Problem of Monty Hall

the problem of Monty Hall is a probabilistic Casse-tête freely inspired by the Jeu televised American Let' S Make has Deal . It is simple in its statement but nonintuitive in his resolution and this is why one speaks sometimes about it about Paradoxe of Monty Hall . The name of this mathematical problem comes from the name of that which has presents this play to the United States during thirteen years, the organizer of Canadian origin Monty Hall .

Statement

The play opposes a presenter to a candidate (the player). This player is placed in front of three closed doors. Behind one of them is a car (or any other splendid price) and behind each of both others are a Chèvre (or any other price of no importance). It must first of all indicate a door. Then the presenter opens a door which is neither that chosen by the candidate, nor that hiding the car (the presenter knows which is the good door as of the beginning). The candidate has the right then or to open the door which it chose initially, or to open the third door.

The questions which arise for the candidate are:

1. Which does it have to make? Is
2. Which its chances to gain the car while acting as well as possible?

History and evolution of the statement of the problem

Below the translation of a famous statement of the problem, resulting from a letter is reproduced that Craig F. Whitaker had made appear in the heading Ask Marylin of Marilyn your Scientist of the Parade Magazine in September 1990:

Suppose that you are on the plate of a television game, vis-a-vis three doors and that you must choose to open only one of them, by knowing that behind one of it is a car and behind the two others of the goats. You choose a door, say number 1, and the presenter, who knows to him what there is behind each door, opens another door, say the number 3, carries which once opened discovers a goat. He asks you then: " do you wish to open the door number 2? ". In is your opinion, with your advantage of changing choice and of opening door 2 rather than initially selected door 1?

The publication of this article in the Parade Magazine had an immediate impact on the assistantship and generated very many discussions among the mathematicians, famous or not, and the anonymous amateurs. Marilyn your Scientist, famous to off appear in Guinness Book Records as being the person with the the Intelligence quotient highest in world (IQ of 228), thus received more than 10.000 letters ( estimate made by itself ) dealing with problem, of which several coming from academics calling in question the relevance of the demonstration reproduced in its heading. In 1991, for a Sunday edition, the One of the NewYork Times opens on this subject. Jerry Pournell celebrates it chronicler of Chaos Manor de Byte, also discussed the problem lengthily as an adversary of the solution of Marilyn, to line up its side finally. Lastly, a discussed discussion took place in connection with the article of the Parade Magazine in the heading The Straight Dope held by Cecil Adams in the weekly magazine The Chicago Reader.

The relevance of the statistical results was sometimes disputed, but what generally posed problem was that the article did not insist on the " contraintes" of the presenter. The results given implied necessarily the following postulates:

• That the presenter cannot open the door chosen by the candidate .

• That the presenter systematically gives the possibility to the candidate of reconsidering his initial choice.

However, as these elements were not proposed in the statement of the problem, and this same if they were implicit, other statistical results that those given in the article became possible. Nothing not indicating that the starting statement must necessarily include these postulates, one should be able to generalize the problem with other cases.

Finally, by considering that these postulates were a condition sine qua non statement of the problem, it proved that the results of the article were actually justified.

However it missed at least an element of size: the question of knowing if the candidate were or not to change his initial decision to be likely more to gain the car had direction only if the statement specified well that the presenter knew precisely what hid behind each door, element precisely omitted in the article of the Parade Magazine. If the presenter did not know it, then the question would have been stripped of direction (as one will vera it in particular in the alternatives).

This said, this statement nothing but does fall under the line of those devoted to this type of paradox.

Indeed, one of the first appearances of this problem goes back to 1898 in Probabilités from Calculation from Joseph Bertrand where it is described like the Paradoxe of the box of Bertrand.

A current statement free from ambiguity

It is thus preferable to be based on an unambiguous statement of the problem, thus including expressly the constraints of the presenter, described by Mueser and Granberg as follows:

• Behind each of the three doors is either a goat, or a car, but only one door gives on a car whereas two doors give on a goat. The door hiding the car was chosen by drawing lot.

