## Don’t Try This With Full Pint Glasses

Posted by Karl Sharman on July 18, 2011 – 7:26 am

Four glasses are placed upside down in the four corners of a square

rotating table.

You wish to turn them all up. You may do so by grasping any two glasses and, optionally, turning either over.

There are two catches:

1. you are blindfolded

2. the table is spun after each time you move the glasses

Assuming that a bell rings when you have all the glasses up, how do you do it?

July 18th, 2011 at 9:18 am

Can you grasp them by placing your palm over the top? That strikes me as an easy solution, since you can always figure out which to turn over, as it would be easy to tell if the glass you were holding was right side up or not.

July 18th, 2011 at 10:37 am

This doesn’t say whether you have to turn all glasses up in a minimum number of spins of the table, or how much time you have to complete the task. Nor whether you can feel the top of the glass to determine if it had already been turned over.

So all you would have to do is feel the edge of the table as you approach it, then feel the two corners at each end of this edge, then feel each glass and turn one over if not already turned over. If both already turned over then move to the other side of the table and feel and turn one of the other two. Repeat until done. Just four visits required.

However, I suspect that you should not be allowed to tell if a glass has been upended by feeling the top of it. In that case this becomes a bit like a random walk: you keep on turning glasses over until you eventually get there – after how many visits?

July 18th, 2011 at 10:48 am

I am going to assume that you can’t feel the top (or bottom) to see if it is open end up or down. I’m also guessing these are straight sided (not tapered).

Here’s how I would do it:

Grab the 2 glasses within reach and hold onto them a while. Flip the one on the right.

Grab the 2 glasses in reach. Iif they are both warm to the touch (from me holding them the first time), flip the left glass. If the right one is warm and not the left, hold them both a while and flip the right one. If the left one is warm, hold them both a while and flip the right one…

extrapolate this concept until you are sure they are all flipped.

**ding**

July 18th, 2011 at 11:43 am

I guess there’s other “smart” ways of determining which glasses have already been turned over. E.g.

1. Move each upended glass away from the corner

2. Smear some of your saliva over each glass as you upend it

3. Pee in it

July 18th, 2011 at 12:23 pm

I suspect that the optimum strategy is to just keep turning both glasses over.

July 18th, 2011 at 12:46 pm

Sorry BearSprite – to clarify, I am looking for an optimum strategy – as soon as you touch the glasses, those are the ones you are flipping, the table then spins round to a random position.

Once you have touched the glasses – yes you can tell whether it is top up or top down, and can then chose what to do, but you can’t change the glasses that you flip on that go.

What’s the best strategy….?

Wizard of Oz – EEEWWWWW – but smart thinking – incentivize yourself not to turn the glass upside down!

July 18th, 2011 at 3:54 pm

Hi Karl, are sure about your last comment? If so, feel the glasses and only flip the upside down ones.

I’d assumed that you couldn’t feel them (before or after) turning them.

July 18th, 2011 at 5:00 pm

Don’t forget Chris, you have to touch two glasses, you can either leave them as they are or turn either, or both the other way up. Each time you have completed your move the table is spun round to a random position. You are blindfolded, so you don’t know which glasses you are reaching out for….

You could be lucky and do it in two moves. Could you do it in two moves every time?what is the least number of moves you could always complete the task in?

July 18th, 2011 at 5:20 pm

Hi Karl. That’s my point. Touch both glasses, feel which way up they are, only turn the upside down ones over, leave the upright ones alone. Being blindfolded seems to be irrelevant. The table being spun simply means you have to wait until all the upside down glasses have been made available. I wouldn’t expect many goes before your finished.

The least number of moves (one move corresponding to one spin) is two. The maximum is infinite – with 0 probability.

July 18th, 2011 at 5:34 pm

If you are sitting on the table, then you can always do it in two “moves”. Turn the two over that are in front of you, then turn the two over that are behind you.

July 18th, 2011 at 6:09 pm

Chris…of course being blindfolded is relevant, are you getting dizzy sitting on that table? if we could see then it would only take 2 moves every time. well, maybe more if we had to empty these pint glasses first.

July 18th, 2011 at 6:46 pm

good one Karl, but i don’t see a strategy that can “guarantee” a win in a number of moves. best i got is to pick two opposite corners (diagonally) and stick with these same two corners (in relation to you) for every move.

after the first flip, you will either pick the right two or the wrong two.

you would have a 50% shot of hearing a bell after move 2.

