Rephrasing die question

Question

Seems I asked my last questions poorly and it got a bad response. Okay so basically here's the problem and I will explain how I am trying to do it. You roll 1 fair die repeatedly until you either [get a 1 followed immediately by a 2] or [you get a 2 followed immediately by a 2]. What is the probability that a [1 immediately followed by a 2] is rolled before a [2 immediately followed by a 2]?

These are my thoughts on the question:

At first I thought they were equally likely. Like I thought "if you roll a 1 then you have a 1/6 chance of rolling a 2 and same thing if you roll a 2 you have a 1/6 chance of rolling another 2, but then I realized that rolling a 1 means that if you roll a 2, that 2 doesn't count towards rolling another 2. I.e. the 1,2 outcome cuts into the 2,2 outcome so there is some dependence going on and I don't know how to account for that.

Maybe I'm over thinking it. Please help!

Answer 1

Your question is easiest to answer if we simplify it slightly. Given that all other numbers [3:6] are irrelevant the question is simply:

"Given that I've rolled a one OR a two what is the probability that I then roll a two."

Which is clearly 1/6. Having identified that the probability of the second roll given the first is ALWAYS 1/6 we need only consider whether we are more likely to have rolled a 1 or a 2 but these probabilities are exactly equal as well.

The answer is therefore that the probability of rolling a 1 followed by a two is exactly equal to the probability of rolling a 2 followed by a 2.

Answer 2

Let's aggregate all irrelevant digits into the digit '3' that occurs with probability $p_3 = \tfrac{2}{3}$, with $p_1 = p_2 = \frac{1}{6}$.

Consider the probability of winning $p(a)$ given the last digit $a$ in the sequence when the game is not yet decided: $$ \begin{eqnarray} p(1) &=& p_1 \cdot p(1) +p_2 \cdot 1 + p_3 \cdot p(3) \\ p(2) &=& p_1 \cdot p(1) +p_2 \cdot 0 + p_3 \cdot p(3) \\ p(3) &=& p_1 \cdot p(1) + p_2 \cdot p(2) + p_3 \cdot p(3) \end{eqnarray} $$ Solving the system:

In[4]:= Solve[{p1 == 1/6 p1 + 1/6 1 + 2/3 p3, 
  p2 == 1/6 p1 + 1/6 0 + 2/3 p3, p3 == 1/6 p1 + 1/6 p2 + 2/3 p3}, {p1,
   p2, p3}]

Out[4]= {{p1 -> 2/3, p2 -> 1/2, p3 -> 7/12}}

The probability of getting 1 immediately followed by 2 before getting 2 immediate followed by 2 equals $$ p_1 \cdot p(1) + p_2 \cdot p(2) + p_3 \cdot p(3) = \frac{7}{12} \approx 0.583 $$

Answer 3

I think I have a simplified answer. The reason why the (1,2) pair is more likely is because it "blocks" the chance of getting (2,2) had more die tosses happened. Since getting a 2 immediately after getting a 1 is a 1/6 chance given we've already rolled a 1, that means that on average, the (2,2) outcome will be "blocked 1/6th of the time by (1,2) (compared to if we were pitting (1,1) against (2,2) instead for example which should be equally likely). So that means that (1,2) should have a 1/6th advantage over (2,2) which means getting (1,2) first would have to be 7/12 prob and getting (2,2) first would have to be 5/12 prob.

To clarify even more, there is nothing "abnormal" about the first (leading) digit. It is only when we get a leading 1 that a trailing 2 causes the "interference" from equiprobable. Since this interference happens 1 out of 6 times (assuming we already have a leading 1), the advantage to (1,2) is as stated above. Note that it doesn't matter how the 1 gets into the leading spot, whether it was rolled first or as a previous trailing digit such as from (3,1). It is only when it moves to the leading position that we should become "concerned" about possible "interference". This assumes that we have a "sliding" window of 2 numbers such that a (3,1) would then become (1,?) meaning we saw a 1 previously and are waiting to see what comes next but the 3 in this example can be discarded (or maybe disdied) because it was part of a no decision.

Answer 4

I first saw this question at its (currently closed) posting, where André Nicolas found the answer ${7\over12}$ by a straightforward setting up and solving of a small system of linear equations. Given the relative simplicity of the result, it felt like there ought to be a solution that gets straight to the answer with less algebra -- something I could do in my head. I finally figured one out, so I'm posting it here.

Suppose you play the die-rolling game a huge number of times, beginning a new game afresh as soon as you've rolled a 12 or a 22. (By "afresh" I mean that each new game ignores the 2 that the previous game just ended with. This will prove to be important.)

When you're done, some number of games, $N_1$, will have ended 12, and some number, $N_2$, will have ended 22. We want to show that $N_1/N_2$ is approximately equal to $7/5$ -- that's the odds ratio that gives the probability ${7\over12}$ for ending 12 -- with the approximation getting better and better, of course, as the number of games goes to infinity.

Now let's simply look at the entire, long list of consecutive rolls, no longer caring where one game ends and the next begins. How many times do we see a 12? That's easy: It's just $N_1$. And how many times do we see a 22? That's a little trickier, but also easy. We certainly see it $N_2$ times. But we also see it every time the next game begins with a 2, which should be approximately one-sixth of the time, or approximately $(N_1+N_2)/6$ additional times.

Now here's the rub: These two counts should be (approximately) equal. That is, when you simply roll a die an enormous number of times, there's no reason to expect one particular string to appear (on average) any more often than any other string of the same length. This tells us that

$$N_1 \approx N_2 + {N_1+N_2\over6}$$

It doesn't take much algebra to see this reduces to ${5\over6}N_1\approx{7\over6}N_2$, which immediately gives the desired ratio, $N_1/N_2 \approx 7/5$. In the limit, you get equality.

Having written all this up, I see that the idea is essentially the same as in the answer by David to the posting here. But the exposition is somewhat different. In particular, the emphasis here on counting things that ought to be (roughly) equinumerous might help in certain more general settings.