User:Martin Hogbin/Monty Hall analysis

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This page is my version of a more complete analysis of the Monty Hall problem The calculated probabilities that the player will win by switching are shown after the headings for each case. The contents therefore serves as a summary of the results. Note that the term 'randomly' is used here to mean uniformly at random.

Contents

Sample space[edit]

The MHP may be described in terms of the stochastic variables C, X and H:

C = number of door with car,
X = number of door of player's initial choice,
H = number of door opened by host.

each of them with values 1,2 and 3. The sample consists of the 27 events: {C=c,X=x,H=h}, for c,x,h = 1,2,3.

Clearly

P(C=1) + P(C=2) + P(C=3) = 1

and

P(X=1) + P(X=2) + P(X=3) = 1

General assumption[edit]

Independence of car placement and initial choice[edit]

The placing of the car and the player's choice are independent:

C and H are independent

Probability[edit]

The probability of any event is given by

P(C=c,X=x,H=h) = P(C=c) P(X=x) P(H=h|C=c,X=x)

The host's door opening probabilities[edit]

We describe the host's possible door opening strategy by a set of parameters P(H=h|C=c,X=x) representing the probability that the host will open door h given that the car is behind door c and the player has chosen door x. So P(H=1|C=2,X=3) is the probability that the host will open door 1 if the car is behind door 2 and the player chooses door 3

>>>>>> Nijdam, I am trying to answer a more general version of this problem at this stage , without any assumptions as to conditions. Conditions can always be added later, as is done below.

Can we discuss this on the talk page please. Martin Hogbin (talk) 09:47, 2 August 2010 (UTC)

Some game rules[edit]

Here we note the effect of various game rules on the possible values of the host door choice parameters.

1 The host must open a door. This means means that for all c and x:

P(H=1|C=c,X=x) + P(H=2|C=c,X=x) + P(H=3|C=c,X=x) = 1

2 The host cannot open the door that the player has chosen. This means that for all c and x:

P(H=x|C=c,X=x) = 0

Therefore:

P(H=2|C=1,X=1) + P(H=3|C=1,X=1) = 1 (if the player chooses door 1 the host must open door 2 or 3)
P(H=2|C=2,X=1) + P(H=3|C=2,X=1) = 1
P(H=2|C=3,X=1) + P(H=3|C=3,X=1) = 1
P(H=1|C=1,X=2) + P(H=3|C=1,X=2) = 1 (if the player chooses door 2 the host must open door 1 or 3)
P(H=1|C=2,X=2) + P(H=3|C=2,X=2) = 1
P(H=1|C=3,X=2) + P(H=3|C=3,X=2) = 1
P(H=2|C=1,X=3) + P(H=1|C=1,X=3) = 1 (if the player chooses door 3 the host must open door 2 or 1)
P(H=2|C=2,X=3) + P(H=1|C=2,X=3) = 1
P(H=2|C=3,X=3) + P(H=1|C=3,X=3) = 1


3 The host must not reveal a car means that for all c and x: P(H=c|C=c,X=x) = 0

Applying the above to the case above rule gives us:

P(H=2|C=1,X=1) + P(H=3|C=1,X=1) = 1
P(H=3|C=2,X=1) = 1
P(H=2|C=3,X=1) = 1
P(H=3|C=1,X=2) = 1
P(H=1|C=2,X=2) + P(H=3|C=2,X=2) = 1
P(H=1|C=3,X=2) = 1
P(H=2|C=1,X=3) = 1
P(H=1|C=2,X=3) = 1
P(H=2|C=3,X=3) + P(H=1|C=3,X=3) = 1

Probability of winning by switching in various cases[edit]

Unconditional problem[edit]

The player decides, before she even has chosen a door, to switch. Then:

P(winning)=P(C≠ X)=1-P(C=X)=1-SUM P(C=x|X=x)P(X=x)=1-SUM P(C=x)P(X=x)

When the car has been placed randomly: P(C=c)=1/3, hence:

P(winning)=1-SUM P(X=x)/3=2/3

Similarly, when the choice is randomly: P(X=x)=1/3, hence:

P(winning)=1-SUM P(C=c)/3=2/3

If on the other hand both are not uniformly, something else may result. For instance

X: 1/2, 1/4, 1/4 and C: 1/2, 1/4, 1/4,

then

P(winning)=1-SUM P(C=x)P(X=x) = 1 - (1/4+1/16+1/16) = 5/8 < 2/3

Or

X: 1/2, 1/4, 1/4 and C: 1/4, 1/4, 1/2,

then

P(winning)=1-SUM P(C=x)P(X=x) = 1 - (1/8+1/16+1/8) = 11/16 > 2/3

The player may even have no chance of winning:

X: 1, 0, 0 and C: 1, 0, 0,

then

P(winning)=1-SUM P(C=x)P(X=x) = 1 - (1+0+0) = 0

Next step[edit]

The player makes her choice and immediately decides to switch. Factually a conditional problem, conditioned by the player's choice. As X and C are independent, often a particular choice is taken as starting point, and considered to be the unconditional situation.