• the player chooses one of the doors, without however what hides behind (goat or car) not being revealed at this stage.
• the presenter knows what there is behind each door.
• the presenter must open one of the two remaining doors and must propose to the candidate the possibility of changing choice as for the door to be opened definitively.
• the presenter will always open a door behind which a goat hiding place itself, indeed:
• If the player chooses a door behind whom a goat is, the presenter will open the other carries where it knows that also a goat is.
• And if the player chooses the door hiding the car, the presenter chooses randomly among the two doors hiding a goat. (one can suppose that a drawing lot before the emission decided if it would be on the right or on the left)
• the presenter must make it possible to the candidate to remain on his initial choice or to return above and to open the door which was selected neither by itself, nor by the candidate.

The question which arises then is to know if:

• the player increases his chances to gain the car by changing its initial choice?

Or formulated differently, that amounts saying: is

• the probability of gaining while changing door larger than the probability of gaining without changing door?

Or:

• Which is the best strategy: To make a new choice or to remain with the initial choice? The chances of profit will increase, decrease or will remain the same ones?

The solution

Discusses

If one asks for a prompt response and intuitive, two incompatible points of view are opposed

• the first affirms that after opening of the door, there remain two doors, each one being likely as many to hide the car. One thus has very as many chances to gain with change without change.

• the second affirms that if one does not change a door, one gains if and only if one had made the good choice at the beginning. However this choice had a chance on three to be good. There are thus 1/3 of chances to gain without changing, 2/3 of chances to gain while changing.

This problem was a long time a probabilistic case of paradox (following the example Problème of Beautiful with wood sleeping) for which it exists two justifiable contradictory solutions (and defended each one by respected mathematicians) without one managing to make triumph an interpretation. But solution 2/3-1/3 ended up being essential (finally, there remain some partisans of assumption 1/2-1/2, as testifies some the discussion page to this article!), in particular after the realization of simulations of a great number of pullings. A roof since mathematics is not an applied science!

Important assumptions

1. the three doors have the same probability of being the gaining door; this assumption is equivalent to both which follows:
2. : (A) the door which is selected in first has a chance on three to be the gaining door
3. : (b) the two doors remaining have an equal probability of being the gaining door
4. the presenter cannot open that a door which is neither that chosen, nor the gaining door (he knows the site of the latter, which enables him to answer this condition without risk of error)
5. When the presenter has the choice between two doors to open, he chooses randomly between the two, with equiprobability

These assumptions all are important: one will see in the alternatives that the modification of any can lead to a different result. But often, the use of several of these assumptions is implicit.

First approach

The answer is yes ; the chance to gain the double car when the player modifies his initial choice systematically.

There exist three possible scenarios, each one being statistically equiprobable (1/3):

By imagining that one is in the case of figure according to: carry 1=chèvre, carries 2=chèvre and carries 3=voiture;

• the candidate chooses door 1 with a goat behind. The presenter will choose the door with the other goat ( door 2 in this example). A change will make it possible to gain the car.

• the candidate chooses door 2 with a goat behind. The presenter will choose the door with the other goat ( door 1 in this example). A change will make it possible to gain the car.
• the candidate chooses the door where it there with the car. The presenter will choose any of the two other doors ( doors 1 and 2 in this example), since it is certain that they hide goats. A change will make lose the candidate.

In the first two scenarios, the player gains if it changes his choice. The third scenario is the only one where the player gains if it does not change. As 2 of the 3 scenarios make it possible to gain when one changes door, the probability of gaining with the change is of 2/3.

The problem would be different if there were no initial choice, or if the presenter chose the door randomly to be opened, or if the presenter could propose to more or less often change according to the original choice of the player. For example, if the presenter opened a door initially hiding a goat, then asked the participant to make his choice (remain two doors), the probabilities would be then of 1/2. In the original problem, it is because the player had 2 chances out of 3 to select one of the two goats that the decision of the Master of play revealed the correct answer 2 times out of 3.

