User:JumpDiscont

This is for Phyti.

I imagine you will either

regard what the host is doing as

or

think computing the resulting probabilities is

at least one of manipulation , rigging ,

but anyway, here is what follows from

The producers choose [where they put the car]

uniformly at random from among the 3 doors.

and

The player makes the player's first choice

uniformly at random from among the 3 doors.

and

Placing the car and player choosing

a door are independent events.

and

The host _must_ open exactly one door.

and

The host cannot open the door from the players 1st guess.

and

The host knows where the car is, and can't open that door.

and

If that door is the door the player guessed, then the host

chooses uniformly at random which goat door the host opens.

:

C = car door

P = player's first choice

H = door host opens

Prob(C=1) = 1/3

Prob(P=2) = 1/3

Placing the car and player choosing a door are independent events, so

Prob(C=1 and P=2) = Prob(C=1) * Prob(P=2) = 1/3 * 1/3 = 1/9 .

When C=1 and P=2 , the host _must_ open door 3,

so Prob(H=3 given (C=1 and P=2)) = 1 .

Prob(C=1 and P=2 and H=3)

=

Prob(C=1 and P=2) * Prob(H=3 given (C=1 and P=2))

=

1/9 * 1 = 1/9

Similarly, Prob(C=i and P=j and H=k) = 1/9 for each choice of

(i,j,k) from { (1,3,2) , (2,1,3) , (2,3,1) , (3,1,2) , (3,2,1) } .

Prob(C=1) = 1/3

Prob(P=1) = 1/3

Placing the car and player choosing a door are independent events, so

Prob(C=1 and P=1) = Prob(C=1) * Prob(P=1) = 1/3 * 1/3 = 1/9 .

When C=1 and P=1 , the host chooses uniformly at random which goat

door the host opens, so Prob(H=2 given (C=1 and P=1)) = 1/2 .

Prob(C=1 and P=1 and H=2)

=

Prob(C=1 and P=1) * Prob(H=2 given (C=1 and P=1))

=

1/9 * 1/2 = 1/18

Similarly, Prob(C=i and P=i and H=k) = 1/18 for each choice

of (i,k) from { (1,3) , (2,1) , (2,3) , (3,1) , (3,2) } .

Both for switching and for staying, the six ways are pairwise disjoint, so

Prob(switching wins)

=

Prob((C=1 and P=2 and H=3) or (C=1 and P=3 and H=2) or (C=2 and P=1 and H=3)

or (C=2 and P=3 and H=1) or (C=3 and P=1 and H=2) or (C=3 and P=2 and H=1))

=

1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 = 6 * 1/9 = 2/3

and

Prob(staying wins)

=

Prob((C=1 and P=1 and H=2) or (C=1 and P=1 and H=3) or (C=2 and P=2 and H=1)

or (C=2 and P=2 and H=3) or (C=3 and P=3 and H=1) or (C=3 and P=3 and H=2))

=

1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18 = 6 * 1/18 = 1/3

.

Now, here is what follows from

The producers always put the car behind door 1.

and

The player makes the player's first choice

uniformly at random from among the 3 doors.

and

The host _must_ open exactly one door.

and

The host cannot open the door from the players 1st guess.

and

The host knows where the car is, and can't open that door.

and

If that door is the door the player guessed, then the host

chooses uniformly at random which goat door the host opens.

:

P = player's 1st choice

H = door host opens

Even if these three probabilities are irrelevant

- because there is no prize for the 1st choice - it _still_ follows from

The player makes the player's first choice

uniformly at random from among the 3 doors.

that Prob(P=1) = 1/3 and Prob(P=2) = 1/3 and Prob(P=3) = 1/3 .

When P=2 , the host _must_ open door 3,

so Prob(H=3 given P=2) = 1 .

Prob(P=2 and H=3) = Prob(P=2) * Prob(H=3 given P=2) = 1/3 * 1 = 1/3

Similarly, Prob(P=3 and H=2) = 1/3 .

When P=2 , the host chooses uniformly at random which goat

door the host opens, so Prob(H=2 given P=1) = 1/2 .

Prob(P=1 and H=2) = Prob(P=1) * Prob(H=2 given P=1) = 1/3 * 1/2 = 1/6

Similarly, Prob(P=1 and H=3) = 1/6 .

The two events [P=2 and H=3] and [P=3 and H=2] are disjoint, so

Prob(switching wins) = Prob( (P=2 and H=3) or (P=3 and H=2) )

= Prob(P=2 and H=3) + Prob(P=3 and H=2) = 1/3 + 1/3 = 2/3

.

The two events [P=1 and H=2] and [P=1 and H=3] are disjoint, so

Prob(staying wins) = Prob( (P=1 and H=2) or (P=1 and H=3) )

= Prob(P=1 and H=2) + Prob(P=1 and H=3) = 1/6 + 1/6 = 1/3

.

User:JumpDiscont

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