Choice: {X=x}

P(winning|X=x) = P(C≠ x|X=x) = 1 - P(C=x|X=x) = 1 - P(C=x)

The producer places the car randomly, the player chooses randomly and the host acts randomly[edit]

If the car is randomly placed and the player chooses randomly

P(C=j)=1/3 and P(X=k)=1/3

Assuming X and C to be independent, the probability distribution in the above sample space is:

P(C=1)= P(C=2)=P(C=3)= 1/3

and

P(X=1)= P(X=2)=P(X=3)= 1/3

Thus P(C=1, X=1, H=2) = P(H=2|C=1,X=1) P(C=1) P (X=1)

As the host acts randomly (within the given rules)

P(H=2|C=1,X=1) = P(H=3|C=1,X=1) = 1/2
P(H=3|C=2,X=1) = 1
P(H=2|C=3,X=1) = 1
P(H=3|C=1,X=2) = 1
P(H=1|C=2,X=2) = P(H=3|C=2,X=2) = 1/2
P(H=1|C=3,X=2) = 1
P(H=2|C=1,X=3) = 1
P(H=1|C=2,X=3) = 1
P(H=2|C=3,X=3) = P(H=1|C=3,X=3) = 1/2

The probabilities of possible events in our sample space are given by:

P(C=c,X=x,H=h) = P(C=c) P(X=x) P(H=h|C=c,X=x)

For example.

P(C=1,X=2,H=3) = P(C=1) P(X=1) P(H=3|C=1,X=2) = 1/3 . 1/3. 1 = 1/9


Probability on the condition that the player has chosen any door and host has opened any door to reveal a goat 2/3[edit]

This means conditioning on {H≠C}, and as P(H≠C)=1, the conditional probabilities will be the same as the unconditional ones.

First we condition the above sample space by removing all events in which the condition is not met. We note that, under the stated game rules, these all already have zero probability thus thus this condition is a null condition - it has no effect. The result is exactly the same as the unconditional case.

The probability that the player wins by sticking may be found by summing the probabilities of for all events in the sample space in which C = X

( P(H=3|C=1,X=1) + P(H=2|C=1,X=1) + P(H=1|C=2,X=2) + P(H=3|C=2,X=2) + P(H=2|C=3,X=3) + P(H=1|C=3,X=3) ) / 9 = 1/3


Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which C <> X

( P(H=3|C=2,X=1) + P(H=2|C=3,X=1) + P(H=3|C=1,X=2) + P(H=1|C=3,X=2) + P(H=2|C=1,X=3) + P(H=1|C=2,X=3) ) / 9 = 2/3

Probability on the condition that the player has chosen door 1 and the host has opened any door to reveal a goat 2/3[edit]

First we condition the above sample space by removing all events in which the condition is not met and renormalizing.


We now have only four events in our sample space

(H=3,C=1,X=1) (H=3, C=2, X=1) (H=2,C=1,X=1) (H=2, C=3, X=1) with probabilities:

with probabilites P(H=3|C=1,X=1)/3, P(H=2|C=3,X=1)/3, P(H=2|C=1,X=1)/3, P(H=2|C=3,X=1)/3 P321/3 P211/3 P231/3 respectively

Probability that the player wins by sticking is found by summing the probabilities for all events in the sample space in which C=X

P(H=3|C=1,X=1)/3 + P(H=2|C=1,X=1)/3 = 1/2


Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which C<>X


P(H=3|C=2,X=1)/3 + P(H=2|C=2,X=1)/3 = 2/3

Probability on the condition that the player has chosen door 1 and the host has opened door 3 to reveal a goat 2/3[edit]

The conditional probability given these events is:

P(A|X=1,H=3)

The only relevant events are: {C=1} and {C=2}, with:

P(C=1|X=1,H=3)=1-P(C=2|X=1,H=3)

Nijdam (talk) 15:39, 24 July 2010 (UTC)


First we condition the above sample space by removing all events in which the conditions (X=1) (H=3) are not met and renormalizing.

???? P(H=3) = 1, P(X=1) = 1

We now have only two events in our sample space these are:

(H=3,C=1,X=1) (H=3,C=2,X=1) with probabilities:

with conditional probabilities P(C=1|H=3,X=1)=1/3 and P(C=2|H=3,X=1)=2/3 respectively

Probability that the player wins by sticking is found by summing the probabilities for all events in the sample space in which C=X

P(C=1|H=3,X=1)=1/3 = 1/3

Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which C<>X

P(C=2|H=3,X=1)=1/3 = 2/3

The producer places the car randomly, the player chooses randomly and the host acts non-randomly[edit]

This is the Morgan scenario

If the car is randomly placed and the player chooses randomly

P(C=j)=1/3 and P(X=k)=1/3

Assuming X and C to be independent, the probability distribution in the above sample space is:

P(C=1)= P(C=2)=P(C=3)= 1/3

and

P(X=1)= P(X=2)=P(X=3)= 1/3

Thus P(C=1, X=1, H=2) = P(H=2|C=1,X=1) P(C=1) P (X=1)

The probabilities of possible events in our sample space are given by:

P(C=c,X=x,H=h) = P(C=c) P(X=x) P(H=h|C=c,X=x)

For example.

