Bayes Theorem on kenji.blog

Sleeping Beauty Paradox: Is the coin probability 1/2 or 1/3? A difficult problem dividing probability theory

Thu, 10 Sep 2026 09:00:00 +0900

1. The Rules of a Strange Experiment

You (Sleeping Beauty) have been selected as a subject for a certain scientific experiment. The experiment runs from Sunday to Wednesday. On Sunday night, you are given a sleeping pill and fall asleep.

After you fall asleep, the experimenter tosses a single fair coin (a coin with an exactly 1/2 probability of landing heads or tails). And depending on the result, you will be awakened according to the following schedule.

[If the coin toss result is “Heads”]

You will be awakened once on Monday and asked a question. After that, you will be put back to sleep and will not awaken again until the experiment ends (Wednesday).

[If the coin toss result is “Tails”]

You will be awakened on Monday and asked a question. After that, you will be given a special drug (an amnesia drug) and put back to sleep.
You will be awakened once more on Tuesday and asked the same question. After that, you will be put back to sleep again, and the experiment will end (Wednesday).

Due to the effects of the amnesia drug, when you awaken, you will not be able to remember “what day of the week it is today” or “whether you have been awakened in the past” at all.

graph TD Sunday["Sunday: Beauty goes to sleep"] --> Toss{"Coin Toss"} Toss -->|Heads (1/2)| Mon_Heads["Monday: Awaken + Question
(Then experiment ends)"] Toss -->|Tails (1/2)| Mon_Tails["Monday: Awaken + Question
(Then amnesia)"] Mon_Tails --> Tue_Tails["Tuesday: Awaken + Question
(Then experiment ends)"] style Toss fill:#ff9999,stroke:#333 style Mon_Heads fill:#aaffaa,stroke:#333 style Mon_Tails fill:#aaffaa,stroke:#333 style Tue_Tails fill:#aaffaa,stroke:#333

Now, it is Monday (or Tuesday), and you have awakened. There are no clocks or calendars in the room, so you do not know what day it is today.

There, the experimenter comes in and asks you this question. “Given that you are currently awake, what do you think is the probability that the tossed coin was ‘Heads’?”

You are a beauty well-versed in mathematics. Now, how do you answer?

2. Two Clashing Factions: 1/2 or 1/3?

This problem was devised in the 1990s and published in an academic journal by philosopher Adam Elga in 2000. The probability of the coin seems obvious, but actually, over this problem, mathematicians, statisticians, and philosophers around the world have split cleanly into two camps: the “1/2 Faction (Halfers)” and the “1/3 Faction (Thirders)”, engaging in fierce debate to this day.

Let’s hear the “perfect logic” of each camp.

The Claim of the “1/2 Faction (Halfers)”

“Since the coin is a fair coin with no cheating, the probability of heads coming up is naturally 1/2. No matter how many times the experimenter wakes me up or erases my memory after I fall asleep, it does not affect the physical outcome of the coin at all. The probability at the time the coin was tossed was 1/2, and my waking up gives me no new information (clues to guess whether it is heads or tails). Therefore, the probability remains 1/2.”

This is a very sound opinion that emphasizes objective physical phenomena and the non-update of information.

The Claim of the “1/3 Faction (Thirders)”

“The very fact that you are ‘awake’ is information that changes the probability. Suppose we repeated this experiment 100 times (100 weeks). The coin should be ‘Heads’ 50 times and ‘Tails’ 50 times.

For the 50 weeks where Heads comes up, you awaken only once on Monday $\rightarrow$ Number of awakenings for ‘Heads’ is 50 times

For the 50 weeks where Tails comes up, you awaken twice, on Monday and Tuesday $\rightarrow$ Number of awakenings for ‘Tails’ is 100 times

In other words, the situation of the moment you wake up has 150 times in total, out of which the ‘pattern of awakening on Heads’ is 50 times, and the ‘pattern of awakening on Tails’ is 100 times. Therefore, the probability that the awakening you are currently experiencing is ‘Heads’ is 50 / 150 = 1/3!”

This is a powerful opinion based on “frequentism” or the “anthropic principle,” which incorporates the very situation that “you currently exist (are observing)” into the calculation as an element of the probability space.

3. Calculating with Bayes’ Theorem

There is also an attempt to unravel this problem using “Bayes’ Theorem,” a tool for mathematically updating probabilities. Let’s organize the logic of the “1/3 faction” from the perspective of conditional probability.

Your state when you awaken is one of the following three:

$E_1$: The coin is “Heads”, and it is now “Monday”
$E_2$: The coin is “Tails”, and it is now “Monday”
$E_3$: The coin is “Tails”, and it is now “Tuesday”

The probability of “Heads” is $1/2$, and the probability of “Tails” is $1/2$. However, in the case of Tails, “Monday” and “Tuesday” are perfectly symmetrical (you cannot distinguish them since you have no memory), so it is thought that $E_2$ and $E_3$ are equally likely to occur.

