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You're about to compete on the standard Monty Hall gameshow. But immediately before the show starts, you overhear the host's wife telling him she wants a divorce. You estimate that with probability 50% the distraction of hearing this will render him unable to remember which door conceals the car - in which case he will simply choose at random which of the two unpicked doors to open. If he reveals the car this way, he will be fired.

If he reveals a goat, and you switch doors, how likely are you to win the car?

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    $\begingroup$ I don't think either answer is answering the question that is asked. If he reveals a goat, and you switch, the probability of winning is the same as the original MHP: ⅔. $\endgroup$ Jun 25, 2023 at 20:44
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    $\begingroup$ @ChrisCudmore That's not correct. If you think fljx's answer contains an error, please leave a comment on it explaining the error. If you think you know the answer, please post it as an answer. $\endgroup$ Jun 25, 2023 at 21:00
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    $\begingroup$ @TasosPapastylianou : any 'accidents' resulting from the distraction being removed as possible outcomes, otherwise stated: host choosing goat even if host sometimes does not know where car is, hints at probability space where probability function on the events in the event space for the sample space has different values than in the case of the original MHP $\endgroup$ Jun 27, 2023 at 0:56
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    $\begingroup$ @FirstNameLastName: No, because there is no clear indication how "host gets fired" related to "you win/lose". Without that, this is a meaningless diversion. $\endgroup$
    – Flater
    Jun 27, 2023 at 3:27
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    $\begingroup$ @Flater "host gets fired" means that the game is off - note that the question asks "If he reveals a goat..." $\endgroup$ Jun 27, 2023 at 8:31

6 Answers 6

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You will win the car if a goat is revealed:

Three fifths (60%) of the time, down from the usual two-thirds.

One third of the time where you initially pick the car, it doesn't matter whether the host is distracted or not, they will open a door revealing a goat, you will switch to the other goat and lose. (1/3 in total)

For the other two-thirds of the time, where you initially pick a goat:
A quarter of the time the host will be distracted and then randomly pick the car and be fired (2/3 * 1/4 = 1/6 in total).
The other three-quarters of the time when the host is not distracted or is distracted and picks a goat by chance, then you will switch to the car and win. (2/3 * 3/4 = 1/2 in total)

So you win 3/6 of the 5/6 of the time that the host reveals a goat = 3/5 in total.

In tabular form:

Every row below has equal probability:

 You pick  |  Host distracted  |  Host reveals  |  Remaining prize
   Car     |        Yes        |      Goat1     |      Goat2
   Car     |        Yes        |      Goat2     |      Goat1
   Car     |        No         |      Goat1     |      Goat2
   Car     |        No         |      Goat2     |      Goat1
  Goat1    |        Yes        |       Car      | n/a host fired
  Goat1    |        Yes        |      Goat2     |       Car
  Goat1    |        No         |      Goat2     |       Car
  Goat1    |        No         |      Goat2     |       Car
  Goat2    |        Yes        |       Car      | n/a host fired
  Goat2    |        Yes        |      Goat1     |       Car
  Goat2    |        No         |      Goat1     |       Car
  Goat2    |        No         |      Goat1     |       Car


Showing that in six out of the ten cases where a goat is revealed, you still win the car.