P(C=1,X=2,H=3) = P(C=1) P(X=1) P(H=3|C=1,X=2) = 1/3 . 1/3. 1 = 1/9

Unconditional problem 2/3[edit]

The player decides, before she even has chosen a door, to switch. Then:

P(winning)=P(C≠ X)=1-P(C=X)=1-SUM P(C=x|X=x)P(X=x)=1-SUM P(C=x)P(X=x)

When the car has been placed randomly: P(C=c)=1/3, hence:

P(winning)=1-SUM P(X=x)/3=2/3

Similarly, when the choice is randomly: P(X=x)=1/3, hence:

P(winning)=1-SUM P(C=c)/3=2/3

Probability on the condition that the player has chosen any door and host has opened any door to reveal a goat 2/3[edit]

This means conditioning on {H≠C}, and as P(H≠C)=1, the conditional probabilities will be the same as the unconditional ones.

First we condition the above sample space by removing all events in which the condition is not met.

The probability that the player wins by sticking may be found by summing the probabilities of for all events in the sample space in which C = X

( P(H=3|C=1,X=1) + P(H=2|C=1,X=1) + P(H=1|C=2,X=2) + P(H=3|C=2,X=2) + P(H=2|C=3,X=3) + P(H=1|C=3,X=3) ) / 9

Noting that P(H=3|C=1,X=1) + P(H=2|C=1,X=1) =1 etc this gives us:

( 1 + 1 + 1 ) /9 = 1/3

Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which C <> X

( P(H=3|C=2,X=1) + P(H=2|C=3,X=1) + P(H=3|C=1,X=2) + P(H=1|C=3,X=2) + P(H=2|C=1,X=3) + P(H=1|C=2,X=3) ) / 9

Here all the terms are equal to 1 so the probability becomes 6/9 = 2/3


Probability on the condition that the player has chosen door 1 and the host has opened any door to reveal a goat 2/3[edit]

First we condition the above sample space by removing all events in which the condition (X=1) is not met and renormalizing.

We now have only four events in our sample space

(H=3,C=1,X=1) (H=3, C=2, X=1) (H=2,C=1,X=1) (H=2, C=3, X=1) with probabilities:

with probabilites P(H=3|C=1,X=1)/3, P(H=2|C=3,X=1)/3, P(H=2|C=1,X=1)/3, P(H=2|C=3,X=1)/3 respectively

Probability that the player wins by sticking is found by summing the probabilities for all events in the sample space in which j=k

P(H=3|C=1,X=1)/3 + P(H=2|C=1,X=1)/3

Noting that P(H=3|C=1,X=1)/3 + P(H=2|C=1,X=1) = 1 this gives us 1/3.

Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which j<>k

P(H=3|C=2,X=1)/3 + P(H=2|C=2,X=1)/3 = 2/3

Probability on the condition that the player has chosen door 1 and the host has opened door 3 to reveal a goat 1/(1+q)[edit]

We condition the above sample space by removing all events in which the conditions (X=1) (H=3) are not met and renormalizing.

P(H=3) = 1, P(X=1) = 1

We now have only two events in our sample space these are:

(H=3,C=1,X=1) (H=3,C=2,X=1) with probabilities:

with probabilites P(H=3|C=1,X=1)N, P(H=3|C=2,X=1)/N respectively

Where N is the normalization factor = P(H=3|C=1,X=1) + P(H=3|C=2,X=1)


P(H=3|C=2,X=1) = 1 and using Morgan's notation in which P(H=3|C=1,X=1) = q

This gives us N = 1 + q

Probability that the player wins by sticking is found by summing the probabilities for all events in the sample space in which C=X

P(H=3|C=1,X=1) / N = q /(1+q)

Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which C<>X

P(H=3|C=2,X=1) / N = 1 /(1+q)

This is Morgan's solution

Probability on the condition that the player has chosen door 1 and the host has opened door 2 to reveal a goat 1/(2-q)[edit]

We condition the sample space by removing all events in which the conditions (X=1) (H=2) are not met and renormalizing.

P(H=2) = 1, P(X=1) = 1

We now have only two events in our sample space these are:

(H=2,C=1,X=1) (H=2,C=3,X=1) with probabilities:

with probabilites P(H=2|C=1,X=1)/N, P(H=2|C=3,X=1)/N respectively

Where N is the normalization factor = P(H=2|C=1,X=1) + P(H=2|C=3,X=1)

P(H=2|C=3,X=1) = 1 and using Morgan's notation in which P(H=3|C=1,X=1) = q and P(H=2|C=1,X=1) = 1 - q

This gives us N = 2 - q

Probability that the player wins by sticking is found by summing the probabilities for all events in the sample space in which C=X

P(H=2|C=1,X=1) / N = q /(2-q)

Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which C<>X

P(H=2|C=3,X=1) / N = 1 /(2-q)

Probability on the condition that the player has chosen door 2 and the host has opened door 3 to reveal a goat 1/ (P(H=3|C=1,X=2)N + P(H=3|C=2,X=2))[edit]

We condition the above sample space by removing all events in which the conditions (X=2) (H=3) are not met and renormalizing.

