A short series that builds intuition for quantum mechanics (without asking you to become a monk)
Series map
- File 001 / Part 1: Geometry, vectors, and the death of “50/50”
- File 002 / Part 2: Claude Shannon, Bayes, and the price of ignoring likelihoods
- File 003 / Part 3: Quantum mechanics and the power of projection
- File 004 / Part 4: Wolfram’s multiway systems (and the Ruliad) in toy form
Part 2 — Claude Shannon and the Valve That Moves 2/3 of the Water
Previously:
In Part 1 we treated probability like a conserved fluid in a three-chamber system.
You chose a door, which split the fluid into two reservoirs:
- 1/3 pooled on your chosen door
- 2/3 pooled in the other two doors as a combined reservoir
- Then Monty acted—not as a random person, but as a valve-and-filter device governed by rules.
- And that’s why “two doors left” did not mean “equal odds.”
Now we do the same story with a different instrument panel:
- Bayes tells you how the fluid must reroute.
- Shannon tells you how much information the valve actually transmitted.
The Core Claim
The Monty Hall paradox is not a paradox. It’s a misdiagnosis.
The 50/50 answer is what you get when you pretend Monty is not a valve.
But Monty is absolutely a valve:
- His output depends on the hidden truth.
- Therefore his output carries information.
- Therefore the fluid must move.
Bayes as Plumbing Law
Bayes’ theorem:
$$
P(H\mid E)=\frac{P(E\mid H)\,P(H)}{P(E)}
$$
Translated into fluid mechanics:
- How much fluid is in the reservoir before the measurement?
- Reservoir: $$H$$
- $$P(H)$$
- The valve openness: “if this world were true, how often would the apparatus output this evidence?”
- $$P(E\mid H)$$
- The post-measurement fluid level after filtering and renormalizing
- $$P(H\mid E)$$
- The total fluid that survives the filter (the normalization constant)
- $$P(E)$$
In one line:
Posterior fluid ∝ prior fluid × valve openness.
Then renormalize (pour it back to sum to 1).
Define the State Space (three chambers, three worlds)
We define three hypotheses:
Car is in Room 1: $$H_1$$
Car is in Room 2: $$H_2$$
Car is in Room 3: $$H_3$$
Your choice: Room 1.
Evidence $$$E$: Monty opens Room 3 and shows a goat.
At the start, the fluid is uniform:
$$
P(H_1)=P(H_2)=P(H_3)=\frac{1}{3}
$$
The Likelihood Function: the valve’s spec sheet
This is the whole problem. The likelihood is not decoration. The likelihood is the apparatus.
| Choice | Result: |
| $$H_1$$ | True -> (car in Room 1) You picked the car. |
| Monty: Can open either goat door (2 or 3). If he chooses uniformly: | $$ P(\text{opens R3}\mid H_1)=\frac{1}{2} $$ |
| $$H_2$$ | True -> (car in Room 2) You picked a goat. |
| Monty: Cannot open your door (1) and cannot open the car door (2), so he must open 3: | $$ P(\text{opens R3}\mid H_2)=1 $$ |
| $$H_3$$ | True -> (car in Room 3) |
| Monty: Cannot open 3 (it’s the car), so: | $$ P(\text{opens R3}\mid H_3)=0 $$ |
Likelihood table (valve openness):
| Hypothesis | $$P(\text{opens R3}\mid H_i)$$ |
|---|---|
| $$H_1$$ | 1/2 |
| $$H_2$$ | 1 |
| $$H_3$$ | 0 |
Interpretation in one sentence:
Under $$H_2$$…the valve is fully open toward “open R3.”
Under $$H_1$$…it’s half open.
Under $$H_3$$…it’s sealed shut.
That difference is the information.
Compute P(E): how much fluid survives the filter?
$$P(E)$$
$$
P(\text{opens R3})=\sum_i P(\text{opens R3}\mid H_i)\,P(H_i)
$$
With priors $$1/3$$:
$$
P(\text{opens R3})
=\frac{1}{2}\cdot\frac{1}{3}+1\cdot\frac{1}{3}+0\cdot\frac{1}{3}
=\frac{1}{6}+\frac{1}{3}
=\frac{1}{2}
$$
- So the filter admits total mass $$1/2$$.
- Renormalization will divide by $$1/2$$
- This is where “2/3” is born.
Compute the Posterior: where the fluid actually goes
$$
P(H_i\mid E)=\frac{P(E\mid H_i)P(H_i)}{P(E)}
$$
Room 1
$$
P(H_1\mid \text{opens R3})
=\frac{\frac{1}{2}\cdot\frac{1}{3}}{\frac{1}{2}}
=\frac{1}{3}
$$
Room 2
$$
P(H_2\mid \text{opens R3})
=\frac{1\cdot\frac{1}{3}}{\frac{1}{2}}
=\frac{2}{3}
$$
Room 3
$$
P(H_3\mid \text{opens R3})
=\frac{0\cdot\frac{1}{3}}{\frac{1}{2}}
=0
$$
Result: $$(1/3,\ 2/3,\ 0)$$
In fluid language:
- You observed “open R3,” so the reservoir drains to zero: $$H_3$$
- The remaining fluid is redistributed—not evenly—because the valve was not symmetric.
And that’s switching: step into the door that inherited the larger reservoir.
The 50/50 error, precisely stated
The “50/50” reasoner silently installs a fake valve.
They act like Monty’s behavior under $$H_2$$ is “coin flip,” when it is forced.
That corresponds to a naive likelihood model:
| Hypothesis | naive $$P(\text{opens R3}\mid H_i)$$ |
|---|---|
| $$H_1$$ | 1/2 |
| $$H_2$$ | 1/2 |
| $$H_3$$ | 0 |
This is literally claiming the valve is half-open under $$H_2$$, when it is fully open.