Another means of having the solution is that by supposing that the player will change, the only means of gaining would be to choose a door behind which there is no car, since the presenter will open the other then carries hiding a goat, eliminating any risk to choose a door hiding a goat after change. Since the full number of doors is of 3 and that the full number of doors with goat is of 2, the probability of gaining a car by changing answer is of 2/3 because it is the probability of choosing a door hiding a goat as of the first choice.

One can still see that the choice which the presenter proposes after having opened a door is in fact equivalent not to open of door, but to propose to give up the door chosen initially to choose two others . Since there is inevitably at least a bad door among the nonselected doors (in fact, if this argument is intuitive for much, it would actually be necessary to resort to the formulas of Bayes to show that the door indicated initially has its probability of hiding the unchanged car, and one will see low than all the preceding assumptions are essential for that).

Refutation of the arguments pro-1/2

A long time the reasoning developed above did not achieve the unanimity. It was reproached to him for considering that the opening of a bad door leaves unchanged the probability so that the initially selected door is the good one (1/3). It is indeed legitimate to wonder why the opening of the third door modifies the probability only of one of the two doors.

Those which refuse this reasoning consider that the situation after opening of a door is equivalent to open a bad front door the choice of the candidate. They affirm consequently that the probability of gaining is the same one while changing or without changing, that is to say 1/2.

The fault of this type of reasoning is to retain only the event " a door was ouverte". If one of the nonselected doors opened by accident and revealed a goat, the probabilities would pass indeed to 1/2 against 1/2 (see in particular on this subject the alternatives). But the presenter not randomly opens in fact the door, but under the constraint.

Let us give a noncalculative argument to maintain the first reasoning: Because of the particular rules of opening: it is known that the open door by the presenter is not good the and is not that selected by the player. That which one chose could ever have been open, therefore the opening does not give any new information on the door chosen at the origin. On the other hand the nonopen nonselected door could have been selected to be open… except if it is the good one. The fact that it was not selected thus increases the suspicions in its case.

One can also notice that it is equivalent to ask " did the presenter have the choice in the door to open? " since there were 2 possible doors only if the initial choice were the good. To make simple, the open door had a chance on 2 to be open if the initial choice were the good, and the certainty to be it if the initial choice were bad. There are thus 2 times more chances so that the initial choice is bad.

But the reasoning of the preceding paragraph is not very rigorous: for a demonstration without fault on the same principle, to see the paragraph " resolution by the formula of Bayes".

Connection of various calculations

To make calculation before opening of the door, it is necessary to reason as follows: one must consider the possibility that the door chosen initially is the good one, and that which each of the two other doors is the good one. It is then necessary to think at the conclusion of each one of these possibilities, i.e. to wonder which door will be opened by the presenter (4 under-cases in all) and what it will be necessary to do then to gain. One sees quickly that the probability of gaining while changing is equal to 1-p, p being the probability for the initially selected door of being the good one, here 1/3 ( Hypothèse 1a ). Thus here, one gains 2 times out of 3 while changing. It is important to remember here that it there forever of handing-over ( Hypothèses 2 ), without what the preceding reasoning is not valid any more.

The probability is it unchanged by the opening (more precisely: by the choice made by the presenter between the two doors which one did consider the opening)? Not inevitably, but such as calculations were made, the situations after opening are under-cases of preceding calculation. Therefore, without affirming immediately that the probability is unchanged, the weighted average of the probabilities corresponding to each open door by the presenter must correspond to preceding calculation. To claim that one has 1 chance out of 2 to gain without changing whatever the open door is thus incoherent.

To evaluate the chances after opening, it is enough in fact to note that there is after choice total symmetry between the 2 nonselected doors ( Hypothèses 1b and 3 ). Since the weighted average must be worth 2/3 and that the under-cases must give the same result, one finds 2 chances well out of 3 to gain by changing whatever the open door. It was thus important to specify that when two doors can be open, the choice is equiprobable.