P(H=3) = 1, P(X=2) = 1

We now have only two events in our sample space these are:

(H=3,C=1,X=2) (H=3,C=2,X=2) with probabilities:

with probabilites P(H=3|C=1,X=2)N, P(H=3|C=2,X=2)/N respectively

Where N is the normalization factor = P(H=3|C=1,X=2) + P(H=3|C=2,X=2)


Morgan's notation is inapplicable to this case

So N = P(H=3|C=1,X=2)N + P(H=3|C=2,X=2)

Probability that the player wins by sticking is found by summing the probabilities for all events in the sample space in which C=X

P(H=3|C=2,X=2) / (P(H=3|C=1,X=2)N + P(H=3|C=2,X=2))

Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which C<>X

P(H=3|C=1,X=2) / (P(H=3|C=1,X=2)N + P(H=3|C=2,X=2))
= 1/ (P(H=3|C=1,X=2)N + P(H=3|C=2,X=2))


The producer places the car non-randomly, the player chooses randomly and the host chooses non-randomly[edit]

If the player chooses randomly

P(X=k)=1/3

Assuming X and C to be independent, the probability distribution in the above sample space is:

P(X=1)= P(X=2)=P(X=3)= 1/3

Thus P(C=1, X=1, H=2) = P(H=2|C=1,X=1) P(C=1) P (X=1)

The probabilities of possible events in our sample space are given by:

P(C=c,X=x,H=h) = P(C=c) P(X=x) P(H=h|C=c,X=x)

For example.

P(C=1,X=2,H=3) = P(C=1) P(X=1) P(H=3|C=1,X=2) = P(C=1). 1/3. 1

Unconditional problem 2/3[edit]

Probability on the condition that the player has chosen any door and host has opened any door to reveal a goat 2/3[edit]

This means conditioning on {H≠C}, and as P(H≠C)=1, the conditional probabilities will be the same as the unconditional ones.

First we condition the above sample space by removing all events in which the condition is not met.

The probability that the player wins by sticking may be found by summing the probabilities of for all events in the sample space in which C = X

( P(H=3|C=1,X=1)P(C=1) + P(H=2|C=1,X=1)P(C=1) + P(H=1|C=2,X=2)P(C=2) + P(H=3|C=2,X=2)P(C=2) + P(H=2|C=3,X=3)P(C=3) + P(H=1|C=3,X=3)P(C=3) ) / 3

Noting that P(H=3|C=1,X=1) + P(H=2|C=1,X=1) =1 etc and (P(C=1) + P(C=2) + P(C=3) ) = 1 this gives us:

( P(C=1) + P(C=2) + P(C=3) ) /3 = 1/3

Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which C <> X

( P(H=3|C=2,X=1)P(C=2) + P(H=2|C=3,X=1)P(C=3) + P(H=3|C=1,X=2)P(C=1) + P(H=1|C=3,X=2) P(C=3) + P(H=2|C=1,X=3) P(C=1) + P(H=1|C=2,X=3)P(C=2) ) / 3

Here all the terms relating to the host choice are are equal to 1 and again (P(C=1) + P(C=2) + P(C=3) ) = 1 so the probability becomes = 2/3

Probability on the condition that the player has chosen door 1 and host has opened door 3 to reveal a goat - indeterminate[edit]

We condition the above sample space by removing all events in which the conditions (X=1) (H=3) are not met and renormalizing.

P(H=3) = 1, P(X=1) = 1

We now have only two events in our sample space these are:

(H=3,C=1,X=1) (H=3,C=2,X=1) with probabilities:

with probabilites P(H=3|C=1,X=1)P(C=1)/N, P(H=3|C=2,X=1)P(C=2)/N respectively

Where N is the normalization factor = P(H=3|C=1,X=1)P(C=1) + P(H=3|C=2,X=1)P(C=2)


Probability that the player wins by sticking is found by summing the probabilities for all events in the sample space in which C=X

P(H=3|C=1,X=1)P(C=1) /(P(H=3|C=1,X=1)P(C=1) + P(H=3|C=2,X=1)P(C=2))

Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which C<>X

P(H=3|C=2,X=1)P(C=2) / (P(H=3|C=1,X=1)P(C=1) + P(H=3|C=2,X=1)P(C=2))

This is not Morgan's solution, in fact it is indeterminate depending on the probabilities with which the car was originally placed

The producer places the car randomly, the player chooses non-randomly and the host acts non-randomly[edit]

Here we see that, provided the car is randomly placed, the choice of the player has no effect. The player could, for example, always choose door 1 without affecting the result.

If the car is randomly placed

P(C=j)=1/3 and

Assuming X and C to be independent, the probability distribution in the above sample space is:

P(C=1)= P(C=2)=P(C=3)= 1/3


Also P(X=1) + P(X=2) + P(X=3) = 1

Thus, for example, P(C=1, X=1, H=2) = P211P(X=k) /3

Unconditional probability 2/3[edit]

The player wins by sticking if X=C otherwise the player wins by switching.

