Imagine three closed doors, one hiding a fictional prize and two hiding nothing. You choose Door 1. A host who knows exactly where the prize is opens one of the two doors you did not choose and reveals an empty room. You can stay with Door 1 or switch to the other unopened door. Which choice is better? In the classic Monty Hall problem, switching wins with probability 2/3, while staying wins with probability 1/3.
This Mathematics in Bukit Timah guide explains conditional probability, sample spaces, the host’s information, switching strategy, decision trees, Bayes’ theorem, simulation, the hundred-door version, and why different host rules produce different answers. Every door, prize, game and participant is fictional. Nothing describes an actual Bukit Timah competition, school programme, local promotion, physical venue or gambling activity. The goal is to learn how additional information changes—and sometimes does not change—probability.
Quick answer: In the standard game the prize location is uniformly random, you initially choose one of three doors, the host always opens an unchosen empty door, and the host always offers you the option to switch. Your first door has a 1/3 chance of hiding the prize. The two doors you did not choose collectively have a 2/3 chance. When the informed host removes an empty door from that pair, the remaining unchosen closed door wins with probability 2/3 if you switch.
Article contents
- Why the Monty Hall problem feels like a paradox
- The Bukit Timah context is a fictional classroom puzzle
- State the original rules completely
- Initial choice probability
- Worked example 1: prize behind Door 1
- Worked example 2: prize behind Door 2
- Worked example 3: prize behind Door 3
- The complete outcome table
- Count the switching wins
- Switching is equivalent to betting your first choice was wrong
- The host cannot open the prize door
- Why the remaining doors are not equally likely
- An important qualification about specific door labels
- A probability tree starts with prize location
- Worked example 4: uniform host tie rule
- Bayes’ theorem for an observed reveal
- Worked example 5: calculate the reveal probability
- Worked example 6: posterior on Door 1
- Worked example 7: posterior on Door 2
- The host’s choice is a likelihood
- Why a host who opens randomly changes the problem
- Worked example 8: random host, initial choice right
- Worked example 9: random host, initial choice wrong
- Conditional answer in the random-host game
- The same visible scene can encode different information
- A host who sometimes offers a switch changes the game too
- Worked example 10: a deliberately biased offer
- The hundred-door version clarifies the intuition
- Worked example 11: one hundred doors
- Many doors do not make every survivor equally likely
- The hundred-door result uses a strong reveal condition
- Four doors with one reveal are more complicated
- Worked example 12: choosing one of two surviving alternatives
- Why the four-door result is not seventy-five percent
- Revealing all but one alternative concentrates the probability
- A simulation can confirm the strategy
- Worked example 13: 3,000 hypothetical rounds
- Simulation cannot replace a proof
- The switching advantage is an average, not a guarantee
- Expected reward provides another comparison
- Worked example 14: changing prize value
- A cost to switch can alter the decision objective
- Worked example 15: a costly switch
- Probability and utility are different ideas
- The host’s tie-break policy can influence a specific observation
- Worked example 16: deterministic host opens Door 3
- Worked example 17: deterministic host opens Door 2
- Why an unconditional strategy can stay optimal
- A host’s randomisation is not always independent
- Conditional probability needs the correct denominator
- Worked example 18: a full posterior check
- The host’s information has value
- Information is not the same as changing the prize location
- A decision tree keeps probabilities and outcomes separate
- Worked example 19: branch probabilities
- Equal-looking branches need not be equally probable
- The initial choice can be any door
- Worked example 20: start with Door 2
- “Always switch” is a strategy, not a new random prize assignment
- Common error 1: call the two closed doors equal
- Common error 2: ignore what the host knows
- Common error 3: assume the host always offers a switch
- Common error 4: count unequal-probability tree leaves equally
- Common error 5: say switching guarantees a prize
- Common error 6: confuse conditional and unconditional probability
- Common error 7: transfer the answer to an arbitrary number of doors
- Common error 8: call simulation a mathematical proof
- Primary readiness: boxes and hidden counters
- Lower Secondary: probability trees and complements
- Advanced probability and curriculum boundaries
- A twenty-minute family diagnostic
- Seven days of independent practice
- Practice questions with checked answers
- Frequently asked questions
- Official context and connected Mathematics in Bukit Timah
Why the Monty Hall problem feels like a paradox
After the host opens an empty door, only two doors remain closed.