Result 2/3 is thus perfectly valid, but it is advisable not to announce it without specifying that he rests on the perfect symmetry of the roles of the nonselected doors. By breaking this symmetry, all the results are possible.

Moreover, the reasoning employed the fact that the play never authorizes the handing-over. If the presenter does not act by exploiting his knowledge of the true door, preceding calculations do not apply.

Keys to include/understand the problem

Diagrams

The probability that the car is behind the remaining door can be calculated below with the diagrams.

After having chosen the door number 3, for example, the candidate has a chance on three to fall directly on the car with two chances out of three that the car is among the two remaining doors.

Note that since there is only one car, there is 100% of chance that there is a goat behind at least one of doors 1 or 2.

The presenter opens door 1 now. Of course the presenter never opens a door giving on the car, therefore without surprised door 1 gives on a goat what causes to transfer the probability of 2/3 from chances to have a car either on the doors 1 and 2 as previously explained, but only on door 2 ( to see graphic below).

In a way even simpler, one can reformulate by saying that so after the initial choice of the candidate it was possible that the car is behind doors 1 and 2 ( with a probability of 2/3 ), it is not more the case after the opening of door 1 by the presenter: only door 2 is still likely to hide the car ( and consequently, always with a probability of 2/3 )

The diagram below watch the same reasoning in a more complete and more formalized way:

Tree of the possibilities of the problem of Monty Hall

Simulation

As shown previously, the theoretical values given by the laws of the probabilities are thus:

- 1/3 of chances to gain the car without changing its initial choice, is approximately 33,3%.

- 2/3 of chances to gain the car by changing its initial choice, is approximately 66,7%.

But one can also practice a Simulation using a reproducing computer program of the ficitves parts and to see whether, on a great number of parts, the simulated result tends towards the result given by the probabilities and confirm them. For that two cases are to be distinguished:

1. the case where there is a change of the initial choice

2. the case where the initial choice is preserved

In each case, it is advisable to carry out a great number of situation to reduce the Margin of error and to note the percentage where the candidate gains the car.

A sample program in Javascript (English ):

Here the result of the program for 25.000 consecutive simulations of parts: Playing 25000 ranges… Win misses with has " don' T switch" strategy: 0.333 Win misses with has " switch" strategy: 0.667 Who could be translated into French as follows: After 25.000 parts… Rate of success ( the candidate gains the car ) without carrying out change ( initial choice ) is of 0.333 ( is 33,3% ) Rate of success by carrying out a change is of 0.667 ( 66,7% )

In addition to this program in Javascript, other equivalent programs in languages C, C++, Java and Perl are available on Wikisource: Simulation programs of the problem of Monty Hall (English '' '')

Here examples of results given by each one of these programs:

In language C (30 000 simulations of parts):

• By changing $\backslash \; simeq\; 66,7067\; \backslash ,\; \backslash \; \%$
• Without changing $\backslash \; simeq\; 32,8733\; \backslash ,\; \backslash \; \%$

In language C++ (1 000.000 of simulations):

• By changing $\backslash \; simeq\; 66,7494\; \backslash ,\; \backslash \; \%$
• Without changing $\backslash \; simeq\; 33,2762\; \backslash ,\; \backslash \; \%$

In language Java (1 000.000 of simulations):

• By changing $\backslash \; simeq\; 66,7032\; \backslash ,\; \backslash \; \%$
• Without changing $\backslash \; simeq\; 33,2968\; \backslash ,\; \backslash \; \%$

In language Perl (3 000 simulations):

• By changing $\backslash \; simeq\; 66,2333\; \backslash ,\; \backslash \; \%$
• Without changing $\backslash \; simeq\; 33,7667\; \backslash ,\; \backslash \; \%$

Simulations above, like others on Internet (University of Rouen (200 iterations), (for versions Internet to explore 4 or +, in English) ( with a margin of error ) the theoretical results of 1/3 and 2/3 confirm and this more especially as the iteration count is important.