Probability that the player wins by sticking is may be found by summing the probabilities PijkP(X=k)/3 for all events in the sample space in which j=k:

Let P211 = a, P322 = b and P133 = c Let P(X=1) = u, P(X=2) = v so P(X=3) = 1 - u - v


(ua + u(1-a) + v(1-b) + vb + (1-u-v)c + (1-u-v)(1-c))/3 =

(u + v + 1 -u - v )/3 =1/3

Probability that the player wins by switching is therefore 2/3

Probability on the condition that the player has chosen door 1 and the host has opened a to reveal a goat 2/3[edit]

First we condition the above sample space by removing all events in which the condition is not met. We note that, under the stated game rules, these all already have zero probability thus thus this condition is a null condition - it has no effect. The result is exactly the same as the unconditional case.

The result is therefore the same as above.

Probability on the condition that the player has chosen door 1 and the host has opened door 3 to reveal a goat 1/(1-q)[edit]

First we condition the above sample space by removing all events in which the condition is not met and renormalizing.

We now have only two events in our sample space these are:

(H=3,C=1,X=1) and (H=3, C=2, X=1)

With probabilities P(X-1)P311/N and P(X=1)P321/N Where N is the normalisation factor

Where P(H=3,C=1,X=1) + P(H=3, C=2, X=1) = 1

Giving N= P(X=1) (P311 + 321)

Probability that the player wins by sticking is found by summing the probabilities for all events in the sample space in which j=k

it is equal to P(X=1)P311 / P(X=1)(P311 + P321)

Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which j<>k

it is equal to P(X=1)P321 / P(X=1) (P311 + P321)

We know that P321 = 1 and if we set P321 = q then

The probability that the player wins by switching is 1 / (q + 1)

Which is the same as Morgan's solution

The producer places the car non-randomly, the player chooses randomly and the host acts non-randomly[edit]

If the player chooses randomly

P(X=j)=1/3 and

Assuming X and C to be independent, the probability distribution in the above sample space is:

P(X=1)= P(X=2)=P(X=3)= 1/3


Also P(C=1) + P(C=2) + P(C=3) = 1

Thus, for example, P(C=1, X=1, H=2) = P211P(C=j) /3

Unconditional probability 2/3[edit]

The player wins by sticking if X=C otherwise the player wins by switching.

Probability that the player wins by sticking is may be found by summing the probabilities PijkP(X=k)/3 for all events in the sample space in which j=k:

Let P211 = a, P322 = b and P133 = c Let P(C=1) = f, P(C=2) = g so P(C=3) = 1 - f - g

(fP211 + fP311 + gP122 + gP322 + (1-f-g)P133 + (1-f-g)P233)/3

(fa + f(1-a) + g(1-b) + gb + (1-f-g)c + (1-f-g)(1-c))/3 =

( f + g + 1 - f - g )/3 =1/3

Probability that the player wins by switching is therefore 2/3

Probability on the condition that the player has chosen door 1 and the host has opened a to reveal a goat 2/3[edit]

First we condition the above sample space by removing all events in which the condition is not met. We note that, under the stated game rules, these all already have zero probability thus thus this condition is a null condition - it has no effect. The result is exactly the same as the unconditional case.

The result is therefore the same as above.

Probability on the condition that the player has chosen door 1 and the host has opened door 3 to reveal a goat - not 1/(1-q) or 2/3[edit]

This is a case that Morgan's analysis could apply to. Morgan do not state that the producer places the car randomly or that the host chooses randomly, although they do mention after the solution the possibility of non-random car placement. For some unexplained reason in their solution they take it that the producer places the car randomly but the host chooses non-randomly.


First we condition the above sample space by removing all events in which the condition is not met and renormalizing.

We now have only two events in our sample space these are:

(H=3,C=1,X=1) and (H=3, C=2, X=1)

With probabilities P(C=1)P311/N and P(C=2)P321/N Where N is the normalisation factor

Where P(H=3,C=1,X=1) + P(H=3, C=2, X=1) = 1

Giving N= P(C=1)P311 + P(C=2)321)

Probability that the player wins by sticking is found by summing the probabilities for all events in the sample space in which j=k

it is equal to P(C=1)P311 / (P(C=1)P311 + P(C=2)321))

Note that this is not equal to 1/3 and is not Morgan's solution. Unsurprisingly it depends also on the probability that the car was placed behind door 1.

Probability that the player wins by switching is the complement of this.

The producer places the car non-randomly, the player chooses non-randomly and the host acts non-randomly[edit]

UNDER CONSTRUCTION

The probabilities of possible events in our sample space are given by:

P(C=c,X=x,H=h)

For example.

P(C=1,X=2,H=3) = P(C=1) P(X=1) P(H=3|C=1,X=2) = P(C=1) P(X=1) . 1


Unconditional problem 2/3[edit]

Probability on the condition that the player has chosen any door and host has opened any door to reveal a goat 2/3[edit]

This means conditioning on {H≠C}, and as P(H≠C)=1, the conditional probabilities will be the same as the unconditional ones.


