Problem Statement

Mario and Luigi are throwing Fireballs. Each one throws a fireball twice.

Let:

• \(M_1, M_2 \sim \text{Exp}(1)\) — Mario’s two fireball throws
• \(L_1, L_2 \sim \text{Exp}(2)\) — Luigi’s two throws

All variables are independent.

Define:
\(M = \max\{M_1, M_2\}, \quad L = \max\{L_1, L_2\}\)

Find the probability: \(\mathbb{P}[M > 2L]\)

Original Problem Link: Click here

Solution

Let’s solve this step by step.

Step 1: PDF of Max(M₁, M₂)

Let’s say we want to find the PDF of \(M = \max(M_1, M_2)\), where each \(M_i \sim \text{Exp}(1)\).

To get this, let’s use a simple idea:
If the max is some value \(m\), then either:

• \(M_1 = m\) and \(M_2 < m\), OR
• \(M_2 = m\) and \(M_1 < m\)

The events are disjoint, so total probability is just the sum of both:

\[f_M(m) = \mathbb{P}[M_1 = m] \cdot \mathbb{P}[M_2 < m] + \mathbb{P}[M_2 = m] \cdot \mathbb{P}[M_1 < m]\]

Now:

• \[\mathbb{P}[M_i = m] = f_{M_i}(m) = e^{-m}\]
• \[\mathbb{P}[M_i < m] = F_{M_i}(m) = 1 - e^{-m}\]

So: \(f_M(m) = e^{-m}(1 - e^{-m}) + e^{-m}(1 - e^{-m}) = 2e^{-m}(1 - e^{-m})\)

Cool — we now have the PDF of Mario’s max distance.

Step 2: PDF of Max(L₁, L₂)

Same reasoning, but Luigi’s variables are \(\text{Exp}(2)\), so:

• \[f_{L_i}(l) = 2e^{-2l}\]
• \[F_{L_i}(l) = 1 - e^{-2l}\]

Using the same logic as above:

\[f_L(l) = 2 \cdot f_{L_1}(l) \cdot F_{L_2}(l) = 2 \cdot 2e^{-2l} \cdot (1 - e^{-2l}) = 4e^{-2l}(1 - e^{-2l})\]

Step 3: Compute \(\mathbb{P}[M > 2L]\)

Let’s now compute the probability \(\mathbb{P}[M > 2L]\).
We’ll condition on \(L = l\), compute \(\mathbb{P}[M > 2l]\), then integrate over all \(l\).

So:

\[\mathbb{P}[M > 2L] = \int_0^\infty \mathbb{P}[M > 2l] \cdot f_L(l)\, dl\]

We already know \(f_M(m) = 2e^{-m}(1 - e^{-m})\), so:

\[\mathbb{P}[M > 2l] = \int_{2l}^\infty f_M(m)\, dm = \int_{2l}^\infty 2e^{-m}(1 - e^{-m})\, dm\]

Let’s compute that inner integral first.

Step 4: Evaluate \(\int_{2l}^\infty 2e^{-m}(1 - e^{-m})\, dm\)

Let’s expand the expression:

\[\int_{2l}^\infty 2e^{-m}(1 - e^{-m})\, dm = \int_{2l}^\infty \left(2e^{-m} - 2e^{-2m}\right)\, dm\]

Break it into two integrals:

\[= \left[ -2e^{-m} + e^{-2m} \right]_{2l}^\infty = (0 + 0) - \left( -2e^{-2l} + e^{-4l} \right) = 2e^{-2l} - e^{-4l}\]

So: \(\mathbb{P}[M > 2l] = 2e^{-2l} - e^{-4l}\)

Step 5: Plug into Total Integral

Now plug this into the full expression:

\[\mathbb{P}[M > 2L] = \int_0^\infty \left(2e^{-2l} - e^{-4l} \right) \cdot 4e^{-2l}(1 - e^{-2l})\, dl\]

Let’s expand this:

\[= \int_0^\infty \left(2e^{-2l} - e^{-4l}\right) \cdot 4e^{-2l}(1 - e^{-2l})\, dl = \int_0^\infty \left[8e^{-4l}(1 - e^{-2l}) - 4e^{-6l}(1 - e^{-2l})\right]\, dl\]

Break into two integrals:

• First: \(\int_0^\infty 8e^{-4l}(1 - e^{-2l})\, dl = \int_0^\infty (8e^{-4l} - 8e^{-6l})\, dl = 2 - \frac{4}{3} = \frac{2}{3}\)

• Second: \(\int_0^\infty 4e^{-6l}(1 - e^{-2l})\, dl = \int_0^\infty (4e^{-6l} - 4e^{-8l})\, dl = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}\)

Now combine both parts:

\[\mathbb{P}[M > 2L] = \frac{2}{3} - \frac{1}{6} = \boxed{\frac{1}{2}}\]

The required probability is:

\[\boxed{\frac{1}{2}}\]

---

💡 Did you enjoy this problem? Check out more puzzles in the Problems section!