Brainteaser OTD - 79: Can You Gerrymander?

[Virgil Showlion]

For EE'ers needing a quick mental workout.

I've amassed quite a collection. I'll try to post one each evening, along with the solution to the previous problem.

For the consideration of your fellow puzzle-solvers, please prefix spoiler posts with (spoiler) and hint posts with (hint).

Also, if EEers have their own favourites, I'd love to see them.

Ignis aurum probat, miseria fortes viros.

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Main Problems:

Find 2 - The Senators at post #20.
Find 3 - Crotchety Ol' Joe at post #24.
Find 4 - The Evil Beer Queen at post #34.
Find 5 - Rockin' the Boat at post #38.
Find 6 - The Mutilated Chessboard at post #41.
Find 7 - The Monty Hall Problem at post #46.
Find 8 - The Missing Dollar at post #60.
Find 9 - Tipsy Turvy at post #73.
Find 10 - Population Control at post #78.
Find 11 - The Mysterious Slide at post #90.
Find 12 - Don't Wine at post #101.
Find 13 - The Hungry Troll at post #115.
Find 14 - Nine Thirsty EE'ers at post #124.
Find 15 - Cabling Labling at post #154.
Find 16 - Forest of Doom at post #159.
Find 17 - Oddly Familiar at post #175.
Find 18 - Colleenz Buys You a Beer at post #185.
Find 19 - Mod Championships at post #200.
Find 20 - What Am I? at post #211.
Find 21 - Rebus Maximus at post #221.
Find 22 - To Number or Not to Number at post #231.
Find 23 - Nuts and Bolts at post #242.
Find 24 - Line Em' Up at post #258.
Find 25 - Rabid Rebus Redux at post #277.
Find 26 - Define 'Tough' at post #310.
Find 27 - May Our Paths Never Cross at post #318.
Find 28 - Avoiding Work at post #326.
Find 29 - No Getting Wet at post #332.
Find 30 - Ch EE'rs! at post #348.
Find 31 - Who Wants to Be a Moddionaire! at post #374.
Find 32 - Think You Can Beat Me? at post #383.
Find 33 - The Wayward Centipede at post #409.
Find 34 - The Drink That Never Was at post #440.
Find 35 - The Best Moneymaking Scheme at post #448.
Find 36 - The Monty Hall Problem... ? at post #466.
Find 37 - Memento Mori at post #480.
Find 38 - Battle of the Sexes at post #488.
Find 39 - So You Think You Know English? at post #501.
Find 40 - Carpet Diem at post #530.
Find 41 - Deceptive Diagnosis at post #548.
Find 42 - The Shortest Path to Victory at post #556.
Find 43 - Tying Up the Earth at post #571.
Find 44 - Oh, The Irony! at post #583.
Find 45 - My Special Number at post #586.
Find 46 - Fluoride, Fluoride Everywhere at post #597.
Find 47 - Robots on a Line at post #606.
Find 48 - The Seven-Hater at post #621.
Find 49 - Those Weird EEers at post #627.
Find 50 - Rapid Rebus Refire at post #654.
Find 51 - From APPLE to LEASH at post #695.
Find 52 - The Tortoise and the Hare at post #738.
Find 53 - All Stacked Up at post #764.
Find 54 - Another Foolproof Get-Rich-Quick Scheme at post #772.
Find 55 - Gone to the Dogs at post #783.
Find 56 - EEers on a Plank at post #816.

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Find 57 - Alien Attack! at reply #212.
Find 58 - Ping Pong Puzzler at reply #221.
Find 59 - Canada's Odd Demographics at reply #226.
Find 60 - Literally a Problem at reply #230.
Find 61 - Logical and Not at reply #234.
Find 62 - Odd Sex at reply #274.
Find 63 - Iggy's Chilli at reply #306.
Find 64 - What's in a Name? at reply #347.
Find 65 - EE-mazing! at reply #381.
Find 66 - sEEgulls and Mods at reply #402.
Find 67 - Love Me Tender at reply #429.
Find 68 - Gridlock at reply #484.
Find 69 - ConfucEEus Say... at reply #487.
Find 70 - Out of Body Experience at reply #555.
Find 71 - Virgil's Password at reply #572.
Find 72 - My Brother's Junk at Reply #619.
Find 73 - Stuck on Purpose at Reply #665.
Find 74 - All Roads Lead to Booze at Reply #679.
Find 75 - Acrobatinyms at Reply #685.
Find 76 - Death by Canadian Taxes at Reply #700.
Find 77 - The Physics Trifecta at Reply #716.
Find 78 - A Puzzle for Rio at Reply #854.
Find 79 - Can You Gerrymander at Reply #889.