The probability that the player wins by sticking may be found by summing the probabilities of for all events in the sample space in which C = X

( P(H=3|C=1,X=1)P(C=1) + P(H=2|C=1,X=1)P(C=1) + P(H=1|C=2,X=2)P(C=2) + P(H=3|C=2,X=2)P(C=2) + P(H=2|C=3,X=3)P(C=3) + P(H=1|C=3,X=3)P(C=3) ) / 3

Noting that P(H=3|C=1,X=1) + P(H=2|C=1,X=1) =1 etc and (P(C=1) + P(C=2) + P(C=3) ) = 1 this gives us:

( P(C=1) + P(C=2) + P(C=3) ) /3 = 1/3


Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which C <> X

( P(H=3|C=2,X=1)P(C=2) + P(H=2|C=3,X=1)P(C=3) + P(H=3|C=1,X=2)P(C=1) + P(H=1|C=3,X=2) P(C=3) + P(H=2|C=1,X=3) P(C=1) + P(H=1|C=2,X=3)P(C=2) ) / 3

Here all the terms relating to the host choice are are equal to 1 and again (P(C=1) + P(C=2) + P(C=3) ) = 1 so the probability becomes = 2/3


This is the most general case possible. We make no assumptions about the original placement of the car, the hosts choice of door to reveal a goat, and the player's original choice of door. We assume that no door numbers are specified in the question

Probability on the condition that the player has chosen door 1 and host has opened door 3 to reveal a goat - indeterminate[edit]

We condition the above sample space by removing all events in which the conditions (X=1) (H=3) are not met and renormalizing.

P(H=3) = 1, P(X=1) = 1

We now have only two events in our sample space these are:

(H=3,C=1,X=1) (H=3,C=2,X=1) with probabilities:

with probabilities P(H=3|C=1,X=1)P(C=1)/N, P(H=3|C=2,X=1)P(C=2)/N respectively

Where N is the normalization factor = P(H=3|C=1,X=1)P(C=1) + P(H=3|C=2,X=1)P(C=2)

Probability that the player wins by sticking is found by summing the probabilities for all events in the sample space in which C=X

P(H=3|C=1,X=1)P(C=1) /(P(H=3|C=1,X=1)P(C=1) + P(H=3|C=2,X=1)P(C=2))

Probability that the player wins by switching is found by summing the probabilities for all events in the sample space in which C<>X

P(H=3|C=2,X=1)P(C=2) / (P(H=3|C=1,X=1)P(C=1) + P(H=3|C=2,X=1)P(C=2))

This is not Morgan's solution, in fact it is indeterminate depending on the probabilities with which the car was originally placed. It is the most general solution given the conditions that the player has originally chosen door 1 and the host has opened door 3

Conclusions[edit]

The solution given by Morgan et al applies only in the case that the producer places the car randomly and the host acts non-randomly and when specific door numbers have been identified for the door initially chosen by the player and the door opened by the host. Whether the player chooses a door randomly or not is not important.

Note that if the car is not initially placed randomly but the player chooses a door randomly, which has the effect of randomising the initial car placement, this does not produce in Morgans result. If the door numbers are not identified then the answer (chances of winning by switching) is always 2/3. If specific doors opened by the player and the host are identified then the answer depends on the probability with which the car was placed behind the chosen door.

Full calculations including the players initial car choice[edit]

Under construction

Sample space[edit]

The MHP may be described in terms of the stochastic variables C, X and H:

C = number of door with car,
X = number of door of player's initial choice,
H = number of door opened by host.

each of them with values 1,2 and 3. The sample consists of the 27 events: {C=c,X=x,H=h}, for c,x,h = 1,2,3.

Clearly

P(C=1) + P(C=2) + P(C=3) = 1

and

P(X=1) + P(X=2) + P(X=3) = 1

General assumption[edit]

Independence of car placement and initial choice[edit]

The placing of the car and the player's choice are independent:

C and H are independent

The host always opens a door different from the door chosen by the player[edit]

This means means that for all c and x:

P(H=1|C=c,X=x) + P(H=2|C=c,X=x) + P(H=3|C=c,X=x) = 1

and

P(H=x|C=c,X=x) = 0

Probability[edit]

The probability of any event is given by

P(C=c,X=x,H=h) = P(C=c) P(X=x) P(H=h|C=c,X=x)

Decision[edit]

One may consider decisions in every stage of the game:

1. From the start: Pws1 = P(C<>X) = 1 - P(C=X) = 1 - P(C=1,X=1) - P(C=2,X=2) - P(C=3,X=3) = 1 - SUM P(C=c)P(X=c)

If either C or X is uniformly distributed this becomes 2/3

2. After the player has picked a door: Pws2(x) = P(C<>x|X=x) = P(C<>x)

If C is uniformly distributed this becomes 2/3 for all x.