Since the sum of the overall probabilities must be $1$, if we assign equal probability to each awakening as an independent “event (observation point)”: $P(E_1) = 1/3$ $P(E_2) = 1/3$ $P(E_3) = 1/3$ Thus, the conclusion is that the “probability it was Heads ($P(E_1)$)” is $1/3$.

On the other hand, the “1/2 faction” argues against this, stating, “Monday and Tuesday when the coin is Tails ($E_2$ and $E_3$) are merely dependent events derived from the single result of one coin toss, and it is wrong to count them as independent probabilities in the first place.”

4. Why Does This Problem Not Get Resolved?

The reason why the “Sleeping Beauty problem” plagues scholars so much is not due to a mere calculation error or illusion. It is because this problem touches upon the deepest and most fundamental question of probability theory: “What exactly is probability?”

For the 1/2 faction, probability is a “physical property of the coin” or an “objective fact”.
For the 1/3 faction, probability is the “degree of belief of the observer (Beauty)” or the “frequency of observation”.

Profound themes that connect to the “measurement problem” in quantum mechanics and the “anthropic principle” in cosmology (the idea of calculating the probability of the universe backward from the fact that we exist) are condensed into this simple coin toss experiment.

5. Summary

If you were a subject in this experiment, would you answer “1/2” or “1/3” when you wake up?

Whichever you answer, world-class mathematicians will stand behind you to defend you. How a seemingly simple mathematical definition collapses the moment it is tied to troublesome concepts like human “subjectivity” and “existence.” The paradox continues to shake our common sense today.

Monty Hall Problem: The Trap of Probability Theory that Betrays Intuition and its Complete Resolution using Bayesian Inference

Thu, 10 Sep 2026 00:00:00 +0900

1. The Stage is a TV Quiz Show: What Would You Do?

In 1990, the following question was submitted by a reader to the column “Ask Marilyn” in the American news magazine Parade.

You are a contestant on a TV game show. In front of you are 3 doors (A, B, C). Behind one door is a new car (the prize), and behind the remaining two doors are goats (the blanks).

First, you choose Door A.

Then, the host, Monty Hall, who knows what is behind each door, opens Door B, which has a goat.

Monty says to you, “You may now change your choice to Door C if you like. What will you do?”

So, should you change your door?

Intuitively, it seems: “The remaining doors are A and C. Since the new car is completely random between the two, the probability of winning is $\frac{1}{2}$ (50%) for each. So it doesn’t matter whether you change or not.”

However, columnist Marilyn vos Savant (recognized by the Guinness Book of Records as having the highest IQ) replied, “You should change. If you change, your probability of winning doubles.”

This answer caused a sensation across the United States, bringing in a storm of harsh criticism with about 10,000 letters of protest (about 1,000 of which were from scholars with PhDs in mathematics), saying things like “You do not understand the basics of probability” and “That’s female logic.” However, to get straight to the conclusion, Marilyn’s answer was mathematically entirely correct.

2. The Gap Between Intuition and Mathematics: Branching Probabilities in Mermaid

Why does our intuition create the illusion that it is “$\frac{1}{2}$”? First, let’s visualize all the patterns of the game.

graph TD Start["Game Start"] --> CarA["Car is behind Door A (Prob. 1/3)"] Start --> CarB["Car is behind Door B (Prob. 1/3)"] Start --> CarC["Car is behind Door C (Prob. 1/3)"] CarA --> PickA1["You pick Door A"] CarB --> PickA2["You pick Door A"] CarC --> PickA3["You pick Door A"] PickA1 --> HostB_or_C["Host opens B or C"] PickA2 --> HostC["Host must open C"] PickA3 --> HostB["Host must open B"] HostB_or_C --> Stay1["Stay: Lose..."] HostB_or_C --> Switch1["Switch: Win!"] HostC --> Stay2["Stay: Lose..."] HostC --> Switch2["Switch: Win!"] HostB --> Stay3["Stay: Lose..."] HostB --> Switch3["Switch: Win!"] style Switch2 fill:#bbf,stroke:#333,stroke-width:2px style Switch3 fill:#bbf,stroke:#333,stroke-width:2px style Stay1 fill:#f99,stroke:#333,stroke-width:2px

Assuming you chose “Door A”, the following three scenarios occur with equal probability ($\frac{1}{3}$).

Scenario 1 (Car is A): The host opens either B or C, both of which have goats. If you change your door, you lose.
Scenario 2 (Car is B): The host can only open C, which has a goat. If you change your door, you win.
Scenario 3 (Car is C): The host can only open B, which has a goat. If you change your door, you win.

In other words, in 2 out of 3 times (Scenarios 2 and 3), you are in a state where “you will definitely win if you change doors”. Therefore, the win rate when you change doors is $\frac{2}{3}$, which is double the win rate of $\frac{1}{3}$ when you do not change.

