3 doors problem - slightly different

[RiverMagic]

Hi everyone, I'm sure most of you are familiar with the Monty Hall Problem. This scenario is slightly different: We have 3 doors, there's a donkey behind 2 doors and behind 1 door is the prize. Contestant selects door 1 by choice. Using a random number generator between door 2 and door 3, door 2 reveals a donkey. What are the odds for door 3 having the prize? 1/2 or 2/3? Does Monty Hall Problem apply? Why or why not?

 

[Dr.Peterson]

Hi everyone, I'm sure most of you are familiar with the Monty Hall Problem. This scenario is slightly different: We have 3 doors, there's a donkey behind 2 doors and behind 1 door is the prize. Contestant selects door 1 by choice. Using a random number generator between door 2 and door 3, door 2 reveals a donkey. What are the odds for door 3 having the prize? 1/2 or 2/3? Does Monty Hall Problem apply? Why or why not?

Tell us your thinking.

Of course the "Monty Hall Problem" can't "apply"; that's a problem, not a theorem, and this problem has different conditions. The question is, how much of the same thinking can you apply? And that depends on what approach you are taking -- your thinking may be different from mine.

 

[lookagain]

We have 3 doors, > > there's a donkey behind 2 doors < < and behind 1 door is the prize.

Look at the highlighted phrase[MATH][/MATH]. This does not mean what you intend. It indicates one donkey,
not two donkeys. The phrase should be closer to "there's a donkey behind each of two doors."

Look at this rewrite: "We have three doors such that there's a donkey behind each of two
doors and behind a third door is the prize."

 

[Cubist]

...This does not mean what you intend...

Love it! If the 3 doors are side-by-side then let's hope for the sake of the one donkey that the "2 donkey doors" are adjacent ? Then, only a fool would pick the middle door. (But maybe they would pick the middle door if they wanted to get "a head", but perhaps an onlooker would simply see an "ass" after the door is opened.)

On a serious note, think of what happens if there are 100 doors such that there's a donkey behind each of 99 doors and behind a remaining door is the prize. The contestant chooses one door, and a random number generator opens 98 of the remaining doors. Very occasionally the prize will still be hidden. What then? And does this give an insight into the 3 door version?

 

[HallsofIvy]

I remember thinking about this in connection when the "Monty Hall Problem" was first popular. There are three doors, donkeys behind two of them and a car behind the third. The contestant chooses one door but before revealing what is behind that door, one of the other two doors, chosen at random (that's how this differs from the original problem- Monty Hall, knowing where the car was, intentionally opened a door without a car) showing there is a donkey. Should the contestant keep the same door or swap?

There is a 1/3 chance that the contestant chose the door with the car. In that case, if the contestants swaps, they lose. If the contestant does not swap they win.

There is a 2/3 chance the contestant did not choose the door with the car. In that case, there is a 1/2 chance the door, opened at random, reveals the car. In this case there is no choice- we igore this possibility. If the door, opened at random, reveals a donkey, swapping wins, not swapping loses.

So, 1/3 of the time they win by not swapping, 2/3 of the time they win by swapping. In this scenario, the contestant has a better chance if he/she swaps.

 

[Dr.Peterson]

As I see it (by making a decision tree), 1/3 of the time they reveal a donkey and we win by staying; 1/3 of the time they reveal a donkey and we win by changing; and 1/3 of the time they reveal a car and we get a donkey regardless. So, given we see a donkey, it's 50:50 whether to change; but we still win only 1/3 of the time.

And that's because, on one hand, we had a 1/3 chance of being right in the first place (so we win by staying 1/3 of the time); but on the other hand, opening the door at random does give us additional information, which Monty would not, so it's more worthwhile to change (we now have a 1/2 chance of being right, if we have any chance at all).

 

[Cubist]

@Dr.Peterson got in there before me, but I'll post this anyway!

So, 1/3 of the time they win by not swapping, 2/3 of the time they win by swapping. In this scenario, the contestant has a better chance if he/she swaps.

I think this post is fully correct right up to the quoted summary line. I did an "ascii art" version of your descriptions to illustrate...

STRATEGY 1, contestant does NOT swap
====================================

    1/3 choose door with car, WIN
  /
 /
<                               1/2  random door has car,invalid game
  \                            /
    2/3 choose door without car
                               \
                                1/2  random door does not have car, LOOSE

1/3 chance win, 1/3 chance loose, 1/3 chance invalid.


STRATEGY 2, contestant DOES swap
================================

    1/3 choose door with car, LOOSE
  /
 /
<                               1/2  random door has car,invalid game
  \                            /
    2/3 choose door without car
                               \
                                1/2  random door does not have car, WIN

1/3 chance win, 1/3 chance loose, 1/3 chance invalid

Probability of winning given that the car was not behind the random door = P(win) / (P(win) + P(loose)) = 1/2 in both strategies


FYI: On reflection I think my 100 door scenario "thought experiment", in post#4, doesn't give any insight into the best strategy.