3. After the host has opened a goat door: The decision asked for is based on the conditional probability given the known events {X=x} and {H=h}. In the example case x=1 and h=3 this becomes:

Pws= P(win the car by switching|H=3,X=1) =

= P(C=2|H=3,X=1) =
= P(H=3|C=2,X=1)P(C=2|X=1) / ( P(H=3|C=1,X=1)P(C=1|X=1) + P(H=3|C=2,X=1)P(C=2|X=1) + P(H=3|C=3,X=1) P(C=3|X=1) )
= P(H=3|C=2,X=1)P(C=2) / ( P(H=3|C=1,X=1)P(C=1) + P(H=3|C=2,X=1)P(C=2) + P(H=3|C=3,X=1) P(C=3) )

In the case where the host never reveals the car, this reduces to: Pws =

= P(H=3|C=2,X=1)P(C=2) / ( P(H=3|C=1,X=1)P(C=1) + P(C=2) + 0)
= 1 / ( P(H=3|C=1,X=1)P(C=1)/P(C=2) + 1 )
= 1/(q c1/c2 + 1)

with: q=P(H=3|C=1,X=1) and c1,c2=P(C=1,2)

If C is uniformly distributed this becomes 1/(q+1).

Analogous for other combination of doors.

Rules of the game[edit]

The host must not reveal a car, means that for all c and x: P(H=c|C=c,X=x) = 0

Applying the above to the case above rule gives us:

P(H=2|C=1,X=1) + P(H=3|C=1,X=1) = 1
P(H=3|C=2,X=1) = 1
P(H=2|C=3,X=1) = 1
P(H=3|C=1,X=2) = 1
P(H=1|C=2,X=2) + P(H=3|C=2,X=2) = 1
P(H=1|C=3,X=2) = 1
P(H=2|C=1,X=3) = 1
P(H=1|C=2,X=3) = 1
P(H=2|C=3,X=3) + P(H=1|C=3,X=3) = 1

The host's door opening probabilities[edit]

We describe the host's possible door opening strategy by a set of parameters P(H=h|C=c,X=x) representing the probability that the host will open door h given that the car is behind door c and the player has chosen door x. So P(H=1|C=2,X=3) is the probability that the host will open door 1 if the car is behind door 2 and the player chooses door 3


>>>>>>What else?

Intermezzo: combined doors[edit]

(>>>>Martin, be so kind to write down the combined doors solution.)Nijdam (talk) 10:24, 14 August 2010 (UTC) >>>>>>until here<<<<<<<

Full Sample set[edit]

P(C=1) P(X=1) P(H=1|C=1,X=1)
P(C=1) P(X=1) P(H=2|C=1,X=1)
P(C=1) P(X=1) P(H=3|C=1,X=1)

P(C=1) P(X=2) P(H=1|C=1,X=2)
P(C=1) P(X=2) P(H=2|C=1,X=2)
P(C=1) P(X=2) P(H=3|C=1,X=2)

P(C=1) P(X=3) P(H=1|C=1,X=3)
P(C=1) P(X=3) P(H=2|C=1,X=3)
P(C=1) P(X=3) P(H=3|C=1,X=3)


P(C=2) P(X=1) P(H=1|C=2,X=1)
P(C=2) P(X=1) P(H=2|C=2,X=1)
P(C=2) P(X=1) P(H=3|C=2,X=1)

P(C=2) P(X=2) P(H=1|C=2,X=2)
P(C=2) P(X=2) P(H=2|C=2,X=2)
P(C=2) P(X=2) P(H=3|C=2,X=2)

P(C=2) P(X=3) P(H=1|C=2,X=3)
P(C=2) P(X=3) P(H=2|C=2,X=3)
P(C=2) P(X=3) P(H=3|C=2,X=3)


P(C=3) P(X=1) P(H=1|C=3,X=1)
P(C=3) P(X=1) P(H=2|C=3,X=1)
P(C=3) P(X=1) P(H=3|C=3,X=1)

P(C=3) P(X=2) P(H=1|C=3,X=2)
P(C=3) P(X=2) P(H=2|C=3,X=2)
P(C=3) P(X=2) P(H=3|C=3,X=2)

P(C=3) P(X=3) P(H=1|C=3,X=3)
P(C=3) P(X=3) P(H=2|C=3,X=3)
P(C=3) P(X=3) P(H=3|C=3,X=3)

Removal of events forbidden under the rules[edit]

The host cannot open a door chosen by the player or a door that hides the car. All P(C=c) P(X=x) P(H=c|C=c,X=x) = 0
All P(C=c) P(X=x) P(H=x|C=c,X=x) = 0

Thus the complete sample set allowed under the game rules is:

P(C=1) P(X=1) P(H=2|C=1,X=1)
P(C=1) P(X=1) P(H=3|C=1,X=1)

P(C=1) P(X=2) P(H=3|C=1,X=2)

P(C=1) P(X=3) P(H=2|C=1,X=3)


P(C=2) P(X=1) P(H=3|C=2,X=1)

P(C=2) P(X=2) P(H=1|C=2,X=2)
P(C=2) P(X=2) P(H=3|C=2,X=2)

P(C=2) P(X=3) P(H=1|C=2,X=3)


P(C=3) P(X=1) P(H=2|C=3,X=1)

P(C=3) P(X=2) P(H=1|C=3,X=2)

P(C=3) P(X=3) P(H=1|C=3,X=3)
P(C=3) P(X=3) P(H=2|C=3,X=3)


Probability of winning by switching[edit]

Consider probabilities of events in which x<>c as a fraction of total probabiliity.

