Friday Puzzler -- Not Monty Hall

Just to Clarify

Alex Y

Alan will pick a curtain. If it contains the car, great -- his turn is over. (He can look at another curtain if he is curious, but cannot communicate what is there to Becky.)

If the first curtain he opens does not contain the car, he may try one more. If he fails, game's over.

The curtains are then all closed and Becky comes out, after Alan is removed. Becky picks a curtain. If it contains the keys, they are winners. If it doesn't, she has one more chance to pick the curtain with the keys.

If they pick at random, Alan has a 2/3 chance of finding the car and Becky has a 2/3 chance of finding the keys. so they have a 4/9 chance of winning.

If they can improve their odds, it is only through their selections, not heel scuffs on the floor, chalk marks on the curtains, etc.

Re: Just to Clarify

Larry Barrett

Suppose they agree that Alan will open curtains 1 or 2. If he finds the car the game continues, he is removed and Becky comes out. If he doesn't find the car the game ends before Becky comes out. So at that point she knows the car was behind 1 or 2 with probability = 1/2 that it is behind either one, and probability = 0 that it is behind curtain 3.

I think this means that from her point of view the keys are behind 3 with P = 1/2 and are behind 1 or 2 with P = 1/2. So her strategy should be to open curtain 3 and then either 1 or 2 and find the keys with probability = 3/4.

So overall strategy has prob= 2/3 x 3/4 = 1/2 of winning.

Hard to believe that the Ask Marilyn version of the Monty Hall problem appeared in 1990, 30 years ago, and the original MH problem dates back to the mid-70s I think.

Nice improvement

Alex Y

on the 4/9 solution.

But it is possible to do even better.

Note that in your solution, both Alan and Becky can state their choices before the game even begins. My "clarification" was to say that is not a requirement.

Wow! Hard to believe the Monty Hall problem is that old!

Hint

Alex Y

Both Alan and Becky consider what was behind the first curtain they picked when deciding what to pick next.

Re: Hint

Larry Barrett

They agree to first open A. Then, for Alan, if not the car he will open B. And if the game continues, Becky will first open A and then,if not the keys, will open C.

So if a goat is behind A (p=1/3), then Alan will open B and if a car the game will continue. Becky will open A and find the goat, then will open C and find the keys.

If there is a car behind A (p=1/3), Alan will find the car. The game continues. Becky opens A, then opens C. The keys will be behind C with p=1/2.

If the keys are behind A (p=1/3), Alan will then open B and find the car with p=1/2. The game continues. Becky opens A and finds the keys.

So the overall probability is 1/3 + 1/3 x 1/2 + 1/3 x 1/2 = 1/3 + 1/6 + 1/6 =2/3.

Getting close

Alex Y

but I think you have overstated the probability of the first case. Adding what I think is a missing piece of what you said:

So if a goat is behind A (p=1/3), then Alan will open B and if a car (p=1/2) the game will continue. Becky will open A and find the goat, then will open C and find the keys.

Looking at the six cases, and upper or lower case letters for Alan and Becky's success with your scheme (if I understand it correctly):

1) CKG A B

2) CGK A b

3) KCG A B

4) KGC a B

5) GCK A B

6) GKC a b

Alan and Becky each get their item in four of the six cases, but they BOTH get their item only in cases 1, 3, and 5.

The hint is that they each make their second choice based on what they see in the first. In what you propose, Alan will always chose curtain B for his second pick and Becky will always pick C.

Hint 2 -- Alan's picks

Alex Y

Alan starts with curtain A.

If it is the car, his second pick is irrelevant.

If it is the keys, he should pick B next

If it is the goat, he should pick C next.

Because of the inherent symmetry of the problem, you can assign any curtains to his picks as long as his second pick is different if his first is the goat vs the keys.

Re: Hint 2 -- Alan's picks

Larry Barrett

If Alan found the car the game continues. Becky then

Opens A, finds the car (p=1/3), then finds the keys behind either B or C (p=1/2). Overall P=1/3 x 1/2 =1/6

Opens A, finds the keys (p=1/3).

Opens A, finds the goat (p=1/3) but then knows the keys are behind B because Alan opened C and found the car.

So overall probability for Becky is 1/6 + 1/3 + 1/3 = 5/6.

Still at 50%, I think

Alex Y

Becky knows that Alan got the car curtain (or equivalently, knows that the only way to win is if he got that, so might as well play it that way). Knowing the moves Alan was going to make, the only combinations in which he will have seen the car are bolded below:

1) CKG

2) CGK

3) KCG

4) KGC

5) GCK

6) GKC

Looking at the logic of her selections, then:

She opens A and finds a car. She then selects B or C, which will be a win for her and the pair of them in one of scenarios 1 or 2.

or

She opens A and finds the keys. That means a win for her if the curtains are arranged as in 3 or 4 above, but that is a win for the couple only if the curtains are as in 3

or

She opens A and finds the Goat, and knowing what Alan will have done when he saw the goat, she will pick curtain B, Which is scenario 6.

So she gets the keys in scenario 1 OR 2 (whichever she chose) and 3,4, and 6, a 2/3 probability. But only 3 of those scenarios match up with ones in which Alan found the car, so we are back to the 50% answer.

Final hint

Alex Y

Picking at random, Alan will find the car 2/3 of the time, i.e., in four of the six possible arrangements of car, keys, and goat.

His strategy for picking the curtains does not improve or hurt his chance of finding the car, but it does allow us (and Becky) to know the four arrangements that Alan might have picked.

So far, the strategies used allow Becky to find the keys in three of the four arrangements Alan might have picked, so they have a 50% chance of winning the car.

Becky can use a strategy that allows her to find the keys in any of the four arrangements Alan might have picked, so they have a 2/3 chance of winning the car.

Solution

Alex Y

Alan starts with curtain 1 (I'm using numbers for the curtains to avoid possible confusion between C for car and C for the third curtain.)

If curtain 1 contains the Goat, he chooses curtain 3; else he chooses curtain 2. He sees the capitalized items in the six equally likely arrangements below:

CKg

CGk

KCg

KGc--fail

GcK--fail

GkC

Note that in two of his successes, the keys are behind the second door, so you might suspect (correctly) that 2 is a good starting place for Becky. If she does not find the keys behind curtain 2, the other two cases where Alan succeeded are cgk and kcg, so selection of curtain 2 points her to the keys--she chooses curtain 3 if she sees the Goat behind 2, else she chooses curtain 1.

So they win 2/3 of the time.