That makes it tempting to call their probabilities 50–50. But the two doors were not created by a fresh random selection from two equivalent possibilities.
One door was chosen before the host revealed information. The other survived an informed elimination rule. Their histories are different, so equal door counts do not automatically imply equal probabilities.
The Bukit Timah context is a fictional classroom puzzle
A learner around Beauty World, King Albert Park or Sixth Avenue can draw three boxes on paper and use counters to represent prizes.
No real local game or contest is being reported. The door labels are abstract choices designed to isolate a probability structure.
A student should first identify the host’s behaviour, not assume every real competition follows the classic probability rules.
State the original rules completely
The standard Monty Hall setup contains exactly three doors, one prize and two nonprizes.
You make one initial selection before any door is opened. The host knows where the prize is and opens one other door that does not contain the prize.
The host then offers a switch to the only other closed door. These conditions are essential to the 2/3 conclusion.
Initial choice probability
Before the host acts, the prize is equally likely to be behind Door 1, Door 2 or Door 3.
If you choose Door 1, your probability of being right is 1/3.
Your probability of being wrong is 2/3. Those probabilities are established by the original equally likely prize placement.
Worked example 1: prize behind Door 1
You choose Door 1, which happens to contain the prize.
The host can open Door 2 or Door 3 because both are empty.
If you stay, you win. If you switch to whichever empty door remains closed, you lose. This prize placement occurs with probability 1/3.
Worked example 2: prize behind Door 2
You again choose Door 1, but the prize is behind Door 2.
The host cannot open Door 2 and cannot open your chosen Door 1. The host must open empty Door 3.
Staying loses. Switching from Door 1 to the only other closed door, Door 2, wins. This placement also has probability 1/3.
Worked example 3: prize behind Door 3
Your initial choice remains Door 1, while the prize is behind Door 3.
The informed host must open Door 2, which is empty.
Staying loses, while switching to Door 3 wins. This third placement has probability 1/3.
The complete outcome table
| Prize location | Your initial door | Host reveals | Stay wins? | Switch wins? |
|---|---|---|---|---|
| Door 1 | Door 1 | Empty 2 or 3 | Yes | No |
| Door 2 | Door 1 | Empty 3 | No | Yes |
| Door 3 | Door 1 | Empty 2 | No | Yes |
Each prize location has probability 1/3. The table reflects the classic informed-host rules.
Count the switching wins
The switching strategy wins for two of the three equally likely prize placements.
Therefore P(switch wins)=2/3.
The stay strategy wins only when your first choice was already correct, giving P(stay wins)=1/3. The result follows without needing a complex formula.
Switching is equivalent to betting your first choice was wrong
In the standard game, after the host reveals an empty unchosen door, switching wins exactly when the initial choice was wrong.
Since the first choice is wrong with probability 2/3, the switch strategy wins with probability 2/3.
That equivalence is a particularly useful mental model because it highlights what the host’s informed reveal has accomplished.
The host cannot open the prize door
The host’s knowledge is not a decorative part of the story.
If your initial door is wrong, the host knows which of the two other doors contains the prize and avoids it.
This means the reveal is selected using information about the hidden prize, rather than being an ordinary random door opening.
Why the remaining doors are not equally likely
Your initial door retains its original one-third chance under the standard uniform host-tie policy.
The surviving unchosen door inherits the two-thirds chance originally associated with the pair of unchosen doors.
The empty-door reveal did not split the initial chance of your first choice into half. It eliminated a known empty option from the alternative pair.
An important qualification about specific door labels
The familiar 1/3 versus 2/3 posterior after observing a particular door opened assumes an appropriate host tie-breaking policy, often equal random choice between available empty doors.
If the host strongly prefers one door when both alternatives are empty, observing exactly which door was opened may affect conditional probabilities.
Nevertheless, before conditioning on the identity of the opened door, always switching still wins with probability 2/3 under the core rule that the host invariably opens a goat and offers the switch.
A probability tree starts with prize location
The first branches are Prize 1, Prize 2 and Prize 3, each with probability 1/3.
Then draw the host’s possible reveals conditional on each prize location.