The resolution by the theorem of Bayes

The statement returns ultimately to a problem of conditional probability and according to the general formulation of the Théorème of Bayes:

• Either $A$ an unspecified event,

• Or $\{B\_1,\; B\_2,\dots ,\; B\_n\}$ is a whole of at the same time exhaustive and mutually exclusive events,

Then for all $i$, one a:

$P\; (B\_i|A)\; =\; \backslash \; frac\; \{P\; (has\; |\; B\_i)\; P\; (B\_i)\}\{\backslash \; sum\_\; \{J\; =\; 1\}\; ^n\; P\; (has|B\_j)\; P\; (B\_j)\}\backslash ,$

An application of the theorem of Bayes to the problem of Monty Hall could be formulated as follows:

Let us consider the case where door 3 was selected and no door is yet open. The probability that the car is behind door 2 p (F2) is of 1/3, probability which would be moreover exactly the same one for each door.

The probability that the presenter opens door 1 p (O1) is then of 1/2. Indeed, the candidate having chosen door 3 and the presenter knowing what mask each door:

• Is the car is behind door 1: the presenter will open door 2.

• Is the car is behind door 2: the presenter will open door 1.
• Is the car is behind door 3: the presenter will open door 1 or the presenter will open door 2 (equiprobability 1/2)

The probability that the presenter opens door 1 knowing that the car is behind door 2 is thus p (O1|F2) = 1. Possibilté that the car is behind door 2 knowing that the presenter opens door 1 is thus:

$$

P (F2|O1) = \ frac {P (O1|F2) P (F2)}{\ sum_ {i=1} ^ {3} P (O1|Fi) P (Fi)} = \ frac {1 \ times \ frac {1} {3}} {0 \ times \ frac {1} {3} + 1 \ times \ frac {1} {3} + \ frac {1} {2} \ times \ frac {1} {3}} = \ frac {2} {3}

Alternatives

Many alternatives were proposed, modifying the parameters. It is often possible to find the solution of each problem by a simple reasoning, as in the principal problem, but the difficulty of seizing the role of each assumption often leads to an erroneous answer, and it is thus preferable initially to solve the problem analytically.

Case of a high number of doors

It is perhaps easier to apprehend the result describes above by regarding 100 doors and either three as previously. When the candidate chooses a door, it has 99% of chances to choose one with a goat of them behind. Conversely, the probability of falling directly on the door hiding the price of value is very weak (1%). Let us imagine now that the presenter does not open any more that only one door, but 98 of only one blow, revealing obviously 98 goats, while always proposing to the candidate to change his initial choice and to choose the last door remained closed. To 99% this door will contain the price of value, just like at the beginning the candidate had 99% of chances to choose a door with a goat. The candidate will thus have any interest has to change his initial choice.

The demonstration is exactly the same one, but the result is more intuitive: it appears so suspect that all the nonselected doors were open except a …

One could also leave more one of the closed not-selected doors, and even authorize to select more than one door at the beginning (of course, the presenter will not be able to open any the selected doors). Generally, on the basis of N closed doors, after the candidate indicated C carries (S), the presenter opens m carries (S), m being an entirety between 0 and n-c-1. The chances to be the good door are of:

• 1/n for one of C initially carries (S) selected (S)
• $\backslash \; frac\; \{1\}\; \{N\}\; *\; (1+\; \backslash \; frac\; \{m\}\; \{n-m-c\})$ for each other door.

Let us note that it is equivalent to open m doors one by one if the candidate refuses between each opening to change one of his choices or to open the m doors at the same time.

These formulas are obtained by calculation with a tree, but one can immediately find them by using the fact that when the doors have in the beginning all the same probability of hiding the car, the opening leaves the unchanged probabilities for the doors chosen in the beginning.

Let us bribe the organizers

Up to now, we always supposed that the doors had in the beginning an equal probability to hide the car. What does it occur if it is not any more the case?