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  • $\begingroup$ Wait, if Monty reveals the car by accident, shouldn't you win it by default? I guess that's not what was asked, but I feel like that should be counted as a technical win by the contestant, along with Monty losing his job... $\endgroup$ Jun 26, 2023 at 14:01
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    $\begingroup$ @DarrelHoffman The question asks for the conditional probability when a goat is revealed, so what happens when a car is revealed does not affect the answer. $\endgroup$
    – fljx
    Jun 26, 2023 at 14:26
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    $\begingroup$ "Three fifths (60%) of the time, down from the usual two-thirds." is slightly amiss to compare to the original game. The correct 60% here relies on a condition: "If he reveals a goat". The original game has no "he will be fired" and all cases play to the end. The contestant originally wins 2/3 of the time, yet now only wins 1/2 the time, 1/3 the time loses and 1/6 of the time standing around hoping for How About a Little Something For the Effort. $\endgroup$ Jun 26, 2023 at 16:29
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    $\begingroup$ Thanks for adding the tabular form as well. It's how I convinced myself that my answer could not possibly be right in the first place. :-) $\endgroup$ Jun 26, 2023 at 20:30
  • $\begingroup$ @chux I agree, but the first sentence could fix that by changing from 'You will the car:' to 'You will win the car if a goat is revealed:'. In the case of the host being fired, we aren't given enough information about what will happen to the game to determine if it is a loss or not. If it is a loss, then 50% is the right conclusion to 'You will win the car:'. $\endgroup$ Jun 27, 2023 at 16:06
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In this scenario, you only win a goat if you pick the car first time, which happens in 1/3 of all cases.
If the probability of Monty remembering is p, then the fraction of all cases where he shows a goat is
$\frac{1}{3} + \frac{2}{3} * (p + \frac{1 - p}{2}) = \frac{2 + p}{3}$.
Among the cases where Monty shows a goat, the fraction where you picked the car the first time is
$\frac{1}{3}/\frac{2 + p}{3} = \frac{1}{2 + p}$,
so your probability of winning the car in this scenario is
$\frac{1 + p}{2 + p}$.
If Monty never forgets, you are playing the normal game, and the probability to win the car is 2/3.
If Monty always forgets, the probability to win the car is 1/2.
If your forget estimate is accurate, the probability to win the car is 3/5.

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    $\begingroup$ Welcome to Puzzling. Nice explanation for the general case $\endgroup$
    – fljx
    Jun 26, 2023 at 16:14
  • $\begingroup$ I do not want to stretch the rope too far per se, but, for really general case, one can, apart from probability p other than 0.5, also find a formulas for N doors more than 3 and O doors opened revealing goat by host other than N-2. $\endgroup$ Jun 28, 2023 at 0:51
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I disagree with the 3/5 solution. I believe the problem as stated reduces to the original MHP, and thus the answer is still 2/3.

I was wrong. The answer is indeed 3/5. See edit below. Leaving the original thought process as it illustrates nicely where the mistake in reasoning was, once corrected :)

I find that the most intuitive way to attack the MHP is to generalise from 3 doors to N doors, where N is large enough to make the statistical properties of the problem more visible.

E.g. in the original problem, you make a choice among N doors (let's say, 1million doors); the host opens N-2 doors to reveal goats, leaving only two doors closed (one of which is the one you chose). The key insight here is that you're not supposed to deal with the probabilities of the doors themselves; the only probability that matters is "what were the chances that your initial guess was correct?". In the general example, this was 1/N (i.e. 1 in a million). Therefore the probability that you were wrong (and therefore the only other remaining door contains the car), was 1 - 1/N, i.e. 999,999/1,000,000.

The same logic can now be applied to this modified problem. You choose a door. The host opens N-2 doors, leaving only 2 closed. If any of the 999,998 doors opened contains a car, the host gets fired. If not (i.e. all the doors contain goats, whether because the host was undistracted or because he got incredibly lucky), you are left with 2 closed doors out of 1,000,000. The problem then asks us to ignore the first scenario where the host is fired, and only consider the second. So the question is still: what is the chance your original guess was correct? The answer is 1 in a million. So switching will get you the car with a probability of 1 - 1/million, i.e. 999,999/1,000,000.

Note that this is "NOT" the same question as "what is your probability of winning the car" more generally, since it disregards the scenario where the host messes up and reveals the car (indeed the problem is undefined at this point, since we don't know the game show's policy in the case of an accidental reveal).

EDIT: Ah I see the error in my logic now. I did not account for the number of games where the Host messes up as reducing the space of possible game outcomes in which "have I chosen the car or not in the first place" competes against.