When the prize is behind your initially chosen Door 1, the host has two permissible empty doors. When the prize is behind Door 2 or Door 3, the host has only one permissible reveal.
Worked example 4: uniform host tie rule
Suppose the host chooses randomly with equal probability between empty Door 2 and Door 3 when your initial Door 1 is correct.
If the prize is Door 1, each specific reveal has conditional probability 1/2.
If the prize is Door 2, the host must reveal Door 3. If the prize is Door 3, the host must reveal Door 2. These are distinct conditional behaviours.
Bayes’ theorem for an observed reveal
Suppose you choose Door 1 and observe the host opening Door 3.
Let H₃ be that host action, and let P₁ or P₂ represent prize locations.
Under uniform tie-breaking, P(H₃|P₁)=1/2, P(H₃|P₂)=1, and P(H₃|P₃)=0.
Bayes’ theorem updates the probabilities of prize locations after the observation.
Worked example 5: calculate the reveal probability
The unconditional probability the host opens Door 3 is
(1/3)(1/2)+(1/3)(1)+(1/3)(0).
That equals 1/6+1/3=1/2.
Every term represents one possible prize location multiplied by the host’s behaviour in that case. Omitting those host action probabilities would distort the posterior.
Worked example 6: posterior on Door 1
By Bayes,
P(Prize 1|H₃)=[P(H₃|Prize 1)P(Prize 1)]/P(H₃).
Substitute [(1/2)(1/3)]/(1/2)=(1/6)/(1/2)=1/3.
Your original Door 1 retains its one-third probability under the specified uniform host policy.
Worked example 7: posterior on Door 2
Likewise,
P(Prize 2|H₃)=[1×(1/3)]/(1/2)=2/3.
Door 3 was revealed empty, so its posterior prize probability is zero. The three posterior probabilities add to one: 1/3+2/3+0=1.
The host’s choice is a likelihood
The probability of observing Door 3 opened depends on where the prize is hidden.
That dependence carries information about the unseen state.
Bayes’ theorem combines the original prize probabilities with these conditional reveal likelihoods. The result is not obtained by simply counting the remaining two closed doors.
Why a host who opens randomly changes the problem
Now alter the rule: the host knows nothing about the prize and chooses one of your two unchosen doors uniformly at random.
The host might accidentally open the prize door. Suppose the host happens to reveal an empty door and only then offers a switch.
This is not the same experiment as the classic informed-host game, even though the visible final scene looks similar.
Worked example 8: random host, initial choice right
If your chosen Door 1 holds the prize, both unchosen doors are empty.
A random opening of either unchosen door is guaranteed to reveal emptiness.
The joint probability of initial correctness and observing an empty reveal is (1/3)×1=1/3.
Worked example 9: random host, initial choice wrong
If your chosen door is wrong, one of the two unchosen doors contains the prize and the other is empty.
The uninformed host has probability 1/2 of opening the empty one.
The joint probability of initial wrong choice and a safe empty reveal is (2/3)×(1/2)=1/3.
Conditional answer in the random-host game
Given that the uninformed host happened to reveal emptiness, the total probability of that event is 1/3+1/3=2/3.
The conditional chance your initial choice was right is (1/3)/(2/3)=1/2.
The chance it was wrong is also 1/2. Therefore, in this altered game, staying and switching are equally likely to win, conditional on observing the empty reveal.
The same visible scene can encode different information
In both versions, you see an empty door opened.
But in the classic game the host was forced to avoid the prize and always makes that reveal. In the random-host game an empty reveal is itself a selective event that did not have to occur.
Conditional probability requires the process that generated the observation, not just the observation’s appearance.
A host who sometimes offers a switch changes the game too
Imagine the host decides whether to offer a switch based on whether you initially chose the prize.
Then being offered the switch may reveal additional information.
Without knowing that offer policy, the standard 2/3 answer might not describe the observed conditional decision. The full rule about when offers occur is part of the probability model.
Worked example 10: a deliberately biased offer
Suppose a fictional host offers a switch only when the participant initially chose the prize.
Receiving an offer would then establish that the initial choice is right under the stated rule.
Switching would lose, not win. This is a different game, showing why host behaviour cannot be omitted from the model.
