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# Alex's mug challenge - no math needed

## Question

It seems, said Alex the other night, that no one likes questions about odds.

They're either too simple to care about or too subtle to believe.

Ian, Davey and Jamie all agreed, and the four of them downed their ale.

Which brings me to tonight's challenge, said Alex.

It has nothing to do with odds -- just this little lazy susan thing.

And he laid it on the center of the table, giving it a spin to show how it worked.

Now here's the deal, boys.

Each of us puts his mug on one corner of this lazy susan thing, upside down or not, doesn't matter.

Then, one of you has to turn them all upside down or all right side up.

There's only three simple rules:

[1] You pick up any two mugs, feel them, and then put them back, whichever way you like - up or down - they don't have to be the same.

[2] Then someone else turns the lazy susan to a new position each time, after the mugs are touched.

[3] And ... you have to do it blindfolded.

Whenever you get them all up or all down, I'll ring this bell here, and you win.

But ... you must do it in 5 moves or fewer.

Davey rolled his eyes and declined; so did Ian.

Jamie thought for a long time and then agreed to try.

Is there any way Jamie can ensure he will win the bet, without just being lucky?

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well, the mugs could be wider at the top than at the bottom.

Or he could touch the top of the mug to see if he is touching the top or the bottom or something....

Whatever, if i'm wrong then we can just wait for writersblock to get it.

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It seems, said Alex the other night, that no one likes questions about odds.

They're either too simple to care about or too subtle to believe.

Ian, Davey and Jamie all agreed, and the four of them downed their ale.

Which brings me to tonight's challenge, said Alex.

It has nothing to do with odds -- just this little lazy susan thing.

And he laid it on the center of the table, giving it a spin to show how it worked.

Now here's the deal, boys.

Each of us puts his mug on one corner of this lazy susan thing, upside down or not, doesn't matter.

Then, one of you has to turn them all upside down or all right side up.

There's only three simple rules:

[1] You pick up any two mugs, feel them, and then put them back, whichever way you like - up or down - they don't have to be the same.

[2] Then someone else turns the lazy susan to a new position each time, after the mugs are touched.

[3] And ... you have to do it blindfolded.

Whenever you get them all up or all down, I'll ring this bell here, and you win.

But ... you must do it in 5 moves or fewer.

Davey rolled his eyes and declined; so did Ian.

Jamie thought for a long time and then agreed to try.

Is there any way Jamie can ensure he will win the bet, without just being lucky?

Since they start out all up or down, couldn't you change nothing and get it in 0 moves?

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Clarifying.

Jamie is blindfolded.

The mugs are placed on the four corners, randomly up or down.

Move #1 - Jamie knows where the four corners of the lazy susan are and can pick up

[1] two adjacent mugs - or -

[2] two diagonal mugs

Then he can feel whether they are up or down. He can flip either, neither, or both and replace them.

Davey randomly turns the lazy susan to a new position. Jamie doesn't know what Davey did.

Move #2 - Jamie [still blindfolded] does the same thing again.

Davey turns the lazy susan again.

Move #3 ... same.

Move #4 ... same.

Move #5 ... same.

At this point [or before] all the mugs must be up or all must be down.

At any time, if all the mugs are up or all the mugs are down, Alex rings the bell, and Jamie wins.

Can Jamie ensure a win - without relying on luck?

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bonanova, your clarification doesn't match your first statement. Which one is right?

Each of us puts his mug on one corner of this lazy susan thing, upside down or not, doesn't matter.

Then, one of you has to turn them all upside down or all right side up.

ALL UPSIDE DOWN or ALL RIGHT SIDE UP

oh wait... do you mean that as BEFORE the game begins that happens... or do you mean the goal of the player is to do that?

nevermind i get it now ;D

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all down

all up

2d, 2u

those 3 scenarios require an even number of moves, while these 2:

1 down, 3 up

3 down, 1 up

requires an odd number of moves

Thus I would feel and act accordingly. The rotations of the Lazy Susan doesn't matter since you can feel the status of all 4 mugs, thus if you REALLY wanted to know how it moved, you could easily know, but it doesnt matter. What matters is what to do for each of those 5 scenarios.

