Friday Puzzler2 -- Monty Hall2

Re: Friday Puzzler2 -- Monty Hall2

Larry Barrett

So far I don't see how to improve the odds to 2/3. But here is a strategy to improve the odds to 1/2.

The possible combinations of car, keys, goat behind doors 1, 2, 3 are as follows:

C K G

C G K

K C G

G C K

K G C

G K C

Suppose the wife's strategy is to first open door 1, then if no C, open door 2.

The probability of finding the C is 2/3 and the game continues. If no C (prob = 1/3) the game ends.

If the husband comes out he know the C is behind doors 1 or 2. His strategy should be to open door 3 first. If K, game ends. If no K, then open door 1. With this strategy the husband will find the keys with prob = 3/4. (If he opens door 3 first, then door 2, the prob of finding the keys is the same).

The overall probability of finding the car and keys is 2/3 x 3/4 = 1/2.

Good start

Alex Y

Knowing the choices his wife will make provides information to guide the husband's choices. Can you think of any way she could provide more information by her choices?

Instead of labeling the doors 1, 2, and 3, try labeling them C, K, and G (which may or may not correspond to what is behind each door).

Re: Friday Puzzler2 -- Monty Hall2

david weaver

Am i correct that he doesn't know if she's opened one door or two?

Re: Friday Puzzler2 -- Monty Hall2

david weaver

I can only think of a situation where they cheat, but it would require a cue (this is too easy to do, though, and makes the problem not that interesting).

If she smiles when he comes out, then it means she opened door 1 and the car was there.

If she doesn't smile, it means she opened two doors (say 1 and 2).

She has a 2/3rd chance of finding a car.

If she smiles, that means the key is in door 2 or 3 (he can pick them in any order).

If she doesn't, he doesn't know if the key is in 1 or 3, but he has to pick those two as the only chance that it's not in 1 or 3 coincides with the car being behind door #3 - that's a non-win, anyway, which leads to a conundrum for the flow of the show (the audience knowing that there will be no win, despite the last two turns being taken by the husband).

If he knows whether she's opened one door or two due to passage of time or being told, then he doesn't need the cue.

If he can't know at all, then conditional probability is out the window and I'm not sure how he can guarantee that he'll pick one of the two correct spots.

Re: Friday Puzzler2 -- Monty Hall2

Alex Y

Correct.

Re: Friday Puzzler2 -- Monty Hall2

Alex Y

Nope; no cues.

Re: Friday Puzzler2 -- Monty Hall2

David Weaver

I have a solution I think. As soon as I get back to a computer.

Solution

David Weaver

Specific solution, but could be generalized.

Wife chooses door 1. If it's a car, she's done.

CGK

CKG

If she chooses position one and it's a key, on choice 2, she chooses the third

KGC

KCG ends the game

If she chooses door 1 and it's a goat, she chooses door 2 as her second choice.

GCK

GKC ends the game

The husband comes into the game and picks door 3 no matter what, knowing that if it's a C, the key is in position 1. If he gets a G, he knows that K is in position 2. if he gets a K, obviously, the game is over with his first pick.

Winning selections are:

CGK

CKG

KGC

GCK

It took me a while to figure this out. I'm not very good at solving problems!!

Correct!

Alex Y

Great job!

Note the difference between this solution and those where the choices for each contestant are fixed. We know which door each person will open first, but we do not know, a priori, which they will choose second.

Allow me to recast your solution, with the doors labeled C, K, and G (to ultimately tie into the second 100 prisoner problem)

The wife, who is looking for the car, initially chooses C. If she finds the car there, great. If not, she next chooses the door corresponding to what was behind door C. That doesn't seem logical, but since either remaining choice is equally likely, there is no harm in following this strategy and this at least lets her pass on some info to her husband if she is successful.

The husband, looking for keys, chooses door K. If the keys are there, great. If not, he does the same thing his wife did, choosing the door corresponding to what was behind door k.

Why does that work?

If he sees the car behind door k, he knows his wife opened k to get the car, and the reason she opened k was that she saw keys when she opened c. If he sees a goat, he knows that his wife did not see the goat behind door c, nor did she check door k and find the car there, so the car must be in c, leaving door g as the only place for the keys.

Re: Correct!

Larry Barrett

Maybe the 100 prisoner problem works the same way. Each prisoner first opens the drawer corresponding to his number and then follows the bunny trail opening up to 49 more drawers. If he finds his own number along the trail he wins. If all 100 find their number they all win.

I tried this one time, with just 10 prisoners, 10 drawers, 10 numbers randomly distributed, each prisoner opening up to 5 drawers. It worked, but I do not know why even after following the C K G problem.

Closing the loop on CKG

Alex Y

We saw how the wife was able to inform the husband's choice, and that was a useful way to develope the strategy. But that's not really what is happening. In fact, we could add a third contestant whose goal is to pick the door with the goat, and if all three of them followed the strategy described here, they can go in any order and even let someone else pick between their two picks, and will all succeed or all fail depending on the way the prizes are distributed behind the doors:

CKG

ckg

cgk

kcg

kgc

gck

gkc

In this small case of three containers and two picks, everyone will succeed or fail, but as Larry observed in the other thread, this gets more complicated when you move to 4 containers.

You might get an insight that will help explain the prisoner puzzle if you pick one of the green cases above and one of the red fail cases and trace the paths each contestant will take to find his prize.