The hundred-door version clarifies the intuition
Imagine 100 closed doors, one prize and 99 empty doors.
You select Door 1. The initial chance your chosen door is correct is 1/100.
The other 99 doors jointly have 99/100 chance of containing the prize. An informed host opens 98 of those unchosen doors, all empty, and leaves exactly one unchosen door closed.
Worked example 11: one hundred doors
Under the same informed-host and compulsory-switch-offer rules, staying wins when the original choice was correct: probability 1%.
Switching to the single remaining unchosen closed door wins when the original choice was wrong: probability 99%.
The much larger gap makes the advantage of switching intuitively clearer.
Many doors do not make every survivor equally likely
Two doors may remain closed after the informed host reveals ninety-eight empty ones.
It would be incorrect to assign them 50% each merely because two remain.
The survivor is not randomly selected from a fresh equal two-door lottery. It is the prize-protecting remainder of a large group of initially unchosen doors.
The hundred-door result uses a strong reveal condition
The host must open 98 empty doors, know where the prize is, avoid your chosen door, and always leave one other closed door.
If the host opens only ten empty doors, there would be many unchosen doors left. The 99% probability belongs to their combined alternatives, not automatically to one arbitrary surviving door.
The precise reveal count matters.
Four doors with one reveal are more complicated
Suppose there are four doors, one prize, and you pick Door 1.
An informed host opens only one empty unchosen door, leaving two other unchosen closed doors.
Your original door has probability one-fourth of holding the prize. The two surviving alternatives collectively carry the other three-fourths, but neither single alternative automatically receives all three-fourths.
Worked example 12: choosing one of two surviving alternatives
Suppose the host always reveals one empty unchosen door and you then choose uniformly at random between the two other closed doors.
If your initial door is wrong, the prize is among those two, so random selection wins with chance 1/2.
Initial wrong probability is 3/4, giving switching success probability (3/4)(1/2)=3/8=37.5%. Staying wins only 25%.
Why the four-door result is not seventy-five percent
The combined two-door alternative set has 75% prize probability.
But the strategy selects only one of those two remaining alternatives, not both.
Averaging uniformly over them leads to a 37.5% win chance under the stated game, rather than the full 75% attached to the pair.
Revealing all but one alternative concentrates the probability
If the informed host instead opens two empty unchosen doors in the four-door game, only one alternative remains.
Then switching wins whenever the initial pick was wrong, giving 3/4.
The general strategy is to exploit the host’s rule that all discarded alternatives are guaranteed empty.
A simulation can confirm the strategy
A basic computational experiment may repeatedly choose a prize door uniformly, hold the participant’s initial choice fixed, and let the informed host reveal an eligible empty door.
Record whether staying or switching wins on each simulated trial.
Across many independent simulated games, the stay fraction should fluctuate around 1/3 and the switch fraction around 2/3.
Worked example 13: 3,000 hypothetical rounds
Under the model, expected staying wins across 3,000 independent rounds are 3,000×1/3=1,000.
Expected switching wins are 3,000×2/3=2,000.
These are expectations, not a promise of exactly those observed counts. Random variation can produce 982 or 1,017 staying wins in a particular simulated run.
Simulation cannot replace a proof
A million simulated trials may provide a convincing numerical check.
But the three-case enumeration already establishes the exact 2/3 probability under the specified assumptions.
Simulation also needs correct programming of the host rules. A program accidentally allowing prize reveals is testing a different game and may produce a different answer.
The switching advantage is an average, not a guarantee
Switching loses when the first choice happens to be correct.
With probability one-third, the switch strategy sacrifices a prize you initially selected correctly.
Therefore “always switch” is the probability-maximising strategy in the classic game, not a method guaranteed to win every round.
Expected reward provides another comparison
Suppose the fictional prize is worth ten utility points and nonprizes are worth zero.
Expected reward from staying is 10×1/3=10/3≈3.33 points.
Expected reward from switching is 10×2/3=20/3≈6.67 points. The advantage is 10/3 expected points per game, with no guarantee in a single trial.
Worked example 14: changing prize value
If the prize value were 100 invented points instead of ten, stay expectation becomes 100/3.
Switch expectation becomes 200/3.
