Modified version of Monty Hall problem?

Question

My friend asked me a so called modified version of Monty Hall problem in his opinion. But I find the description a bit spooky and maybe someone here can enlighten us with what is the problem with the description of the problem, or maybe it is me missing something.

Imagine exactly same setting as in monthy hall problem with one difference. Imagine I randomly pick one card. Now, as opposed to the original problem, the show host doesn't know where the car is, and of the remaining two cards, he randomly picks a card: if it happens to be a goat the host shows us this card (as in original game), if it happens to be a car however, my friend said you "quit" the game (or don't consider such case). Now the question is same as in original problem, is it better for me to change my initial choice?

One problem with this description is mainly with the "quit" part. Does not this hinder the process of calculating whether we should switch or not? How do you model the case the host quit the game (in calculating whether you should switch or not)?

He claimed that in such case, switching my initial choice doesn't give me any advantage anymore and made a simulation program as follows. He made 10000 experiments, where he was skipping the part where the host chose a car, so of course he was remained with 2/3 of the experiment test cases, and he claimed that now, since the number of matches the user made (1/3rd of 1000) is half of the number of remaining test cases(2/3rd of 1000- the ones where host didn't choose a car) - it is not advantageous anymore to change the card.

I am failing to find a flaw either in description in problem or in the implementation I mentioned above. Can someone help figure what is wrong with either the problem description or implementation?

I would appreciate some help because I got confused overall with the whole thing now (whereas I understand original monthy hall problem well).

PS. Here is the Java implementation actually: http://codepad.org/rt7fOqei, where he deduces that since match is one half of count, then it makes no sense to change my initial choice.

Answer 1

If the host has no inside information, then with no new information there should be nothing to be gained by switching.

However you can model the outcomes explictly.

scenario 1. You choose the right card (P = 1/3). The host reveals a goat probability (always).

scenario 2. you choose the wrong card P = 2/3 the host reveals the car (P=1/2) game over (net P = 1/3)

scenario 3. you choose the wrong card (P = 2/3) the host reveals the goat (P=1/2) game continues P = 1/3.

If you have survived scenario 2. the conditional probability is that there is a 50% you have the right card and a 50% chance you have the wrong card.

Here is a picture

Suppose the car is on card A (you don't know that).

You choose, A,B or C. (the black lines). The host makes the red line choice. You make the final keep or switch choice.

All of the keep/switch choices are equally likely. Half win, half lose.

Answer 2

In the modified game, identify the 3 cards as (1) the card you choose at random (2) the card the host chooses at random from the remaining two cards (3) the other card. It is assumed that the two random selections are done without any information about which card is the winning card, unlike in the actual Monte Hall problem.

Prior to showing (2), all have equal probability of being the winning card, by symmetry. If (2) is shown to not be the winning card, the remaining cards (1) and (3) continue to have equal probability of being the winning card, again by symmetry. This is also the case if (2) is shown to be the winning card, but since the probabilities equal zero, you might as well quit in that case.

Answer 3

The "quit" is simply throwing away one third of the games and not considering them. This is because you are "viewing" the problem at a specific point of time; after the host says his card and before you the player decide to switch.

Imagine instead of quitting the host takes out an anti-matter vial and destroys the entire universe if he gets the car. So before the game what is probability earth will continue to exist the next morning? The answer is $\frac 23$. So the host turns over a card at random and ... thank god ... it's a goat! NOW given that, what is the probability of anything happening in this timeline after an event with $\frac 23$ probability of happening did happen?

This is conditional probability and a very simple case.

There were three equally likely outcomes for where the car was: $1,2,3$.

There are three equally likely outcomes for what card you pick for $9$ equally likely outcomes.

In those there are $2$ equally likely outcomes for that what card the host pick for $18$ equally likely event.

In $1/3$ of the cases we picked the car and the probability the host picks the car as well is $0$. But in $2/3$ of the cases we didn't pick the car and host and a $1/2$ probability of picking the car (and the universe ends), this is a $\frac 12*\frac 23= \frac 13$ probability of us all ending in oblivion.

Here are the $18$ cases spelled out:

$[1:1:2], [1:1:3], \color{red}{[1:2:1]}, [1:2:3], \color{red}{[1:3:1]},[1:3:2], \color{red}{[2:1:2]}, [2:1:3], [2:2:1], [2:2:3], [2:3:1],\color{red}{[2:3:2]}, [3:1:2],\color{red}{[3:1:3]}, [3:2:1], \color{red}{[3:2:3]}, [3:3:1], [3:3:2]$.

In $6 $ of them the universe ends. Of the $12$ equally likely ones that remain, it's equally likely that we switch. The possibilities spelled out are:

$\color{blue}{[1:1:2:\text{stay with car #1}]},[1:1:2:\text{switch to goat #3}] \color{blue}{[1:1:3:\text{stay with car #1}]},[1:1:3:\text{switch to goat #2}], \color{red}{[1:2:1:\text{the universes ended. We don't have a choice}]}, [1:2:3:\text{stay with goat # 1}]\color{blue}{[1:2:3:\text{switch to car # 2}]}, \color{red}{[1:3:1:]\text{the universes ended. We don't have a choice}},[1:3:2:\text{stay with goat # 3}]\color{blue}{[1:3:2:\text{switch to car # 1}]}, \color{red}{[2:1:2:\text{the universes ended. We don't have a choice}]}, [2:1:3:\text{stay with goat # 2}]\color{blue}{[2:1:3:\text{switch to car # 1}]}, \color{blue}{[2:2:1:\text{stay with car #2}]},[2:2:1:\text{switch to goat #3}], \color{blue}{[2:2:3:\text{stay with car #2}]},[2:2:3:\text{switch to goat #1}], [2:3:1:\text{stay with goat # 2}]\color{blue}{[2:3:1:\text{switch to car # 3}]}, \color{red}{[2:3:2:\text{the universes ended. We don't have a choice}]}, [3:1:2:\text{stay with goat # 3}]\color{blue}{[3:1:2:\text{switch to car # 1}]}, \color{red}{[3:1:3:\text{the universes ended. We don't have a choice}]}, [3:2:1][3:2:1:\text{stay with goat # 3}]\color{blue}{[3:2:1:\text{switch to car # 2}]}, \color{red}{[3:2:3:\text{the universes ended. We don't have a choice}]}, []\color{blue}{[3:3:1:\text{stay with car #3}]},[3:3:1:\text{switch to goat #2}], \color{blue}{[3:3:2:\text{stay with car #3}]},[3:3:2:\text{switch to goat #1}]$.

The red cases could have happened but they didn't. They are off the table. Of the 24 remaining equally likely cases, 12 of the blue ones have favorable outcomes and 12 of the black ones have unfavorable outcomes.