The Monty Hall problem explained with a table

The rules matter

The puzzle only has the 2/3 answer under the standard rules: the car is placed at random, the host always opens a door you did not pick, he knows where the car is and always reveals a goat, and he always offers the switch. Change any of these and the answer can change. Most arguments about the problem are really arguments about which rules apply.

Why it is not 50/50: enumerate the cases

Say you pick door 1. The car is behind door 1, 2 or 3 with probability 1/3 each. If it is behind door 1, the host opens either goat door and staying wins. If it is behind door 2, the host must open door 3, and switching takes you to door 2, the car. If it is behind door 3, he must open door 2 and switching wins again. Staying wins in one case of three, switching in two. In Excel the winning chances are =SUMPRODUCT(B2:B4,D2:D4) for staying and =SUMPRODUCT(B2:B4,E2:E4) for switching: 33.3% and 66.7%.

The "two doors left, so 50/50" intuition fails because the two remaining doors are not alike. Your door was chosen before the host acted and keeps its 1/3. The other door survived a host who was not allowed to open the car, so it collects the 2/3 that belonged to the two doors you did not pick.

Worked example: Monty Hall in three tables

Top: every position of the car when you pick door 1. Middle: Bayes' theorem after the host opens door 3. Bottom: more doors and the ignorant host.

Worked example: Monty Hall in three tables

	A	B	C	D	E
1	Car behind	P(car)	Host opens	Stay wins	Switch wins
2	Door 1 (your pick)	33.3%	door 2 or 3	1	0
3	Door 2	33.3%	door 3	0	1
4	Door 3	33.3%	door 2	0	1
5	P(win)	100%		33.3%	66.7%
6
7	Bayes: host opened door 3	Prior	P(host opens 3)	Prior × likelihood	Posterior
8	Car behind door 1	33.3%	0.5	0.1667	33.3%
9	Car behind door 2	33.3%	1.0	0.3333	66.7%
10	Car behind door 3	33.3%	0.0	0.0000	0.0%
11	Total			0.5000	100%
12
13	Doors (n)	Stay	Switch: host opens all but one	Switch: host opens one door
14	3	33.3%	66.7%	66.7%
15	4	25.0%	75.0%	37.5%
16	5	20.0%	80.0%	26.7%
17	10	10.0%	90.0%	11.3%
18	100	1.0%	99.0%	1.0%
19
20	Ignorant host: opens one of the other two doors at random				Value
21	P(car shown by accident)				33.3%
22	P(goat shown)				66.7%
23	P(switch wins | goat shown)				50.0%

Standard rules: car placed at random, host knows where it is, always opens a goat door you did not pick and always offers the switch. Educational example.

The same answer with Bayes' theorem

Suppose the host opens door 3. How likely is that under each position of the car? If the car is behind your door 1, he picks door 2 or 3 at random, so the likelihood is 1/2. If it is behind door 2, he must open door 3: likelihood 1. If it is behind door 3, he cannot open it: likelihood 0. Multiply each prior of 1/3 by its likelihood (1/6, 1/3, 0), add them up (1/2) and divide: the posterior is 1/3 for door 1 and 2/3 for door 2. The host's choice of door is information, and Bayes' theorem measures exactly how much. The same table structure works for medical tests with false positives.

More doors make it obvious

With 10 doors, you pick one and the host opens 8 goat doors, leaving yours and one other. Staying wins only if your first pick was right, 1 in 10, so switching wins =(n-1)/n = 90%, and with 100 doors 99%. If the host opens just one goat door out of the others, switching still helps, but less: with n doors it wins (n − 1) ÷ (n × (n − 2)), 37.5% with 4 doors against 25% for staying.

When switching does not help: the ignorant host

If the host does not know where the car is and opens one of the other two doors at random, he reveals the car one time in three and the game is spoiled. In the games where a goat happens to appear, staying and switching are equally good: P(switch wins | goat shown) = (1/3) ÷ (2/3) = 1/2. The doors on screen look identical, but the host's knowledge changes the probability. That is the heart of the paradox: what you learn depends on how the information was produced.

Simulating it in Excel

For a quick experiment, put =RANDBETWEEN(1,3) for the car and another for your pick in 1,000 rows, mark stay wins with =IF(A2=B2,1,0) and switch wins with =IF(A2<>B2,1,0), and average both columns. The averages land near 0.33 and 0.67 and move a little every time the sheet recalculates (F9). A simulation persuades, but the table above proves the answer exactly.

Common mistakes

• Treating the two remaining doors as equal. Your door was chosen blind, the other one survived a host who could not open the car.
• Ignoring the host's knowledge. A host who opens doors at random and happens to show a goat gives a 50/50 game. The rules decide the answer.
• Forgetting that the host picks at random when you hold the car. That 1/2 likelihood is what makes the Bayes table give 1/3 and 2/3.
• Dividing by the wrong total in Bayes. The posterior divides each prior × likelihood by their sum (1/2 here), not by 1.
• Trusting a small simulation. With 20 rows the share of switch wins can easily land at 50%. Use hundreds or thousands of rows, or enumerate the cases exactly.