In the last 2, all it takes is 1 move to finish em. So you want to reach a 3-1 status by the 4th move

oops i have to go now. I will finish this after lunch

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i was making it too complicated:

Initial Setups in red

4d

4u

flip any two over to reach a 2/2 status

2d, 2u

2u, 2d

flip any one over to reach a 3/1 status

1d,3u

3d,1u

flip the outcast over to reach a 4/0 status

4d,0u

0d,4u

flip NOTHING over to reach a 4/0 status

guaranteed in 3 moves, as little as 0 moves

if you HAVE to move at least once do the same thing except for 4/0 flip one cup then back again the next move

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Pshhh.

I GUESS that makes sense.

My idea was SO much cooler though.

you're just mad cuz you didn't think of it

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Further clarification.

This differs a little from the OP for clarity - it does not change the solution.

Jamie is blindfolded.

The four mugs are placed on the corners, randomly up or down.

Each move, Jamie touches only two mugs [not all of them].

Here is what constitutes one move:

Davey turns the lazy susan an arbitrary multiple of 90 degrees.

Touching only the lazy susan, Jamie chooses any two adjacent corners or any two diagonally opposite corners.

He picks up those two mugs.

He puts them back on the same corners, individually flipped or unflipped, i.e. individually up or down, as he chooses.

Jamie gets 5 moves.

At the end of any move: if the mugs on the lazy susan are all up or all down, Alex rings the bell and Jamie wins.

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Yeah, see, I was totally right.

PWNED

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I believe this works:

We have two positions - up and down. Lets say in his minds he marks them positions P1 and P2, regardless which is which (solely for the convinience of the solution).

Move 1. He takes 2 diagonally opposite mugs and flips them to P1.

Move 2. He takes 2 adjacent mugs and flips them to P1 again. That means that he certainly took 1 of the mugs he turned in move 1, and 1 which he didn't turn in move 1. Therefore, he now has 3 mugs in P1. If the 4th mug was originally in P1, he would win already. If not, then that means it is in P2. Which leads him to:

Move 3. He takes 2 diagonally opposite mugs. If one of them is in P2, he flips it and he wins. If both are in P1, he flips just one of them to P2. Now he certainly has 2 pairs of adjacent mugs in the same positions (P1 next to P1 and P2 next to P2).

Move 4. He takes 2 adjacent mugs. If they are in the same position, he flips them both around and he wins. If they are in different positions, then he flips them both around anyway. Now he knows that the pairs of diagonally opposite mugs are in the same positions, but every two adjacent mugs have different positions. That makes it easy in the next move:

Move 5. He takes either pair of diagonally opposite mugs and flips them both around.

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Blindfolds on?

1. Turn two adjacent glasses up.

2. Turn two diagonal glasses up.

3. Pull out two diagonal glasses.

[a] if one is down, turn it up and you're done.

if not, turn one down and replace.

4. Take two adjacent glasses. Invert them both.

5. Take two diagonal glasses. Invert them both.

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I believe this works:

We have two positions - up and down. Lets say in his minds he marks them positions P1 and P2, regardless which is which (solely for the convinience of the solution).

Move 1. He takes 2 diagonally opposite mugs and flips them to P1.

Move 2. He takes 2 adjacent mugs and flips them to P1 again. That means that he certainly took 1 of the mugs he turned in move 1, and 1 which he didn't turn in move 1. Therefore, he now has 3 mugs in P1. If the 4th mug was originally in P1, he would win already. If not, then that means it is in P2. Which leads him to:

Move 3. He takes 2 diagonally opposite mugs. If one of them is in P2, he flips it and he wins. If both are in P1, he flips just one of them to P2. Now he certainly has 2 pairs of adjacent mugs in the same positions (P1 next to P1 and P2 next to P2).

Move 4. He takes 2 adjacent mugs. If they are in the same position, he flips them both around and he wins. If they are in different positions, then he flips them both around anyway. Now he knows that the pairs of diagonally opposite mugs are in the same positions, but every two adjacent mugs have different positions. That makes it easy in the next move:

Move 5. He takes either pair of diagonally opposite mugs and flips them both around.

Bravo!

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if he uses the hadels of the mugs as an aditional indicator of which they are he should be able to do it

he would also need to pick up alternating both on one side and diagnal

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