The underlying win probabilities stay one-third and two-thirds. Scaling payoffs does not change the probability structure of the original game.
A cost to switch can alter the decision objective
Suppose switching requires a fixed cost c in fictional utility points, paid whether or not it wins.
Then its expected net reward is (2/3)V−c, while staying earns expected reward (1/3)V for prize value V.
Switching has greater expected net reward if c<V/3. This illustrates how a decision rule combines probabilities with payoffs rather than only comparing win chances.
Worked example 15: a costly switch
Let the prize value be V=12 points and switching cost c=3.
Expected switch net is (2/3)12−3=8−3=5.
Expected stay reward is (1/3)12=4. Switching still has the higher expected score. If c=5, switching expectation would be three and staying would be better under the fictional payoff rule.
Probability and utility are different ideas
The standard Monty Hall theorem concerns which strategy maximises win probability.
If actions have different fees, emotional costs or other consequences, maximising expected utility may require additional modelling.
Our toy payoff calculations are mathematical examples only, not advice about real promotions, betting or commercial decisions.
The host’s tie-break policy can influence a specific observation
Return to the three-door informed-host game.
Suppose the host always opens Door 3 whenever your initially selected Door 1 contains the prize, rather than choosing equally between empty alternatives.
The host still never reveals the prize and still always offers a switch. The overall switching advantage remains 2/3, but knowing which particular door was opened can change the posterior for that case.
Worked example 16: deterministic host opens Door 3
Choose Door 1. Suppose Door 3 is opened.
This observation is possible when the prize is Door 1 (host deliberately opens 3) or Door 2 (host must open 3), each with initial probability 1/3.
The observed reveal gives 1/2 posterior probability to Door 1 and 1/2 to Door 2. Conditional on this reveal, switching and staying tie.
Worked example 17: deterministic host opens Door 2
Under the same deterministic tie rule, the host opens Door 2 only when the prize is Door 3.
If the host opens Door 2, the prize must therefore be Door 3.
Switching wins with conditional probability one. The mixture of conditional results still gives unconditional switching success 2/3 before learning which door opens.
Why an unconditional strategy can stay optimal
The probability that your initial pick is wrong is 2/3.
The informed host’s mandatory reveal always leaves the prize behind the only unchosen closed door when your initial pick is wrong.
Thus the unconditional always-switch success probability remains 2/3, regardless of how the host chooses between empty doors when both are available.
A host’s randomisation is not always independent
When the host selects a door to reveal, the possible choices depend on the hidden prize location.
Therefore the host’s reveal and prize location are not generally independent random variables.
Using P(host opens Door 3)=1/2 without specifying the tie rule can be unjustified, even though it happens to hold in the symmetric standard version.
Conditional probability needs the correct denominator
For events A and B, P(A|B)=P(A∩B)/P(B) when P(B)>0.
In the door problem, B is the observed host action. The denominator includes every prize location that could lead to that same host action, weighted by its probability.
Leaving out the correct-initial-choice case or treating all host actions as equally likely changes the answer.
Worked example 18: a full posterior check
Under uniform tie-breaking and the observation that Door 3 opens, posterior probabilities are:
Prize behind Door 1: 1/3. Prize behind Door 2: 2/3. Prize behind Door 3: 0.
They sum to one. This check catches many errors involving missing or duplicated cases.
The host’s information has value
The host’s refusal to reveal the prize ensures that an initially wrong choice can be corrected by switching.
The participant does not need to know the hidden prize location directly. It is enough to know that the host’s reveal was constrained by that location.
This illustrates why a deliberately selected observation can convey information beyond the fact that one empty door was seen.
Information is not the same as changing the prize location
Opening an empty door does not physically move the prize.
The initial prize was already behind one of the three doors. The host reveals information about that fixed hidden arrangement.
The probabilities describe uncertainty about its location. The door opening changes what the participant knows, not where the prize is.
A decision tree keeps probabilities and outcomes separate
First branch on the true prize location. Then branch on any random host tie choice.
Finally mark the payoff for stay versus switch.
Keeping chance nodes separate from participant decisions prevents confusion between the probability of an event and the reward from a chosen action.