3. Strict Proof Using Bayes’ Theorem

To strictly solve this problem mathematically, we use “Bayes’ Theorem” to calculate conditional probabilities.

$$ P(H|E) = \frac{P(E|H) P(H)}{P(E)} $$

Here, we define the events as follows:

$C_A, C_B, C_C$ : The events that the new car is behind doors A, B, and C, respectively. The prior probabilities are $P(C_A) = P(C_B) = P(C_C) = \frac{1}{3}$
Suppose you initially selected Door A.
$M_B$ : The event that the host opens Door B, which has a goat.

What we want to find is “the probability that the new car is behind Door C given that the host opened Door B”, i.e., the posterior probability $P(C_C|M_B)$.

First, let’s consider the probability $P(M_B|C_X)$ that the host opens Door B depending on where the new car is.

When the new car is behind Door A ($C_A$) The host can open either B or C randomly.
$$ P(M_B|C_A) = \frac{1}{2} $$
When the new car is behind Door B ($C_B$) The host cannot open the door with the new car, so the probability of opening B is zero.
$$ P(M_B|C_B) = 0 $$
When the new car is behind Door C ($C_C$) The host cannot open A (which you picked) or C (where the new car is), so they must inevitably open B.
$$ P(M_B|C_C) = 1 $$

Next, we find the total probability $P(M_B)$ that the host opens Door B using the “Law of Total Probability”.

$$ P(M_B) = P(M_B|C_A)P(C_A) + P(M_B|C_B)P(C_B) + P(M_B|C_C)P(C_C) $$ $$ P(M_B) = \left(\frac{1}{2} \times \frac{1}{3}\right) + \left(0 \times \frac{1}{3}\right) + \left(1 \times \frac{1}{3}\right) = \frac{1}{6} + 0 + \frac{1}{3} = \frac{1}{2} $$

Finally, we apply Bayes’ Theorem to calculate the posterior probabilities for Door A and Door C.

Probability that the new car is behind Door A (if you stay):

$$ P(C_A|M_B) = \frac{P(M_B|C_A) P(C_A)}{P(M_B)} = \frac{\frac{1}{2} \times \frac{1}{3}}{\frac{1}{2}} = \frac{1}{3} $$

Probability that the new car is behind Door C (if you switch):

$$ P(C_C|M_B) = \frac{P(M_B|C_C) P(C_C)}{P(M_B)} = \frac{1 \times \frac{1}{3}}{\frac{1}{2}} = \frac{2}{3} $$

The mathematical proof also clearly demonstrates that “changing doors doubles your probability of winning (2/3)”.

4. Cognitive Bias: The Value of Information as “Conditioning”

Why did even many genius mathematicians intuitively get this problem wrong? The answer lies in the “equiprobability bias” and the “failure to update information” built into the human brain.

4.1. Equiprobability Bias

When presented with unknown options, humans have a tendency to unconsciously assign that “the probabilities of the remaining options are always equal.” The moment we see two doors remaining, our brain automatically labels them as “$50\%$ : $50\%$”.

4.2. The Information of the Host’s “Intent”

The biggest reason intuition goes wrong is overlooking the fact that the host’s actions are not random. If the rule was “the host opens a door randomly without knowing where the car is, and it just happened to be a goat” (known as the Monty Fall problem), then the probabilities for Door A and Door C would both be $\frac{1}{2}$.

However, in the actual Monty Hall problem, the host operates under the following strict constraints:

They cannot open the door chosen by the contestant.
They cannot open the door with the new car.

Because of these constraints, the very act of the host “opening Door B” gives us massive information about Door C. It contains the unspoken message, “I couldn’t open Door C (because the new car is there).”

5. Correcting Intuition with an Extreme Example

If you’re still not convinced, try increasing the number of doors to 1,000,000.

You pick Door 1 out of 1,000,000 doors. (Probability of winning is $\frac{1}{1,000,000}$)
The host, who knows everything, opens all 999,998 doors with goats behind them out of the remaining 999,999 doors.
The only doors closed are “Door 1” which you picked, and “Door 777,777” which the host deliberately left closed.

Now, do you change? In this case, unless you believe you pulled off a “one in a million” miracle right at the start, you should change. Realistically, it should be intuitively clear that the probability of the new car being behind “the single door the host absolutely could not open” is $\frac{999,999}{1,000,000}$.

The Monty Hall problem (with 3 doors) is simply a scaled-down phenomenon of this “1,000,000 doors” scenario.

pie title "Effect of Switching Doors (100 Simulations)" "Win by switching (approx. 66.7%)" : 67 "Win by staying (approx. 33.3%)" : 33

6. Conclusion: Life and Business Lessons from Probability Theory

The Monty Hall problem goes beyond a mere quiz and teaches us important lessons.

Intuition is often wrong: The human brain has not evolved to intuitively process complex conditional probabilities. In important decision-making, relying solely on intuition is dangerous.
Update probabilities with new information (Bayesian updating): When situations change and new information (such as which door the host opened) is provided, the key to success is whether you can flexibly update your probabilities and strategies without clinging to existing beliefs.

The small decision to “change your door” just might double the probability of getting a “new car” in your life.