My Initial Approach

Like many mathematicians, I immediately dived into the calculations. I concluded that the toad could either cross on her first, second, third, or subsequent attempts. Each observation is an independent event, so I had to sum the probabilities for all attempts.

• If she crosses on the first look, the probability is 0.1.
• For the second look, the sequence is unsafe on the first and safe on the second, yielding a probability of 0.9 * 0.1.
• Following this pattern, if she crosses on the third attempt, it is unsafe on the first two looks and safe on the third, giving a probability of 0.9² * 0.1, and so forth.

Thus, the probability that the toad ultimately crosses the street can be formulated using the geometric series with a first term of 1 and a common ratio of 0.9. Therefore, it is certain that she will cross the street eventually.

However, did I genuinely need to undertake such extensive calculations? The setup clearly indicated that the toad only crosses when it is safe, suggesting there is no risk of an incident. Hence, it is certain she will cross without incident. The answer was implicit in the question; I simply failed to recognize that before diving into the math.

Doh!

The Next Challenge

The subsequent part of the problem required genuine mathematical reasoning. Let’s analyze it.

An immortal frog also attempts to cross the same street. He examines the street every minute, but he’s more impatient than the toad. If the street is unsafe, he might still attempt to cross with a probability of n/3 (where n is the number of minutes since his first look), applicable for n ≥ 3. If he crosses when it’s unsafe, there’s a 0.8 probability of an incident occurring. Determine the likelihood that he crosses the street safely.

To tackle this, we must systematically organize our information. We can outline a series of events at Minutes 0, 1, 2, and 3, each with four possible outcomes:

• Safe: It is safe to cross, and the frog proceeds.
• Unsafe Good: It is unsafe, but the frog crosses successfully.
• Unsafe Bad: It is unsafe, and the frog crosses but faces an incident.
• Unsafe Neutral: It is unsafe, and the frog opts not to cross.

Using this information, we can create a comprehensive table:

After constructing this table, we can verify that each column sums to one, serving as a useful checksum.

The first two rows of the table correspond to crossing the street without incident, and it’s clear the frog will either succeed or face an incident within three minutes.

Now, we can compute the probability that the frog crosses without incident. The scenarios are: 1. He succeeds right away (minute zero) — probability of 0.1. 2. He waits until minute one but succeeds then — probability of 0.9 * 0.16 = 0.144. 3. He delays until minute two but succeeds — probability of 0.9 * 0.6 * 0.22 = 0.1188. 4. He waits until minute three but succeeds — probability of 0.9 * 0.6 * 0.3 * 0.28 = 0.04536.

Adding these probabilities gives us a total of 0.40816.

The Next Question

Now, let’s address the seemingly tougher question: What is the probability that both the toad and the frog cross the street without incidents, with the frog crossing in less time?

At first glance, this seems complicated. I started to enumerate the possibilities and calculate their probabilities. However, a more intuitive approach reveals that the frog can only cross faster if he crosses unsafely yet successfully (otherwise, he either faces an incident or matches the toad’s crossing time).

Therefore, we can return to our table and sum the probabilities of the frog crossing unsafely but without incident (the “Unsafe Good” option). This can occur at: - Minute 1 (probability 0.9 * 0.06 = 0.054) - Minute 2 (0.9 * 0.6 * 0.12 = 0.0648) - Minute 3 (0.9 * 0.6 * 0.3 * 0.19 = 0.02916).

The cumulative probability amounts to 0.14796, showcasing that no additional calculations were necessary.

The Final Inquiry

If the frog hasn’t reached the other side after 2 minutes, what is the probability that an incident has occurred?

Again, I could adopt the mathematician's method by applying Bayes’ Theorem. However, relying on common sense using our table is more straightforward. We can easily compute the likelihood of an incident occurring, which is 0.9 * 0.24 + 0.9 * 0.6 * 0.48 + 0.9 * 0.6 * 0.3 * 0.72 = 0.59184.

Next, we calculate the probability that the frog hasn’t crossed after 2 minutes by subtracting the probability of crossing from one: 1 - (0.1 + 0.9 * 0.16 + 0.9 * 0.3 * 0.22) = 0.6372.

Consequently, the probability that an incident occurred given the frog hasn’t crossed after 2 minutes is 0.59184 ÷ 0.6372 = 0.9288136. It doesn’t bode well for our frog!

What are your thoughts on this probability problem and my demonstration of the value of utilizing common sense approaches? I invite your comments!