$.