75% after move 3

87.5% after move 4

anyway…getting to move 11 would be like flipping a coin heads up 10 times in a row

July 18th, 2011 at 8:10 pm

Hi Karl. I’ve only just noticed that you can grab any two glasses – I thought you could only touch the two nearest you.

July 18th, 2011 at 8:33 pm

I’d turn over a diagonal pair first. Then after the spin, I’d randomly touch one glass, if it was the wrong way up I’d then turn it and it’s diagonal partner the right way up. If the first one was the right way up, I’d turn one of the other two glasses over. From then I’d just touch the nearest two, and do nothing until I find one that is the wrong way up. I’d then turn it the right way up DING.

If you have to touch two simultaneously, then I’ll go along with Knightmare’s answer.

But I can’t see how to guarantee a maximum number of moves.

July 19th, 2011 at 8:09 am

Well you would grab one glass then do not grab the glass from left or right side of the first glass grab the glass that is 90 degrees from the first glass then flip them the table rotates you grab the same location the you did first then all glasses should be upright.

July 20th, 2011 at 1:43 am

Begin by always flipping both glasses until you hear the bell ring. Then on your next turn, flip only the glass to your right. Then continue turning both glasses. Repeat until win.

Because it is random, there is no discrete technique for n-moves till win, just stick to a system and persevere till the stars align for you.

The flipping of only 1 glass on the next move after the bell rings is a rule that must be adhered to if you are to win, flipping 2 glasses after the bell rings will only keep you at 3 glasses if you were lucky, or more probable, reduce you back to 1. Assuming that when flipping only 1 glass you will always choose the same side, each time the bell rings you will have 25% chance of winning on the next move.

Not sure if this is better or not but you could alternate between both glasses and single glass each move. Once the bell rings, be sure to flip only one glass on the next move.

July 20th, 2011 at 1:50 am

Disregard last post, I misread and thought that the bell rings at 3 glasses flipped, not all 4.

In that case I would go for a pattern like 2-2-2-1-1-1-1-1-1-1-1-1… where ‘2′ represents both flipped and ‘1′ represents only the glass to my right.

July 20th, 2011 at 3:02 am

I just created a quick model in excel, using its psuedo-random number functions I plugged in the above sequence of flips and it solves it in an average of approx. 15 rounds from a sample of 100 games.

July 20th, 2011 at 4:33 am

Here are the averaged values from 25 samples of 100 games each.

Sequence: 2-2-2-2-2-2-2-2-2-2-2-2…

Rounds to win.

Mean: 7.98

Median: 6

Mode: 2

Games won in 2 rounds: 26.16%

Sequence: 2-2-2-1-1-1-1-1-1-1-1-1…

Rounds to win.

Mean: 16.59

Median: 11

Mode: 2

Games won in 2 rounds: 25.52%

Sequence: 1-1-1-1-1-1-1-1-1-1-1-1…

Rounds to win.

Mean: 21.00

Median: 15

Mode: 6

Games won in 6 rounds: 11.5%

Sequence: 2-1-1-1-1-1-1-1-1-1-1-1…

Rounds to win.

Mean: 19.49

Median: 15

Mode: 3

Games won in 3 rounds: 14.29%

Sequence: 2-1-2-1-2-1-2-1-2-1-2-1…

Rounds to win.

Mean: 16.76

Median: 12.5

Mode: 5

Games won in 5 rounds: 15.45%

Based on my simulations, it seems that simply flipping both glasses each round provides the best results.

July 20th, 2011 at 6:25 am

The above is based on the assumption that there is no knowledge of the glass’s orientation before the decision to flip is made.

From Karl’s further comments I gather that you are able to check orientation, then flip both, either or none as you please based upon this knowledge. This removes any need for strategy and leaves the entire game to chance.

Assuming that you are not a complete fool, each round you would simply turn each glass right-way-up as required and move on. This means that you would only need to land on 3 out of the 4 sides in order to have access to and flip each glass.

The probability of solving within 3 rounds is as follows:

On the 1st round you have p1.0 (4/4) of landing on a new side.

On the 2nd round you have p0.75 (3/4) of not landing on a previously visited side.

On the 3rd round you have p0.5 (2/4).

1.0 * 0.75 * 0.5 = 0.375

So 37.5% chance of success in 3 rounds.

Winning in 2 rounds has a probability of 25%.

Using my model and applying the logic of only flipping glasses that need flipping, the largest number of rounds I saw from 100 samples of 100 games was 16 which happened once only. 1/100000 = 0.0001 (0.01%).

I’m crap at probability so I’ll just go ahead and use my data to predict with 99.99% certainty that I could solve the game in less than 16 rounds.