Worked example 19: branch probabilities
Under uniform tie-breaking and Door 1 initially chosen:
Prize 1, host opens 2: probability 1/6. Prize 1, host opens 3: probability 1/6. Prize 2, host opens 3: probability 1/3. Prize 3, host opens 2: probability 1/3.
The four path probabilities sum to one. In three of the four paths? Count probabilities, not branches: switching wins on the last two paths, whose total is 2/3.
Equal-looking branches need not be equally probable
The preceding probability tree has four terminal paths, but their probabilities are 1/6,1/6,1/3 and 1/3.
Simply saying switching wins in two of four paths and therefore has probability one-half is incorrect.
The correct total uses path probabilities, not an unweighted count of unequal-probability branches.
The initial choice can be any door
Door 1 was chosen only for convenience.
If the player instead chooses Door 2 or Door 3, the same symmetry and informed-host rules apply under uniform prize placement.
Staying wins with probability one-third; switching wins with two-thirds. The label of the initially chosen door does not create an advantage.
Worked example 20: start with Door 2
If the prize is behind Door 2, staying wins and switching loses.
If the prize is behind Door 1 or Door 3, the host reveals the other unchosen empty door, so switching wins.
Two of three equally likely prize locations favour switching, exactly as before.
“Always switch” is a strategy, not a new random prize assignment
The participant chooses a rule to follow after a valid host reveal.
The rule does not change the original prize placement probabilities.
Its advantage arises because a wrong initial choice is more common than a right one, and the informed reveal gives a way to move from that wrong initial door to the surviving prize door.
Common error 1: call the two closed doors equal
Two closed doors remain, but they have different selection histories.
The original door was chosen before the host’s informed reveal. The surviving alternative was protected by the host’s empty-door rule.
Their probabilities should be derived from the full experiment, not from the number of closed doors alone.
Common error 2: ignore what the host knows
If the host opens a random unchosen door without knowing the prize, the conditional odds can become 50–50 after an empty reveal.
That is a different model.
Write down whether the host knows the prize and whether prize reveals are allowed before choosing a probability formula.
Common error 3: assume the host always offers a switch
The standard theorem assumes a switch opportunity is always offered after an eligible empty reveal.
If the offer is selective based on hidden information, the fact an offer occurred can alter the posterior.
The offer rule is as important as the reveal rule.
Common error 4: count unequal-probability tree leaves equally
In the uniform-tie probability tree, two paths have probability 1/6 each and two paths 1/3 each.
A count of two winning branches out of four gives the wrong probability.
Multiply along each path, then add mutually exclusive winning paths.
Common error 5: say switching guarantees a prize
Switching still loses when the first chosen door was already correct.
That initial-correct case occurs with probability one-third under the standard model.
Switching optimises the probability of winning; it does not ensure success in every round.
Common error 6: confuse conditional and unconditional probability
The unconditional probability of always switching successfully is 2/3 under classic informed-host rules.
A conditional probability after observing a specific host action can depend on the host’s tie-breaking behaviour.
Specify which observation is conditioned on, if any, before quoting a posterior.
Common error 7: transfer the answer to an arbitrary number of doors
With 100 doors and the host opening 98 empty alternatives, switching wins with 99%.
But with 100 doors and only one opened, many unchosen possibilities remain. Switching to one arbitrarily selected door does not inherit the entire 99% chance.
The number of doors left closed is part of the model.
Common error 8: call simulation a mathematical proof
A computer run may produce results near the predicted win rates, but finite samples vary.
A simulation that accidentally permits the host to reveal the prize changes the rules.
The complete three-placement proof and conditional calculation explain why the result holds for the specified experiment.
Primary readiness: boxes and hidden counters
Primary learners can put a small object beneath one of three opaque cups and use an adult host who knows its location.
They can play several rounds using a consistent rule and record whether switching succeeds.
The aim is understanding what information the host supplies, not teaching students to pursue any real prize game.
Lower Secondary: probability trees and complements
Students can write the three equally likely prize placements and use a table to track stay and switch outcomes.
They can then add the unequal-probability host branches and distinguish conditional from unconditional probabilities.
A clear sample-space table is more reliable than memorising “switch wins twice as often” without knowing when that rule applies.
Advanced probability and curriculum boundaries
The Monty Hall problem connects sample spaces, conditional probability, Bayes’ theorem, information and decision theory.
