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Every engineering student who learns probability and statistics have bumped into this famous problem. Refer to to know the problem statement. (Don't look at answers yet!!)
The *wrong * solution which many of us might bump into:
Probability of getting a car in any of the remaining two doors is 1/2, since one has a goat and the other has a car.
The correct approach:
Events:
Hx: Host selects door x to put a car.
Py: Player selects door y as his guess.
Space:
[(H1 and P1), (H1 and P2), (H1 and P3),
(H2 and P1), (H2 and P2), (H2 and P3),
(H3 and P1), (H3 and P2), (H3 and P3)]
All of the events are equiprobable. i.e. 1/9.
We break down these events space into two:
[x=y]: [(H1 and P1), (H2 and P2), (H3 and P3)]
[x!=y]: [(H1 and P2), (H1 and P3),
(H2 and P1), (H2 and P3),
(H3 and P1), (H3 and P2)]
Events that win a car are: (H1 and P1), (H2 and P2) and (H3 and P3).
P([x=y]) = 1/9 + 1/9 + 1/9
= 3/9
= 1/3
P([x!=y]) = 1 - P([x=y])
= 1 - 1/3
= 2/3
The interesting part starts from here.
The host opens a door which was not selected by the player and which is not a car. And then, asks to switch his guess.
_The question: What is the probability of winning if the player switches his guess? _
The events will look like below when player switches his guess:
Events x=y will be extended like this:
(H1 and P1), then the player switches and loses.
P(win after switch | x=y) = 0.
Events x!=y will be extended like this:
H1 and P2, player switches from door 2 to door 1 and wins.
Note that, the host can only reveal door 3. And the player cannot switch to door 3. Only option is door 1 to switch
Hence, [x!=y] events followed by a switch will always win.
P(win after switch | x!=y) = 1.
P(win after switch) = P(x!=y)*P(win after switch | x!=y)
+ P(x=y)*P(win after switch | x=y)
= 1*2/3 + 0*1/3
= 2/3
P[x!=y] = 2/3.
Hence, probability of winning after switching the guess is 2/3.
In each of the x!=y cases, if the player switches, player wins.
Conclusion
In every problem related to probability, define the sample space well and try to break down the events space.
Follow up question:
If a player plays 100 rounds of this game, what should be the player's tendency to switch his guess to maximize wins?
The *wrong * solution which many of us might bump into:
Probability of getting a car in any of the remaining two doors is 1/2, since one has a goat and the other has a car.
The correct approach:
Events:
Hx: Host selects door x to put a car.
Py: Player selects door y as his guess.
Space:
[(H1 and P1), (H1 and P2), (H1 and P3),
(H2 and P1), (H2 and P2), (H2 and P3),
(H3 and P1), (H3 and P2), (H3 and P3)]
All of the events are equiprobable. i.e. 1/9.
We break down these events space into two:
[x=y]: [(H1 and P1), (H2 and P2), (H3 and P3)]
[x!=y]: [(H1 and P2), (H1 and P3),
(H2 and P1), (H2 and P3),
(H3 and P1), (H3 and P2)]
Events that win a car are: (H1 and P1), (H2 and P2) and (H3 and P3).
P([x=y]) = 1/9 + 1/9 + 1/9
= 3/9
= 1/3
P([x!=y]) = 1 - P([x=y])
= 1 - 1/3
= 2/3
The interesting part starts from here.
The host opens a door which was not selected by the player and which is not a car. And then, asks to switch his guess.
_The question: What is the probability of winning if the player switches his guess? _
The events will look like below when player switches his guess:
Events x=y will be extended like this:
(H1 and P1), then the player switches and loses.
P(win after switch | x=y) = 0.
Events x!=y will be extended like this:
H1 and P2, player switches from door 2 to door 1 and wins.
Note that, the host can only reveal door 3. And the player cannot switch to door 3. Only option is door 1 to switch
Hence, [x!=y] events followed by a switch will always win.
P(win after switch | x!=y) = 1.
P(win after switch) = P(x!=y)*P(win after switch | x!=y)
+ P(x=y)*P(win after switch | x=y)
= 1*2/3 + 0*1/3
= 2/3
P[x!=y] = 2/3.
Hence, probability of winning after switching the guess is 2/3.
In each of the x!=y cases, if the player switches, player wins.
Conclusion
In every problem related to probability, define the sample space well and try to break down the events space.
Follow up question:
If a player plays 100 rounds of this game, what should be the player's tendency to switch his guess to maximize wins?