One can for example imagine that the candidate picked up the assistant (E) of the presenter, who revealed to him that the door of right-hand side hiding place a goat. Badly warned, the candidate chooses initially the door of the medium. The presenter opens the door of left then. Which are the probabilities?

• If there is fully confidence in the assistant (E), 1 for the central door, 0 for the door of right-hand side
• If the assistant (E) were unaware of all and claimed to know to make itself interesting (E), the problem is equivalent to the initial problem, 1/3 for the central door, 2/3 for the door of right-hand side
• If the assistant (E) sought to induce the candidate in error, 0 for the central door, 1 for the door of right-hand side

One finds a formula general by applying the theorem of Bayes: one numbers 1 arbitrarily the door chosen in the beginning, 2 the open door by the presenter and 3 the last door; one notes $p\_i$ the probability in the beginning so that door I hiding place the car

the chances are after opening of $p\_3\text{'}=\; \backslash \; frac\; \{p\_3\}\; \{p\_3+p\_1\; /2\}$ for door 3 (with change) and of $p\_3\text{'}=\; \backslash \; frac\; \{p\_1\; /2\}\; \{p\_3+p\_1\; /2\}$ for door 1 (without change).

In fact, it is correct to say that the door chosen initially has its probability of hiding the unchanged car if it is null or if $p\_3=p\_2$, or more generally for N doors if the average probability of the open doors to hide the car were equal to the average probability of the nonselected doors to hide the car.

Let us recall that it is equivalent to say that the door chosen initially has its probability of hiding the unchanged car or that the nonopen nonselected doors inherited the probability of the open doors.

Let us change the rules of opening

One owes with Jean-Paul Delahaye two alternatives which of course clarify the importance of the rules of the opening of the door. In article of For Science, it proposed that the presenter randomly opens a selected door among the two doors not selected by the candidate (he can have possibly decided if the door would be on the left or on the right before that the candidate does not indicate a door), the starting again play with zero if he opened the door hiding a car. One second alternative proposes to randomly open a selected door among the two doors hiding a goat, the starting again play if it opened the door selected by the candidate.

Delahaye affirmed that the results of these alternatives were equivalent to those of the original problem. But the letters to the Editor made him be begun again: the probabilities are of 1/3 so that the change is gaining, 1/3 so that the maintenance of the initial choice is gaining, 1/3 so that it gave there… Maybe on the whole of the play (after as many handing-over as it will have been necessary) 1 chance out of 2 to gain whatever the strategy adopted at the time of the first not cancelled sleeve.

One seizes here the importance of the rules of the game which condition the opening of a door at the same time to the choice of the player and the position of the good door!

Still let us introduce some alternatives for including/understanding well:

The open door by the presenter is selected among the 3 doors without taking account neither of the choice nor of the place of the car: one finds this time 5 chances out of 9 of handing-over, 2 chances out of 9 to gain while changing, 2 chances out of 9 to gain without changing. There still, probability equalizes to gain ultimement with or without change.

The presenter opens one of the two doors nonselected by the candidate, by choosing 3 times out of 4 the door hiding the car if it forms part of it: this time there is a chance on 2 of handing-over, 1 out of 3 to gain without changing and 1 out of 6 to gain while changing! Ultimement, 2 chances out of 3 to gain without changing. Eh yes, by skewing the rules, one can reverse the effectiveness of the strategies.

More generally, if there is a probability p so that the door hiding the car is open (cancelling the sleeve) when it was not selected, one 1/3 chance to gain while changing, 2* (1-p) /3 chance to gain while changing, 2p/3 chance to give concerned. Ultimement, there is 1 (3-2p) chance to gain without changing, 1-1/(3-2p) while changing.

Last alternative: the presenter opens a door not hiding the car and nonselected by the candidate, but not randomly: on the contrary, it opens systematically on the right of the doors answering the preceding criteria. This time, if the door of the medium is chosen:

• If the good door is one of two on the left, the presenter opens the door of right-hand side
• If the good door is that of right-hand side, the presenter opens the door of left.

