Quant Interview Question - Die Rolls

Question

You have a fair 6 sided die. You roll the die and keep track of the outcomes as they appear. You roll until you see two 5s (not necessarily consecutive) or both 4 and 6. Find the probability you stop rolling due to seeing two 5s (this is on the site QuantGuide).

My approach was to set up some recursive equations and then solve for them. You start of in position A.

Here are the description of the positions

A - Neutral.

B - Rolled a 5.

C - Rolled a 4 (No 5 Rolled)

D = Rolled a 4 having rolled a 5 previously.

$A = \frac46 \cdot A + \frac16 \cdot B + \frac16 \cdot C$

$B = \frac16 \cdot D + \frac46 \cdot B + \frac16 $

$C = \frac16 \cdot C + \frac16 \cdot B + \frac36 \cdot A$

$D = \frac16 \cdot D + \frac36 \cdot B + \frac16$

I solved this and got $A = \frac{36}{49}$ but the site is saying that is not correct. Anyone have any approaches to this problem or a better way to solve it.

Answer 1

$\mathtt{ One\; interpretation\; of\; the\; question}$

$4$ and $6$ may occur in any order consecutively to "win"

With this interpretation. and using $S$ (for start and neutral state}, and $A,B,C$ for states $4,6,5$ respectively and probabilities moving between states with identical but small case letters, we have

$s = \frac36 s + \frac16 a + \frac16 b + \frac16 c \tag1$

$a = \frac36 s + \frac16 a + \frac16*0 + \frac16 c \tag2$

$b = \frac36 s + \frac16* 0 + \frac16 b + \frac16 c \tag3$

$c = \frac36 s + \frac16 a + \frac16 b + \frac16 * 1 \tag4$

which yields P(see two $5's$ before a $ 4$ and a $6$) $\Large = s = \frac1{3}$

Answer 2

The interpretation that makes the most sense to me is that consecutivity of rolls is never required. So the process stops once you have seen a ⚃ and ⚅ previously or a 2 ⚄'s previously. For shortness I'll use game language and say you win or loose depending on if the stop happens on a 5 or not.

At any given time, since the die is fair, the 6 potential orders in which you will see the first next 4,5 or 6 are all equally likely.

It turns out that looking at when the first 5 occurs is enough, so there are 3 cases, each with probability $\frac13$.

1. You see the 5 first.

Again, from now on the 6 orders of first 4,5 or 6 are equally likely. You win if the first 5 from now on (which is the second 5 overall) is on position 1 or 2, you loose if it is on position 3. That means you win in this case with probability $\frac23$

2. You see the 5 second.

Now it depends which number comes next, 5 or the other number from $\{4,6\}$ you haven't seen yet. Both outcomes are equally likely, so you win with probability $\frac12$ in this case.

3. You see the 5 last.

You already lost, the game stopped once the second of $\{4,6\}$ was rolled. So the probability to win is $0$ in the case.

That means the overall probability to stop on a 5 is

$$\frac13\left(\frac23 + \frac12 + 0\right)= \frac7{18}$$