Consult the SEAB 2027 G3 Mathematics and Additional Mathematics syllabus directory for actual school examination scope. This full host-policy analysis is enrichment rather than a claim of compulsory content for every learner.
A twenty-minute family diagnostic
Minutes 1–5: Write prize locations 1,2,3 with Door 1 chosen; identify which locations make switching win.
Minutes 6–10: Compute unconditional switch probability 2/3 and stay probability 1/3.
Minutes 11–15: Add an informed host choosing uniformly between two empty doors when the initial choice is correct; derive a 2/3 posterior on Door 2 after observing Door 3 opened.
Minutes 16–20: Change the host to an uninformed random opener; condition on an empty reveal and calculate the new 1/2 versus 1/2 result.
Seven days of independent practice
Day one lists sample-space outcomes. Day two distinguishes prior probabilities from information revealed later.
Day three calculates the standard strategy success rates. Day four creates a probability tree with host tie-breaking.
Day five derives a Bayes posterior. Day six solves a many-door version. Day seven deliberately changes the host rules and explains why the correct answer changes.
Practice questions with checked answers
A. Initial chance of selecting the prize from three uniform doors? Answer: 1/3.
B. Initial chance of selecting an empty door? Answer: 2/3.
C. Standard informed-host switching success? Answer: 2/3.
D. Staying success? Answer: 1/3.
E. A hundred doors with 98 empty alternatives revealed: switching success? Answer: 99%.
F. Uninformed random host, conditional on an empty reveal: switching success? Answer: 1/2.
G. Four doors, host opens only one empty, then player randomly chooses one of two remaining unchosen closed doors: success? Answer: 3/8.
H. Does switching guarantee success? Answer: no.
Frequently asked questions
Should I switch in the classic Monty Hall problem?
Yes, if the objective is to maximise win probability and the classic informed-host rules hold. Switching wins with probability 2/3.
Why are the two remaining closed doors not 50–50?
Because the host deliberately removed an empty alternative using knowledge of the prize location, and the doors have different selection histories.
Does the host always need to know the prize location?
Yes for the standard informed-host model; otherwise the observed reveal can have a different probabilistic meaning.
Does it matter how the host chooses between two empty doors?
The overall always-switch win rate remains 2/3 under the compulsory informed reveal, but specific conditional posteriors can depend on the tie-breaking rule.
What if the host accidentally reveals the prize?
That belongs to a different experiment, not the standard Monty Hall setup.
Is switching certain to win?
No. It loses whenever the first selection was already correct.
Can the problem be solved with Bayes’ theorem?
Yes. Bayes updates the prize-location probabilities using the likelihood of the host’s observed reveal.
Why does the hundred-door version help?
It makes the original 1% chance and 99% alternative-group chance more visually apparent when the host removes 98 known empty alternatives.
Are these actual Bukit Timah prize or classroom results?
No. Every door, host and prize is a fictional Mathematics example.
How can parents recognise understanding?
The learner can specify the host rules, list the three prize cases, distinguish different conditional games and explain why switching wins exactly when the first pick was wrong.
Official context and connected Mathematics in Bukit Timah
The SEAB 2027 G3 Mathematics syllabus directory provides Singapore school-level probability context. The extended host-rule analysis is not claimed to be a universal SEC G3 examination requirement.
Continue with Bayes’ Theorem and Base Rates, Probability and Dependent Events, Expected Value and Repeated Trials and Birthday Paradox and Collision Probability for complementary mathematical questions.
The immutable Clementi Secondary 1 Mathematics small-group teaching reference anchors explicit assumptions, worked case tables, checked probability arithmetic and independent explanation.
The lasting habit is to define the experiment completely, distinguish an informed reveal from a random one, identify the original probability mass and recompute conditional probabilities whenever the rules change.
Companion Mathematics in Bukit Timah articles from this November batch: Birthday Paradox, Shared Birthdays and Collision Probability · Spherical Geometry, Great Circles and Shortest Surface Distances · Mathematical Invariants, Parity Puzzles and Impossible Moves. These four original articles distinguish mathematical probability from certainty, conditioned decisions from naive guessing, surface geometry from flat distance, and structural invariants from trial-and-error searches.