Use that fake valve and you’ll get $$P(H_1\mid E)=1/2$$ and $$P(H_2\mid E)=1/2$$.
So now the issue is clean:
50/50 is what you get when you model the apparatus incorrectly.
2/3 is what you get when you respect the valve.
- Notice how “two doors remain” never entered the math.
- Because the fluid doesn’t care about vibes. It obeys constraints.
Shannon Entropy: measuring uncertainty as a volume
Now we measure “how uncertain we are” before and after the valve acts.
Shannon entropy:
$$
H=-\sum_i p_i\log_2 p_i
$$
Before Monty speaks
Uniform over 3 states:
$$
H_{\text{before}}
=-3\cdot\frac{1}{3}\log_2\frac{1}{3}
=\log_2 3
\approx 1.585\ \text{bits}
$$
This is the “spread” of the fluid across possibilities.
After Monty opens Room 3:
Posterior $$(1/3,\ 2/3,\ 0)$$:
$$
H_{\text{after}}
=-\frac{1}{3}\log_2\frac{1}{3}
-\frac{2}{3}\log_2\frac{2}{3}
-0\cdot\log_2 0
$$
By convention, $$0\log 0=0$$.
Numerically:
$$
H_{\text{after}}\approx 0.918\ \text{bits}
$$
Information gained
$$
I=H_{\text{before}}-H_{\text{after}}
=1.585-0.918
=0.667\ \text{bits}
$$
- Monty’s valve, in this branch, transmits about two-thirds of a bit.
- Not a full bit.
- Not nothing.
- A fractional, protocol-shaped amount.
Monty as a Channel
In Shannon’s language:
- Input: Car Location: $$X$$
- Output: Which door Monty opens (given your choice):$$Y$$
That mapping $$X\to Y$$ is a channel—a rule-governed mechanism that turns hidden truth into an observation.
And it’s a weird channel:
- If you picked the car, Monty’s output is partially random (coin flip)
- If you picked wrong, Monty’s output is forced (deterministic)
- So the channel is informative in a structured way.
- It doesn’t emit facts. It emits lawful constraints.
- Which is exactly what measurement is…in Quantum Mechanics too.
Conditional Independence: the incorrect assumption behind 50/50
The wrong model assumes Monty’s action is conditionally independent of the car location given your choice:
$$
P(M\mid C,\ \text{Choice}) = P(M\mid \text{Choice})
$$
- Meaning: “once I know what I picked, Monty’s reveal tells me nothing about the car.”
- That is false.
- The correct causal sketch is:
Car Position ---> Monty’s Action <--- Your Choice- Monty’s action depends on both variables.
- So observing Monty is observing a channel output that reacts to the hidden input.
- The 50/50 reasoner deletes the arrow:
Car Position → Monty’s Action
…and then wonders why reality refuses to become symmetric.
Mutual Information: the clean formal statement
Mutual information here is:
$$
I(\text{Car};\text{Monty}\mid\text{Choice})
=H(\text{Car}\mid\text{Choice})-H(\text{Car}\mid\text{Choice},\text{Monty})
$$
Your choice does not change your uncertainty about the car:
$$
H(\text{Car}\mid\text{Choice})=\log_2 3
$$
After Monty opens R3, you have:
$$
H(\text{Car}\mid\text{Choice},\text{Monty})\approx 0.918
$$
So:
$$
I\approx 0.667\ \text{bits}
$$
Which is exactly the entropy drop we computed.
- The valve spoke.
- The fluid moved.
- The uncertainty shrank.
Calibration: a quick reality check (Brier score)
If you switch, you win with probability $$2/3$$.
If you stay, you win with probability $$1/3$$.
A Brier score punishes bad probabilistic predictions.
For switching ($$p=2/3$$ on a win event that happens $$2/3$$ of the time):
$$
BS_{\text{switch}}
=\frac{2}{3}\left(\frac{2}{3}-1\right)^2+\frac{1}{3}\left(\frac{2}{3}-0\right)^2
\approx 0.222
$$
For a 50/50 believer who effectively predicts $$p=1/2$$ on a win event that happens $$1/3$$ of the time:
$$
BS_{50/50,\text{stay}}
=\frac{1}{3}\left(\frac{1}{2}-1\right)^2+\frac{2}{3}\left(\frac{1}{2}-0\right)^2
=0.25
$$
- Calibration is not a vibe. It’s a number.
- And the numbers prefer agents who respect valves.
Log-odds: the whole update in one stroke
Compare $$H_2$$ vs $$H_1$$.
Likelihood ratio (Bayes factor):
$$
\frac{P(E\mid H_2)}{P(E\mid H_1)}=\frac{1}{1/2}=2
$$
So the evidence adds a single unit of push toward:
$$H_2$$
So,…,
$$
\log 2
$$
Odds become $$2:1$$
Which is probability $$2/3$$
That’s Monty’s entire informational contribution in one constant.
Summary: what the valve actually did
- You started with three equal reservoirs.
- You observed a valve-controlled outcome (“opens R3”).
- The outcome had unequal likelihood across worlds $$(1/2,\ 1,\ 0)$$.
- So posterior fluid became $$(1/3,\ 2/3,\ 0)$$
- Entropy dropped by about 2/3 bits $$0.667$$
- Switching wins because it captures the larger remaining reservoir.
And now the serious closing line:
Two doors remaining is a statement about possibility.
The likelihood model determines the weights.
In a lawful universe, the weights move when the valve speaks.
Under $$H_2$$, the valve is fully open toward “open R3.”
Under $$H_1$$, it’s half open.
Under $$H_3$$, it’s sealed shut.