If the contestant chooses the car with p=1/N, then the host has a chance of messing up equal to (N-1)/N * 1/2. This leaves the space of possible outcomes to 1/N for picking the car, vs 1 - (N-1)/2N - 1/N corresponding to the number of remaining outcomes minus invalids and correct car choice.

In the N=3 example, this means 1/6 probability of messing up, 1/3 of having chosen the car, and therefore 1 - 1/6 - 2/6 = 3/6 for the remaining outcomes. This makes the total space 2/6 for winning if you don't switch, vs 3/6 for winning if you do switch, giving rise to the 2:3 odds (i.e. prob of winning = 3/5).

I stand corrected :)

The above became much clearer (as is usually the case) once coded. Here's a simulation in octave:

1; % forces octave to consider this file as a script

function Outcome = experiment( NumDoors )

    Doors           = zeros( 1, NumDoors );
    DoorIndices     = 1 : NumDoors;
    CarDoorIndex    = randi( [1, NumDoors] );
    GoatDoorIndices = DoorIndices; GoatDoorIndices(CarDoorIndex) = [];   % remove cardoor index
    Doors(CarDoorIndex)  = 1;
    HostDistracted  = randi( 2 ) - 1;   % i.e. boolean with 50% probability
%    HostDistracted  = 0;   % classic MHP

  % Contestant selects a door
    ContestantDoorIndex  = randi( [1, NumDoors] );
    RemainingDoorIndices = DoorIndices; RemainingDoorIndices(ContestantDoorIndex) = [];

  % Host opens a door
    if   HostDistracted, HostDoorIndex = RemainingDoorIndices(randi( [1, NumDoors - 1] ));
    else if   ContestantDoorIndex == CarDoorIndex, HostDoorIndex = GoatDoorIndices(randi( [1, NumDoors - 1] ));
         else HostDoorIndex = CarDoorIndex;
         end
    end

  % Contestant switches to HostDoor.
    if (ContestantDoorIndex != CarDoorIndex) && (HostDoorIndex != CarDoorIndex); Outcome = "Invalid";
    elseif HostDoorIndex == CarDoorIndex; Outcome = "Win";
    else Outcome = "Lose";
    end

    return % Outcome
end

Outcomes = zeros(1, 3); % Wins, Losses, Invalids (in that order)

NumExperiments = 1000000;
NumDoors = 3;
for i = 1 : NumExperiments
   Outcome = experiment( NumDoors );
   if   strcmp( Outcome, "Win" ),  Outcomes += [1, 0, 0];
   elseif strcmp( Outcome, "Lose" ), Outcomes += [0, 1, 0];
   else Outcomes += [0, 0, 1];
   end
end

fprintf( "Outcomes: Wins(%d), Losses(%d), Invalids(%d)\n", Outcomes(1), Outcomes(2), Outcomes(3) );
fprintf( "Probability of Win, after ignoring invalids: %d\n", Outcomes(1) / (Outcomes(1) + Outcomes(2)) );

Output:

Outcomes: Wins(499481), Losses(333440), Invalids(167079)
Probability of Win, after ignoring invalids: 0.599674

Thank you for the interesting puzzle! :)

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  • $\begingroup$ @cdude gave answer for any p and N=3 ... I guess answer for any p and any N is also not so hard to find. $\endgroup$ Jun 27, 2023 at 11:43
  • $\begingroup$ @FirstNameLastName yep. But it's less about showing the general case; I just find reasoning for large N's to be easier to visualise (like in the original MHP). $\endgroup$ Jun 27, 2023 at 11:47
  • $\begingroup$ right, but the general case might give general insight, and perhaps unexpected results (much deviating from original MHP for big N, and, say p=1/2), but, indeed as you say, still easier to convince people in arguments that some MHP variants differ. $\endgroup$ Jun 27, 2023 at 14:42
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    $\begingroup$ Also you make the, indeed nice, classical visual point with "host opens N-2 doors to reveal goats, leaving only two doors closed", but this can also be generalized to host opening less doors, see for example question 114230 $\endgroup$ Jun 27, 2023 at 15:39
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You win the car if you switch and are shown the goat:

50% of the times.

