Is Switching the Best Option in the Door Game Show Dilemma?

[DrGreg]

OK, there's a glimmer of understanding starting here... Because the magician is allowed to see all the cards, he has a 51/52 chance of finding the ace so he has skewed the probability in his favor. I only have a 1/52 chance based on my initial random selection. So at this point I'm offered a choice between two cards which are unknown to me, but the magician's card has a high probability of being the ace since he was able to eliminate all but my one card. Is that it?

Yes. The two cards are not both random. One is always random, but 51 times out of 52 the other is not random at all.

 

[pallidin]

Good job, denny!

 

[Indefinite]

Yes. I understand all the explanations presented and I understand the mathematics behind the calculation. Each one who has attempted an explanation has presented what I would consider a reasonably good explanation (for whatever that's worth). My big "hang-up", and I'm not sure I agree, at least yet, that it's possible for a player to have a 1/2 chance of winning while another player at the same point in time with exactly the same set of conditions can have a 2/3 chance of winning.

Either something wrong with the theory or there is something wrong with the application that doesn't consider (or account for) the "hypothetical" new player (3rd observer).

I think to achieve 2/3 chance of wiinning requires and depends on an event that has a 100%chance of occurring --> "mechanically" requiring the original player to "switch" everytime which takes away from, in my opinion, some of the "randomness". An event which must occur 100% of the time is, in my opinion, not random.

So bottom line, yes... I've understood your explanations and appreciate them yet I'm just having a hard time "wrapping" my thoughts 100% around them.

While it is true that the second player has a 50% probability of winning, and the first player has a 2/3 probability of winning if he switches doors, this is not a contradiction, because the probability values describe different events. The .5 probability describes the probability that the second player will win if he selects from the two doors at random. The 2/3 probability describes the probability that the first player will win if he switches doors. Note that it does not actually matter who is playing in order for these probability values to hold. The second player also has a 2/3 probability of winning if he selects the door that player 1 can switch to, and the first player has a 1/2 probability of winning if, after being asked if he wishes to switch, he makes his selection at random.

No inconsistencies arise from the fact that the probability values are not equal, because they describe events occurring under separate conditions.

The 2/3 probability of winning applies only to a player who chooses to select the door that was not selected initially by the first player, and that was not opened by the host. The 1/2 probability of winning applies only to the player that selects from the two remaining doors at random after the host has opened the third door. Similarly, the 1/3 probability of winning applies only the player who chooses to stay with the door that was selected before the losing door was opened.

To summarize, let door 1 represent the door that is first opened, door 2 represent the door that is opened by the host, and door 3 represent the door that player 1 has an opportunity to switch to.

There is a 1/3 probability that either player will win if he selects door 1.
There is a 2/3 probability that either player will win if he selects door 3.
There is a 1/2 probability that either player will win if he selects a remaining door at random after the host has revealed one of the goats.

Note that each distinct probability value is associated with a distinct condition. So you see that there is internal consistency between them. Does this clarify things?

Personally, I would vote for the host's chances as 0 since the host is generally not allowed to play.

I note the qualifications of "long run" or "repeated many times" and respectfully submit that the theory is based on the "long run" assumption. In this particular case, that assumption is not met. This experiment is setup as a one time chance of winning. If the player had hundreds of chances, in the long run, his chances would approach a limit of 2/3. However, the player only get's 1 chance and that invalidates any use of the "repeated many times" assumption. The player only has one chance.

Applying any theory that is based on certian assumptions being met to the solution of a problem where those assumptions are not being met does not seem cosistent

The derivation of the solution to this problem is not predicated on the assumption of repeated trials. I only broached the topic of repeated trials to bring a deeper understanding of the implications of the asserted probability value, because we know that the theoretical probability value represents the frequency of occurrence that we will converge to in the limit as the number of trials approaches infinity. The theoretical probability value represents the frequency that a hypothetical experiment would converge to in the limit of arbitrarily many trials, regardless of whether or not such an experiment is actually conducted. My discussion of a large number of trials was only meant as another means of interpreting theoretical probability values.

