The Puzzle Thread (nee Enigma Puzzle #24)

[Radii]

Originally Posted by Maple Leafs

On the off chance that any of the people who used to like puzzles on this forum are still around, this may be the most satisfying YouTube video I've seen in months.

(I strongly recommend watching it straight through from the beginning, but you can skip the firs two minutes of housekeeping if you must. But once the british dude starts explaining the rules, just sit back and watch him.)

Seconded. Once he got started on it and I grasped it, I considered trying to solve it myself, but honestly, just watching him puzzle it out was a thing of joy.l

[rjolley]

Yeah, that was definitely kool to watch. I wouldn't mind doing one like that.

[QuikSand]

I don't think there's anything "like that," really. His fascination connects to it being such a total unicorn.

[QuikSand]

And I'd chide cuervo to "get over it" but I know a windmill when I see one.

[cuervo72]

If it ever turns up in a LearnedLeague question, you'll thank me.

[QuikSand]

That's true

[Maple Leafs]

It's not the same without the wry brittish dude, but here's a version of the puzzle you can try to work through yourself. My son and I just finished it; in theory it's harder than the first one because it doesn't start with 1 and 2, but using the techniques from the first video it an be done.

Cracking the Cryptic

[britrock88]

A pretty terrific puzzle, that.

[Maple Leafs]

From The Guardian...
==

Today’s puzzle is a new adaptation of the Monty Hall problem. You will be set what seems to be an impossible challenge. Yet the solution is simple and ingenious.

Here we go. Again, the set-up involves a game show and three doors. Behind one door is a car. Behind a second door are the car keys. Behind the third door is a goat. The car, the keys and the goat were placed there randomly, meaning that each item has a 1/3 chance of being behind any particular door.

Twins Timmy and Tammy, the contestants, are backstage on the game show. They are told the rules:

1) Timmy will be taken on stage first. He will be asked to open two of the doors, and then shut them. Timmy will then be led off stage to a holding room on his own.

2) Tammy will then be taken on stage. She will be asked to open two of the doors.

If Timmy opens the door with the car, and Tammy opens the door with the keys, then they both get to keep the car. In all other outcomes, they leave with nothing.

The twins are given 10 minutes to think up a door-opening strategy before Timmy goes on stage. What strategy gives them the best chance of winning the car?

Just to be clear: The twins do not know what is behind any of the doors before they ask for a door to be opened. When one door is opened, all they can see is what is behind that door. The car, car keys and goat stay behind the same door for the duration of the programme. When Timmy is on stage opening his two doors, Tammy cannot see or hear what is going on. Thus when Tammy is choosing her two doors, she has no idea what was behind the two doors that Timmy opened.

If the twins had no strategy, that is, if both of them choose two doors at random, the probability Timmy gets the car is 2/3, and the probability Tammy gets the keys is also 2/3. The probability they get to keep the car is thus 2/3 x 2/3 = 4/9 = 44 per cent.

Yet, rather incredibly, there is a strategy that gives them well over 50 per cent chance of keeping the car. What is it?

[albionmoonlight]

Fun.

OK. The seemingly most obvious piece of relevant information is that the car and the keys cannot be behind the same door.

So, what to do with that . . .

[QuikSand]

ok, the spirit of the puzzle seems to be there can be no communication between the first reveal and the second selection... which to me makes the puzzle totally impossible

so, I'm looking for some wording that might allow a split between that impression and the solution... some way he can leave behind some information for her to glean for her selections... but I'm at a loss thus far

it seems like the solution must lie in wording, rather than math

[QuikSand]

by just agreeing in advance to make different selections, does that help us enough?

T1 = doors number 1 and 2
T2 = doors number 2 and 3

CKX = win
CXK = win
KCX
KXC
XCK = win
XKC

...I think that gets us TO 50%, but not "well over"

[albionmoonlight]

OK

What about a plan where they both open door 1 to start. And then their next moves change based on what is behind it

[QuikSand]

"when Tammy is choosing her two doors, she has no idea what was behind the two doors that Timmy opened" this sounds absolute... so it seems (to me) to rule out most of the trickery I have been contemplating... things like altering the order of his selection based on what he sees behind his first door, or marking something on the stage to leave her information... if she truly "has no idea" then it seems all those baloney answers are off limits

[albionmoonlight]

getting to 50% is easy, using Q's method or having them make sure to pick different doors if they start with door 1 and it is a goat.

How to get over 50 . . .

[QuikSand]

Originally Posted by albionmoonlight

OK

What about a plan where they both open door 1 to start. And then their next moves change based on what is behind it

ok, a conditional plan makes the most sense... for him, it really won't matter, right? he's got a 2/3 chance of hitting... but her knowing what he would do based on what he sees behind D1 is the one piece of information she can carry (despite the absolute wording above)

...spoiler-tagging this for now, in case this is actually "it" (I think it is)

Spoiler

So, they agree he opens D1 first, and:
CAR = D2
KEY = D2
GOAT = D3

then the six setups:

CKX = they are in play, she needs to pick door 2
CXK = they are in play, she needs to pick door 3
KCX = they are in play, she needs to pick door 1
KXC = they have lost
XCK = they have lost
XKC = they are in play, she needs to pick door 2

...this seems to only get us to 50% as well.