If the host is not distracted (50% of the times):

The usual calculations apply. That is, you have a chance of 2/3 of winning the car if you switch.

If the host is distracted (50% of the times):

Suppose you initially chose the goat (2/3 of the cases). Then, if the host shows the goat by chance (50%) and you switch, you win the car. Suppose you initially chose the car (1/3 of the cases). Then, the host will always show the goat, and if you switch, you lose the car.

Overall:

With 50% probability the host remembers and your chance of winning the car by switching is 2/3. With 50% probability the host forgets. In 1/3 of the cases he shows the goat and you win the car. 50%(2/3) + 50%((50%*1/3)+(50%*1/3)+0) = 50%

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    $\begingroup$ See Carmeister's comment on FirstName LastName's answer for why this isn't correct reasoning. $\endgroup$
    – Sneftel
    Jun 26, 2023 at 10:29
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wrong answer (see comments)

You will win the car:

approximately $0.5833$% of the time, down from the usual two-thirds.

Call $W$ event 'you win car' and $D$ event 'host is distracted', then requested probability of $W$ is

$$P(W) = P(W \cap D) + P(W \cap \lnot D)$$

being

$$P(W) = P(W|D) * P(D) + P(W|\lnot D) * P(\lnot D)$$

but

$$P(D) = P(\lnot D) = 1/2$$

so

$$P(W) = 1/2 * (P(W|D) + P(W|\lnot D))$$

but

$P(W|\lnot D)$ is 'standard' 'Monty Hall Problem' probability $2/3$ (where host is always revealing a goat) whereas $P(W|D)$ is 'Monty Fall Problem' probability $1/2$ (where host happens to reveal a goat)

so

$$P(W)=1/2*(2/3+1/2) = 1/2*(4/6+3/6)=1/2*7/6=7/12$$

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    $\begingroup$ see also puzzling.stackexchange.com/questions/116850 for Monty Fall Problem $\endgroup$ Jun 25, 2023 at 20:02
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    $\begingroup$ This answer isn't quite correct unfortunately - the problem is that all the probabilities must be conditioned on the event of the host revealing a goat, including $P(D)$. The conditional probability of $D$ is then no longer $1/2$. $\endgroup$
    – Carmeister
    Jun 25, 2023 at 20:48
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    $\begingroup$ Thanks @Carmeister for correcting. $\endgroup$ Jun 26, 2023 at 11:08
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Choosing from 2 possible options would render you a 50% chance of picking the car. Under all circumstances of the equation you are always presented with two doors, one choice, and one prize. The other present variables are there to make it interesting for television. Explanation: Eliminate the third door immediately as you are already 100% guaranteed that Monte will not choose that door. This leaves you with 2 doors, one choice, and one prize. Your next option would be to alter your original choice and pick the opposing door from your initial selection. What changed? Please note I am answering this as if the rules for the contest are as follows: That what is behind each door remains behind each door with no rotation of entities leaving no room for variable change that holds any value to the question being asked. The only possible variable change would be switching of doors and that would only entail the markings of the A,B, or 1,2. However they labeled them for the contest. Thus leaving no real absolute value from one to the other. Their only value comes in the form of recognizable resource for the sake of the contest but doesn't alter the percentage of a possible victory whatsoever. That is how I perceive it to be correct.

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  • $\begingroup$ When you first pick a door, you are getting one of three random doors and have 1/3 chance of winning. If you could "switch" to pick both of the two other doors before the reveal, you would have 2/3 chance of winning. The reveal (in the original Monty Hall problem) always removes a non-picked losing door - so the last door must have 2/3 chance of being the winner. $\endgroup$ Jun 27, 2023 at 8:37

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