Therefore, the probability of gaining while changing is not more than 2/3 but of 1 or 1/2 according to the cases, the total Espérance remaining of 2/3.

Still let us generalize: if the door which was open had a probability p of being open if there were the choice, the probability of gaining is of 1 (1+p) while changing and of p (1+p) without changing. Always favors with the change.

Total formula

Let us give a total fomule for all the alternatives of the two preceding paragraphs. One will note:

concerning the probabilities before opening (these probabilities can translate for example the fact that the candidate is almost sure to have heard a goat behind a door):

• $p\_c$ the probability for the initially selected door is good the
• $p\_o$ the probability for the door which is then open either good the
• $p\_t$ the probability for the third door or good the

And there is of course $p\_c+p\_o+p\_t=1$ (and even $p\_c=p\_o=p\_t=1/3$ in the traditional problem)

when the door chosen initially is bad:

• $o\_c$ the probability so that the presenter chooses to open the door initially selected by the candidate (cancels the sleeve, $o\_c=0$ in the traditional problem)
• $o\_v$ the probability so that the presenter chooses to open the door hiding the car (cancels the sleeve, $o\_v=0$ in the traditional problem)
• $o\_t$ the probability so that the presenter opens the third door ($o\_t=1$ in the traditional problem)

And there is of course $o\_c+o\_v+o\_t=1$

when the door chosen initially is good the

• $o\_i$ the probability so that the presenter opens the door " interdite" , chosen and hiding the car (cancels the sleeve, $o\_i=0$ in the traditional problem)
• $o\_o$ the probability that the door had to be open which it precisely opened when he can choose (represents for example the preference of the presenter for the left, $o\_o=1/2$ in the traditional problem)
• $o\_a$ the probability that the other door had to be open (even notices that previously, $o\_a=1/2$ in the traditional problem)

And there is of course $o\_f+o\_o+o\_a=1$

The probability are:

• P (given) = $p\_c*\; (o\_i+o\_t-1)\; +1-o\_t$
• P (the play finishes with this sleeve) = $P\_f=p\_c*\; (1-o\_i-o\_t)\; +o\_t$
• P (profit while changing) = $\backslash \; frac\; \{o\_t*p\_t\}\; \{o\_t*p\_t+o\_o*p\_c\}\; *P\_f$
• P (profit by maintaining the choice initial) = $\backslash \; frac\; \{o\_o*p\_c\}\; \{o\_t*p\_t+o\_o*p\_c\}\; *P\_f$

for a sleeve

And ultimement (while having given concerned as much of time than necessary):

• P (profit while changing) = $\backslash \; frac\; \{o\_t*p\_t\}\; \{o\_t*p\_t+o\_o*p\_c\}$
• P (profit by maintaining the choice initial) = $\backslash \; frac\; \{o\_o*p\_c\}\; \{o\_t*p\_t+o\_o*p\_c\}$

Let us note that the probabilities of final profit for each one of the strategy is equal to the probability of profit on a sleeve knowing that this sleeve will end.

These formulas are obtained without too many difficulties by calculation with a tree.

Quantum problem

A daring alternative consists in transposing the problem in the world of the quantum physics. This time it is not a question any more of opening doors but of carrying out measurements on a system.

Except that this time as a vector of measurement is chosen, the possibilities are unlimited.

The situations which result from it are variable. According to whether one only authorizes the player, the presenter only or both to use with their advantage the phenomena, the probabilities is more or less in favor of the player.

But in a way a little similar to the traditional problem, the bad strategy is to preserve its initial vector and the good one is to choose an orthogonal vector with the initial vector.

http://xxx.lanl.gov/abs/quant-ph/0202120

You can play yourself at this address: http://www.imaph.tu-bs.de/qi/monty/MontyS.html

References in the literature

The writer and British scenario writer Mark Haddon exposes and shows the Problem of Monty Hall in its novel the Odd Incident of the dog during the night .

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