In a similar sense, I might say that there is a .5 probability of landing heads on a coin flip, and expand on what this means by asserting that if we conduct many trials, we can reliably expect to obtain heads approximately 50% of the time. However, the fact that we do not actually conduct these trials does not change the fact that there is a .5 probability of obtaining heads in a single trial. The notion of many trials simply furnishes us with another perspective for understanding what a theoretical probability value means.

Because we know that the results of an experiment of many trials will tend to converge toward the theoretical probability value for a single trial, we can use our expectations of the results of such an experiment to determine whether or not our theoretical value seems intuitively reasonable.

 

[Indefinite]

I've read all these comments and theories and I understand what's been said, but to me it boils down to this:

When you make your initial choice, you would of course have a 33% chance of winning. But, no matter which door you choose, the host is going to open one of the losing doors, then present you with a new choice. So based on these facts, your chances of winning are 50% from the start.

You chances of winning will only be 50% if you select from the two remaining doors at random.

The solution to the "paradox" does not state that the player has a 2/3 probability of winning from the start of the game. The solution states that the player has a 2/3 probability of winning if he selects the second door. In order to expect to win approximately 2/3 of the time, the player must switch doors every time.

The crucial error in reasoning that continues to be made here is that the 1/2 probability of winning when selecting between two doors only applies when one selects between those 2 doors at random. The 2/3 probability does not apply to the player who selects between the two doors at random, it applies to the player who switches doors.

If it is decided at the start of the game that the player will always switch, we can expect in a large number of trials that he will win approximately 2/3 of the time.

If it is decided that the player will always select between the two doors at random after the host reveals the goat, then his frequency of winning will converge to 1/2 as the number of trials becomes arbitrarily large.

Your confusion arises from the fact that you know that if there are two doors, and one contains the prize, and you select between those doors at random, you will win approximately half of the time. However, the conditions of the 2/3 probability value do not state that the two doors are selected from at random; they state that the player switches.

The 2/3 probability of winning applies only under the conditions in which the player switches doors, not under the conditions in which the player selects between the remaining two doors at random.

Denny stated that it is a 50% probability if it is a choice between two unknowns, and this is true. The problem is that the 2/3 probability is not intended to describe a choice between two unknowns. It describes only the probability of winning if the player switches every time.

I think most of the confusion here derives from the fact that people do not fully understand the conditions that are specified in the problem. The problem does not state that the player has a 2/3 probability of winning if he has the opportunity to select between the two doors after the host opens one with a goat behind it. It states that the player has a 2/3 probability if he switches doors. The player must switch doors in order for this value to describe his probability of winning.

 

[mbro009]

An attempt to explain in fewest words possible:

Put yourself in the shoes of the showhost who knows where the car is.

For "one third" of the experiments the contestant will initially choose the car door, in which case the contestant will switch to a goat upon changing doors.

For "two thirds" of the experiments the contestant will initially choose a goat door in which case the showhost is forced to pick the other goat door that the contestant did not choose, "leaving the car door as the single remaining door".

So, one third of door changes the showhost will see her giving up the car for a goat, and two thirds of the time the showhost will see her giving up a goat for the car.

So, probabilities are: 1/3 a goat, 2/3 a car.

This can be made to be intuitive by realising the advantage for the contestant arises as a result of the showhost being forced into revealing the location of the car 2/3 of the time.

 

[Wakey7]

Not sure if this has already been understood but I think I can explain this more understandably.

If you are given a choice between 100 boxes, and you are told by a host that one box contains £100, you have a 1/100 chance of picking the money. Every other box is empty, so this translates that you have a 1% chance of picking the box with the money.

So...let's say that you pick box 37 and do not yet open it. The host then opens all other boxes except one. All the boxes he opens are empty, he knew this and so you are now left with two boxes. One box MUST contain the money. You are now given the option to switch.

