Last Updated on February 20, 2026 by Rajeev Bagra
When learning probability, many people feel confused when they see expressions like:
[latex]\frac{1}{3}p + \frac{1}{3}[/latex]
The natural reaction is:
“Why are we adding these numbers? Where do they come from?”
If you’ve ever asked this question, you’re not alone.
In this article, we’ll explain this idea using simple reasoning, everyday thinking, and the Monty Hall problem as an example — without heavy mathematics.
The Setup: A Biased Monty Hall Scenario
Let’s start with the context.
- There are three doors.
- One door has a car.
- Two doors have goats.
- You choose Door 1.
- Monty opens one goat door.
- If Monty has a choice, he prefers Door 2 with probability p.
Our goal is to find:
The probability that Monty opens Door 2.
In math form:
[latex]P(M=2)[/latex]
This is where the expression
[latex]\frac{1}{3}p + \frac{1}{3}[/latex]
comes from.
Let’s understand why.
Step 1: Think in Terms of Many Games
Probability becomes easier when we imagine repeating the game many times.
Suppose we play 300 games.
Because the car is placed randomly:
- About 100 games → Car behind Door 1
- About 100 games → Car behind Door 2
- About 100 games → Car behind Door 3
So each case happens about one-third of the time:
[latex]P(\text{Car behind each door})=\frac{1}{3}[/latex]
Step 2: When Can Monty Open Door 2?
Now we analyze each case separately.
Case 1: Car Behind Door 1
Probability:
[latex]\frac{1}{3}[/latex]
In this case:
- Door 2 has a goat
- Door 3 has a goat
- Monty has a choice
Because Monty prefers Door 2 with probability p, he opens Door 2 with:
[latex]P(M=2 \mid \text{Car=1})=p[/latex]
So this case contributes:
[latex]\frac{1}{3}\times p[/latex]
This is the first part of our formula.
Case 2: Car Behind Door 2
Probability:
[latex]\frac{1}{3}[/latex]
Here:
- Door 2 has the car
- Monty cannot open it
So:
[latex]P(M=2 \mid \text{Car=2})=0[/latex]
This case contributes nothing.
Case 3: Car Behind Door 3
Probability:
[latex]\frac{1}{3}[/latex]
Here:
- Door 3 has the car
- Door 2 has a goat
- Monty must open Door 2
So:
[latex]P(M=2 \mid \text{Car=3})=1[/latex]
This contributes:
[latex]\frac{1}{3}[/latex]
Step 3: Add All Door-2 Openings
Now we count all situations where Door 2 is opened.
From Case 1:
[latex]\frac{1}{3}p[/latex]
From Case 3:
[latex]\frac{1}{3}[/latex]
Add them:
[latex]P(M=2)=\frac{1}{3}p+\frac{1}{3}[/latex]
That’s where the formula comes from.
Why Do We Add These Terms?
We add them because these cases never overlap.
In one game, the car is behind:
- Door 1
- OR Door 2
- OR Door 3
Never more than one.
So we are counting results from separate situations.
When counting separate situations, we add them.
This idea is known in statistics as the Law of Total Probability, which says:
[latex]P(A)=\sum P(A\mid B_i)\cdot P(B_i)[/latex]
In simple words:
Total probability = Add the probability from each case.
A Simple Analogy
Imagine three online stores.
Each gets one-third of your customers.
You want to count how many customers use a special coupon.
- Store A → Some use it
- Store B → None use it
- Store C → All use it
To find the total, you add:
Store A + Store C
Same logic.
Why This Matters in Real Life
This way of thinking appears everywhere:
Business
- Revenue from different channels
- Sales from different products
Web Development
- Errors from different modules
- Speed from different scripts
SEO
- Traffic from different keywords
- Conversions from different pages
In every case:
Break → Analyze → Add.
Final Takeaway
The expression
[latex]\frac{1}{3}p+\frac{1}{3}[/latex]
is not magic.
It simply means:
Some Door-2 openings happen when the car is behind Door 1.
Some happen when it is behind Door 3.
We count both.
Once you understand this counting logic, probability becomes much easier.
And that’s the real lesson behind this example.
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