Pws =

( P(C=1) P(X=2) P(H=3|C=1,X=2) + P(C=1) P(X=3) P(H=2|C=1,X=3) + P(C=2) P(X=1) P(H=3|C=2,X=1)

+ P(C=2) P(X=3) P(H=1|C=2,X=3) + P(C=3) P(X=1) P(H=2|C=3,X=1) + P(C=3) P(X=2) P(H=1|C=3,X=2) )

/

( P(C=1) P(X=1) P(H=2|C=1,X=1) + P(C=1) P(X=1) P(H=3|C=1,X=1) + P(C=1) P(X=2) P(H=3|C=1,X=2) + P(C=1) P(X=3) P(H=2|C=1,X=3)

+ P(C=2) P(X=1) P(H=3|C=2,X=1) + P(C=2) P(X=2) P(H=1|C=2,X=2) + P(C=2) P(X=2) P(H=3|C=2,X=2) + P(C=2) P(X=3) P(H=1|C=2,X=3)

+ P(C=3) P(X=1) P(H=2|C=3,X=1) + P(C=3) P(X=2) P(H=1|C=3,X=2) + P(C=3) P(X=3) P(H=1|C=3,X=3) + P(C=3) P(X=3) P(H=2|C=3,X=3) )

Applying the game rules that the host must open an unchosen door to reveal a goat gives:

Pws = ( P(C=1) P(X=2) + P(C=1) P(X=3) + P(C=2) P(X=1) + P(C=2) P(X=3) + P(C=3) P(X=1) + P(C=3) P(X=2) )

/

( P(C=1) P(X=1) P(H=2|C=1,X=1) + P(C=1) P(X=1) P(H=3|C=1,X=1) + P(C=1) P(X=2) + P(C=1) P(X=3) +

+ P(C=2) P(X=1) + P(C=2) P(X=2) P(H=1|C=2,X=2) + P(C=2) P(X=2) P(H=3|C=2,X=2) + P(C=2) P(X=3) +

+ P(C=3) P(X=1) + P(C=3) P(X=2) + P(C=3) P(X=3) P(H=1|C=3,X=3) + P(C=3) P(X=3) P(H=2|C=3,X=3) )

The following facts still need to be applied:

P(X=1) + P(X=2) + P(X=3) = 1

P(C=1) + P(C=2) + P(C=3) = 1

P(H=2|C=1,X=1) + P(H=3|C=1,X=1) = 1, P(H=1|C=2,X=2) + P(H=3|C=2,X=2) = 1, P(H=2|C=3,X=3) + P(H=1|C=3,X=3) = 1

These are best applied after the distributions and conditions.

This is a general solution to the MHP. Specific solutions may be obtained by applying distributions and conditions to the above. Note that in the general case the probability of winning by switching is indeterminate and can be anything from 0 to 1. For example:

If the car is always behind door 1 and the player always picks door 1 then Pws = 0
If the car is always behind door 1 and the player always picks door 2 then Pws = 1

The player initially chooses a door uniformly at random[edit]

This means that P(X=1) = P(X=2) = P(X=3) = 1/3 thus

Pws = ( P(C=1) + P(C=1) + P(C=2) + P(C=2) + P(C=3) + P(C=3) )

/

( P(C=1) P(H=2|C=1,X=1) + P(C=1) P(H=3|C=1,X=1) + P(C=1) + P(C=1)

+ P(C=2) + P(C=2) P(H=1|C=2,X=2) + P(C=2) P(H=3|C=2,X=2) + P(C=2)

+ P(C=3) + P(C=3) + P(C=3) P(H=1|C=3,X=3) + P(C=3) P(H=2|C=3,X=3) )

As P(H=2|C=1,X=1) + P(H=3|C=1,X=1) etc = 1

Pws = ( P(C=1) + P(C=1) + P(C=2) + P(C=2) + P(C=3) + P(C=3) ) / ( P(C=1) + P(C=1) + P(C=1) + P(C=2) + P(C=2) + P(C=2) + P(C=3) + P(C=3) + P(C=3) )

= 2/3

Note that this result is independent of the host goat door policy and the initial placement of the cars.

No uniform distributions - the player has chosen door 1 and the host has opened door 3[edit]

The solution must be conditioned by removing all events in which the above conditions do not hold.

Pws = ( P(C=2) P(X=1) P(H=3|C=2,X=1) ) / ( P(C=1) P(X=1) P(H=3|C=1,X=1) + P(C=2) P(X=1) P(H=3|C=2,X=1) )

The value of Pws is indeterminate. If P(C=2) = 0 the Pws = 0, if P(C=2) = 1 (and therefore P(C=1) = 0 then Pws = 1 The results clearly show that Morgan's solution only applies in the case where:

  1. The producer has acted randomly in that the car is initially randomly placed.
  2. The host has not acted randomly in that he may choose non-randomly which door to open.
  3. The problem identifies the door that the host has, in fact, opened

Combining doors[edit]

Nijdam, please can we have these discussions on the talk page. I have moved your comment there. The purpose of this page is to provide a general solution to the MHP on an agreed mathematical basis, so that it is clear to everyone involved in discussion exactly what he effect of any given condition or assumption is.