Because the host knows which box has the money, it is always best to switch. Remember, when you picked your box, you had a 1% chance of getting the money, translating to a 99% chance that the money is in the box the host has left you with.

 

[xonuzofa]

Switching is NOT the best solution for this question.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1 [but the door is not opened], and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

By switching, you do not have a 2/3 chance. If the question was, before I show you a door, do you want to switch AFTER I show you a door? or Is it to your advantage to switch? In this case most definitely. As seen before, CGG, GCG, GGC, by choosing one you only have a 1/3 chance and by always switching you will receive a 2/3 chance. There are 3 variables here. BUT, the question is, AFTER I show you a door, which eliminates a choice, is it in your best interest to change. In the problem, if contestant chooses Door #1 and shows Door #3 having a goat, the only 2 possible choices are CGG and GCG. One must not be confused and include GGC because this is not possible, the 3rd door MUST be a goat. There are only 2 variables here, thus it is a 50/50 chance. Essentially this would be no different than the game show host opening a door initially. He is removing a variable from the equation. Or the same as opening the door you choose (if you obviously didn't pick the car) and allowing you to switch to either of them.

 

[xonuzofa]

C: the number of the door hiding the car,
S: the number of the door selected by the player
H: the number of the door opened by the host

P(C=2|H=3,S=1) = Probability of the car in the other door =
P(H=3|C=2,S=1)*P(C=2|S=1)/(Summation(i=1,2,3) of
P(H=3|C=i,S=1)*P(C=i|S=1).

Wiki shows this:
(1 * 1/3) / (1/2 * 1/3 + 1 * 1/3 + 0 * 1/3) = 2/3. INCORRECT!
(1/2 * 1/3) / (1/2 * 1/3 + 1/2 * 1/3 + 0 * 1/3) = 1/2. CORRECT

There is no reason P(H=3|C=2,S=1) should equal 1. C could be in EITHER 1 or 2,
so this is obviously a 50/50 chance or 1/2. Why would they say:
P(H=3|C=2,S=1) = 1 BUT P(H=3|C=1,S=1) = 1/2? Makes absolutely no sense.
When they are both equal to 1/2, the probability of the car in the other door is 1/2,
just like it should be.

 

[Robert1986]

Keep this in mind: The host CANNOT show you what is behind the door you picked. And the Host CANNOT show you what is behind the door with the car (this is really the most important part.)

There is no reason P(H=3|C=2,S=1) should equal 1.

P(H=3|C=2,S=1) is the probability that the host will show you door 3 GIVEN that you picked door 1, and the car is behind door 2. The host CANNOT show you the door with the car and the Host CANNOT show you the door you picked. This leaves exactly 1 choice for him to open: Door3:

There is no reason P(H=3|C=2,S=1) should equal 1. C could be in EITHER 1 or 2,
so this is obviously a 50/50 chance or 1/2. Why would they say:
P(H=3|C=2,S=1) = 1 BUT P(H=3|C=1,S=1) = 1/2? Makes absolutely no sense.

This: P(H=3|C=1,S=1) is the probability the the host will show you Door 3 GIVEN that you picked Door 1 AND the car is behind Door 1. In this case, the host can show you Door 2 OR Door 3 because you picked neither of the doors, and there is a goat behind both of them.

 

[xonuzofa]

Appreciate ya

Keep this in mind: The host CANNOT show you what is behind the door you picked. And the Host CANNOT show you what is behind the door with the car (this is really the most important part.)


P(H=3|C=2,S=1) is the probability that the host will show you door 3 GIVEN that you picked door 1, and the car is behind door 2. The host CANNOT show you the door with the car and the Host CANNOT show you the door you picked. This leaves exactly 1 choice for him to open: Door3:


This: P(H=3|C=1,S=1) is the probability the the host will show you Door 3 GIVEN that you picked Door 1 AND the car is behind Door 1. In this case, the host can show you Door 2 OR Door 3 because you picked neither of the doors, and there is a goat behind both of them.