July 22nd, 2011 at 4:30 pm

Knightmare is correct. First round, simply grab a glass and grab the glass diagonal from it, and flip them both. Next round, grab a glass, if it is upside-down, grab the glass diagonal and flip them both, 50% chance to win on round 2. If it is already the right way, leave it up-side down, and precede to round 3. Each round you have 50% chance to win.

Your chance to win by round n is:

100-100x(1/2)^(n-1)

n(1)=0%

n(2)=50%

n(3)=75%

n(4)=87.5%

n(5)=93.75%

July 22nd, 2011 at 4:32 pm

Meant to say “If it is already the right way, leave it alone, and precede to round 3.” This way you keep your 50% chance to win each round.

July 26th, 2011 at 9:37 am

Sorry for not posting earlier – senility is setting in, and I forgot…

Knightmare has come up with the best solution, however, as he has also noted there is the element of luck that swings between being able to do it in a minimum of 2 moves and 4 or 5 moves to be almost guaranteed doing it….almost!

July 26th, 2011 at 8:44 pm

thanks Karl

“next to nothing” is as far away from zero as “the largest number” is away from infinitely.

‘almost’ dosen’t count.8)

July 26th, 2011 at 8:45 pm

.8) should have been

July 26th, 2011 at 9:16 pm

A couple simple steps can solve that problem:

1-

grab the two glasses nearest you

2-

flip each one over

(table turns)

3-

grab the tops of the two glasses nearest you

4-

a)if one is up, flip the other

b)if both are up, flip neither

c)if neither are up, flip them both

(in the case of “a”, you have one glass left)

(in the case of “b”, you are half way there)

(in the case of “c”, you are finished and the bell will ring)

a:

1-grab the two glasses nearest you

2-option1-if one is up, flip the other and you are done

2-option2-if neither are up, repeat option1 until done

b:

1-grab the two glasses nearest you

2-option1-if neither are up, flip both and you are done

2-option2-if one is up, flip the other

3-either hear the bell or repeat option2

July 27th, 2011 at 3:45 am

Hi rhiannon, Knightmare’s solution is much better than that – it’s quicker and simpler.

Your solution takes more moves on average. After the first move it’s 1/4 that’ll you’ll get the already upright glasses, 1/2 that’ll you get an one upside down glass, and 1/4 that you’ll get both of the upside down glassess. So it’s only 1/4 that you’ll do it in two goes, and 3/4 that you’ll need 3 or more goes.

If I’ve done the calculations right, Knightmare’s method will take an average of 3 turns, yours will take an average of 4 [edit: that should be 3

^{2}/_{3}turns]. There is no upper limit to the number of turns required for either solution.July 27th, 2011 at 11:23 am

Turn one glass over at a time; if the table only spins in one direction.

July 27th, 2011 at 11:37 am

With Knightmare’s method, we have average number of turns to go, N = 1 + 0/2+ N/2. That’s because we must make at least 1 turn, then half the time we’re finished, and half the time we still have n turns to go. So we get N = 2. Add the initial move and we get 3 turns, average.

Rhiannon’s method requires a more involved analysis. The initial move turns two adjacent glasses the right way up. After the table rotates, we could have both of the right way up glasses (prob=1/4), both of the wrong way up glasses (prob=1/4) or one of each (prob=1/2).

Let N

_{2}be the number of turns to go after the initial turn. If we got one of each glass, then we’d turn the wrong way up glass over and then we’d have N_{3}turns to go.So N

_{2}= 1 + 0/4 + N_{2}/4 + N_{3}/2and N

_{3}= 1 + 0/2 + N_{3}/2 => N_{3}= 2Then N

_{2}= 1 + N_{2}/4 + 2/2 => N_{2}= 8/3Adding the initial move => 11/3 = 3

^{2}/_{3}turns on average.So I did goof in my last post

July 27th, 2011 at 11:56 am

Hi JL. It makes no difference which way the table turns. 90° anti-clockwise has the same effect as 270° clockwise.

You haven’t defined what your method actually is. Do you mean you will only test one glass and turn it over if it is upside down, or do you mean only turn one the right way up, even though both might be the wrong way up, or do you mean something else?

August 3rd, 2011 at 12:02 pm

1: Take two adjacent glasses, turn both up.

2: Take two adjacent glasses. If both are down, turn both up and done. If both up, repeat turn 2. If one up, flip both. Once you don’t have to repeat step 2, you are either done or have diagonals up and down.

3: Take two diagonal glasses. If both up, repeat step 3. If both down, flip both and done.