Hmmmm... a conditional based on his first reveal has to be part of this.

Right now, I'k simplifying her selections into one sweep, but maybe we break it down based on what she sees with her first guess.

Say she starts with Door 2.
K = they win
C = she picks 1, since the other option means they have already lost
X = she picks 3, since the other option means they have already lost

overall chance of winning = 2/3

as it had to be

[AnalBumCover]

Timmy needs to open a door with the car first, in order for the prize to be valid. Then Tammy needs to open a door with the keys. Once that order of operations is completed, they win the car.

I think the solution lies on whether the doors that each twin selects are opened simultaneously, or one after the other. If Timmy opens Door #1, and it's a car, then he'll request to open Door #2. All else, open Door #3...

And Tammy will then have a similar, but different selection criteria, depending on what appeared behind her first choice.

[edit] Wow, I missed a lot of posts while typing this out.

[Maple Leafs]

Originally Posted by albionmoonlight

What about a plan where they both open door 1 to start. And then their next moves change based on what is behind it

I think this has to be it. This one reminds me of the old one from years ago where the three players have lights behind them that only the others can see.

The key is that you can't change your odds or success or failure, so you need to come up with a strategy that lines up all the failures so that when you fail, you *really* fail (and the other odds go up).

If they pick at random, here's what the outcomes will look like (where "Y" means the person picked the right option for them, and "N" means they didn't):
YN YN W
NY NY W
NN NN L
YN NN L
NY YN W
NN NY L
YN NY W
NY NN L
NN YN L

We need a strategy where all "NN" outcomes always line up, which would mean we win 2/3 of the time.

[NobodyHere]

Maybe when Timmy opens the door and finds a goat, he lets the goat out? It sounds like he has control over when he shuts the door.

[Maple Leafs]

OK, I think I've got it, although I feel like I kind of mentally brute-forced it rather than really "solving" it in a thoughtful way.

This might be helpful if you're working on it: the six possible combinations for how the three options are arranged.

GKC
GCK
KGC
CGK
KCG
CKG

You want to get to a strategy where two of these are double-losses, and the other four are double-wins.

[NobodyHere]

Actually I'm going to have to agree with Quiksand on this one.

[QuikSand]

only if you absolutely have to

[AlexB]

I think I got it, or at least a solution that is not explicitly against the rules

Spoiler

I focussed on the wording ’He will be asked to open two of the doors, and then shut them’. Not open one door, close it, then open another, but open two doors, then close them both.

So Timmy opens door 1 and door 2. Hopefully the car is in one of them. There’s no way of knowing, so we are at a 2:3 chance

Assuming he found the car, if the second door has the goat, he closes them both up and moves on. If the second door has the keys, he moves the keys behind door 1, then closes them both.

Then all Tammy has to do is open doors 1 and 3 and if Timmy found the car, she is guaranteed to find the keys

So the overall odds remain 2:3

Edit - actually it is explicitly against the rules

Double edit - I agree with Quik. Probably should have just written that and not tried to be clever

[Maple Leafs]

Here's my solution, which I think is the same as Quik's in different words:

Spoiler

First player (who needs to find the car):
Open door 1.
If they see a goat, open door 2
If they see a key, open door 3
If they see a car, it doesn't matter what they do

Second player (who needs to find the key):
Open door 3.
If they see a goat, open door 2
If they see a car, open door 1
If they see a key, it doesn't matter what they do

This gives us the following possible outcomes, based on the original layout:

Goat/Key/Car: Both players lose
Goat/Car/Key: Both players win
Key/Goat/Car: Both players win
Key/Car/Goat: Both players lose
Car/Goat/Key: Both players win
Car/Key/Goat: Both players win

Same number of wins and losses as random chance, but now they're all lined up to maximize the double-wins they need to get the car.

The key to the strategy, weirdly enough, is for Player 2 to try to lose when they know that player 1 has already lost. Which is possible, if you know which doors player 1 has opened. It's the times that one player wins and the other loses that kill your chances.

[albionmoonlight]

Apparently, this puzzle/thought experiment has been around for a couple of decades, but I had not heard of it before: https://www.youtube.com/watch?v=XeSu9fBJ2sI

[Maple Leafs]

Love that guy's channel.

[QuikSand]

So, with sleeping beauty... it's new to me (and I have paused about 90s in to type this), but it has obvious parallels with another familiar puzzle (two coins, one double-headed, you look down and see one of them heads-up, what is the chance that it's the double-headed coin?).

In my mind, the difference is that the two-coins puzzle allows for an outcome that does not fit the evidence - namely, the ordinary coin was flipped and came up tails, so because that could have happened and did not, you now have new information that can alter the probability to two-thirds, which it indeed does.

Here, no matter what happened with the coin, the experienced outcome is the same - she wakes up with no information about the time it has taken. The true parallel to the two-coin problem would be something different... "she wakes up and is told that it's Monday?" There you'd have the comparable new information, and a familiar 1/3 v 2/3 split again.

Here, I start in the 1/2 camp. But given that spread of scholarly-looking papers, I fear I will look at least partially foolish soon.