What is more likely? Part 2

First, some intuition: Observe when the first time Event A happens, and assume it occurs earlier than Event B. Then before the three heads, there should be a tails (otherwise it's not the earliest occurrence), and before that there should be a heads (otherwise Event B would have occurred one toss ago). Thus Event A is actually HTHHH instead of only HHH! This should be suspicious enough that "Event B" now suddenly looks very probable as the answer. (In fact I answered this by using only this intuition, before making the rigorous solution below.)

---

We use Markov chain to denote the possible states of the past few coin tosses. For the states, we will give an ordered pair $(a,b)$ where $1 \le a \le 3, 1 \le b \le 4$ , indicating that this state is $a$ steps away from reaching Event A and $b$ steps away from reaching Event B. Of course, some of the ordered pairs are impossible; for example, $(3,1)$ is impossible, as $3$ steps away from Event A implies that the last toss was a tails (no heads to build the streak of three), while $1$ step away away from Event B implies that the last toss was a heads (since only one more heads for $TTHH$ to happen). In addition to these, we also have the states $A, B$ , indicating that Event A or B respectively has occurred earlier than the other.

First, the ordered pairs.

Thus the valid states are $(2,1), (3,2), (3,3), (1,4), (2,4), (3,4)$ , along with $A, B$ .

Then, state transitions:

• $(3,4)$ transitions to $(2,4)$ on heads or $(3,3)$ on tails.
• $(2,4)$ transitions to $(1,4)$ on heads or $(3,3)$ on tails.
• $(1,4)$ transitions to $A$ on heads or $(3,3)$ on tails.
• $(3,3)$ transitions to $(2,4)$ on heads or $(3,2)$ on tails.
• $(3,2)$ transitions to $(2,1)$ on heads or $(3,2)$ on tails.
• $(2,1)$ transitions to $B$ on heads or $(3,3)$ on tails.
• $A$ always stays in $A$ .
• $B$ always stays in $B$ .

In the following diagram, a red arrow is a transition on heads and an orange arrow is a transition on tails. $A$ and $B$ don't have arrows; once the game goes to either state, it will stay there forever.

---

We can use intuition again here. Note how many arrows lead to the chain leading to $B$ , while only $(3,3) \to (2,4)$ leads to the chain to $A$ , so this further strengthens the possibility that "Event B" is the correct answer. But still, we'll proceed further.

---

Let $P(E|X)$ be the probability that some event $E$ occurs (usually "arrive to a given state"), given that we are currently in state $X$ . Suppose $S$ is a state, and it can transition to $S_1, S_2, \ldots$ with $w_1, w_2, \ldots$ probability respectively. Then $P(E|S) = \sum w_i P(E|S_i)$ .

Now let $E=A$ , the probability that Event A occurs before B; that is, the probability that we arrive in state $A$ . Then $P(A|A) = 1, P(A|B) = 0$ . For the rest, we have a system of six linear equations in six variables, which technically is solvable...

Let $P(A|(3,4)) = a, P(A|(2,4)) = b, P(A|(1,4)) = c$ , $P(A|(3,3)) = d, P(A|(3,2)) = e, P(A|(2,1)) = f$ as shown in the diagram above. Then,

$\begin{cases} a = \frac{1}{2} b + \frac{1}{2} d \\ b = \frac{1}{2} c + \frac{1}{2} d \\ c = \frac{1}{2} \cdot 1 + \frac{1}{2} d \\ d = \frac{1}{2} b + \frac{1}{2} e \\ e = \frac{1}{2} e + \frac{1}{2} f \\ f = \frac{1}{2} d + \frac{1}{2} \cdot 0 \\ \end{cases}$

Fifth equation implies $e=f$ . Fourth and sixth equations, together with that, imply $\frac{3}{4} d = \frac{1}{2} b$ .

Second and third equations imply $b = \frac{1}{4} + \frac{3}{4} d$ . Combining with above gives $b = \frac{1}{4} + \frac{1}{2} b$ , or $b = \frac{1}{2}$ , so $d = \frac{1}{3}$ . Thus first equation gives $P(A|(3,4)) = a = \frac{5}{12}$ .

Similarly, we can compute $P(B|(3,4)) = \frac{7}{12}$ as well. This shows that Event B is more likely , with probability of $\frac{7}{12}$ over Event A's $\frac{5}{12}$ .

Since the difference in probabilities is not at all subtle it's easy to check with a Monte-Carlo experiment.

Let head is a sucess event with P= 0.5 In 1st case probability of happening = 3C3 (0.5)^3 =0.125 In 2nd case probability = 4C2 (0.5)^4=0.375

First we assume that the coin is fair, that is, the probabilities of heads and tails appearing is 0.5 apiece. Since the probability of each outcome is equal, we simply look at the number of tosses required to obtain each event in the problem.

Now imagine that we simultaneously toss two coins A and B repeatedly, producing the corresponding events in the problem. By the third toss of each coin, Event A will already have occurred in Coin A, while Coin B needs one more toss for Event B to occur.

Simply put, Event A will happen first because it is "shorter" than Event B.