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Contributor Problems:

Find Instamate by whiskmav at post #169.
Find Full Court Press by whiskmav at post #177.
Find Newly Knighted by whiskmav at post #189.
Find Cornered Royal by whiskmav at post #206.
Find Lady in Waiting by whiskmav at post #219.
Find An Orgy of Sacrifice by whiskmav at post #226.
Find Harder than Banking Reform by whiskmav at post #237.
Find Prolonging the Inevitable by whiskmav at post #255.
Find King Wiggle by whiskmav at post #273.
Find Got Your Back by whiskmav at post #301.
Find Oldie but Goodie by synergyo1 at post #313.
Find Multipurpose Knight by whiskmav at post #338.
Find Wiggle Room by whiskmav at post #546.
Find Frog Leap Test by kent at reply #219.

[jan]

I have no idea what you mean Virgil.

[ՏՇԾԵԵʅՏɧ_LԹՏՏʅҼ]

Oh, boy...a nerd thread. I've been waiting for these puzzles, even though they make my brain bleed.

[Virgil Showlion]

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VirgilBot TP v0.01 Started Thread Port on 12/22/2010

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Virgil Syonid
Message #2 - 04/20/10 02:02 AM

From where comes this hole?



Official solution posted at #13.

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beagleowner
Message #3 - 04/20/10 02:06 AM



it is quite obvious where the hole comes from and why.

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Scottish_Lassie
Message #4 - 04/20/10 02:25 AM

Structural Engineering for Dummies. It's too easy. Here's the Engineer's Bible----->

[Virgil Showlion]

Virgil Syonid
Message #5 - 04/20/10 03:33 AM

I can never tell if you guys are being sarcastic.

Maybe <sarcasm></sarcasm>?

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Solstice Tesla_DC-Meme
Message #6 - 04/20/10 03:50 AM

From where comes this hole?

[HINT]



pitch switch causes brain itch?



now that is unpossible

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beagleowner
Message #7 - 04/20/10 02:17 PM

Virgil, I wasn't being sarcastic. I just figured it out right away. But didn't want to post the answer and ruin it for others.

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laterbloomer
Message #8 - 04/20/10 02:40 PM

I don't understand how it happens. The triangles are the same height and width so there should not be an empty square.

[Virgil Showlion]

SlowbutSure
Message #9 - 04/20/10 02:45 PM



(Spoiler Alert)









Later - The slopes of the red and blue triangles are different. Count five squares in from the right and look at the number of squares you have to go up.

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PaladinHG
Message #10 - 04/20/10 02:51 PM



Spoiler Alert!!





Later, it wasn't obvious for me, either, but SBS is right. The result is that the four shapes together don't really total a triangle. It's an opitical illusion.

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laterbloomer
Message #11 - 04/20/10 02:59 PM



Spoiler alert









So the sloped line isn't straight? In one it is concave, the other convex?

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David Lee Roth
Message #12 - 04/20/10 03:05 PM

I got a headache just reading the word brain teaser in the title thread...please, don't make me think that hard again...

[Virgil Showlion]

IHearYou2
Message #13 - 04/20/10 03:58 PM



(Spoiler Alert)



The ratios are different for the red triangle it is 3:8 and for the teal it is 2:5. In other words three height to 8 width. The ratio is 2.67 across for every one up for the red and for the teal 2.50 resulting in a .5 difference for these shapes. Triangles are 1/2hw which equals one square exactly.

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whiskmav
Message #14 - 04/20/10 06:30 PM

Keep the brainteasers coming!

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SantaSpydah
Message #15 - 04/20/10 07:21 PM

Good explanation IHY!!!

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JennysBeenAGoodGirl
Message #16 - 04/20/10 07:28 PM

do you smell that? It's my brain smoking!!!

[Virgil Showlion]

SantaSpydah
Message #17 - 04/20/10 07:50 PM



Jenny,

All I can focus is on your question, not on you answer.

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JennysBeenAGoodGirl
Message #18 - 04/20/10 07:55 PM

very funny.....

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Virgil Syonid
Message #19 - 04/20/10 10:50 PM

We'll consider IHY's post #13 as the 'official' solution to Problem 1.

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Virgil Syonid
Message #20 - 04/20/10 10:52 PM

Problem 2 - The Senators

Consider five senators in a room with you:

A B C D E


One of these senators (you don't know which one) is an INDEPENDENT. A independent will always answer truthfully when asked a question.

The other senators are either REPUBLICANs or DEMOCRATs.

• REPUBLICANs: Will respond to question #1 with a lie, question #2 with the truth, question #3 with a lie, etc., toggling each time.
  
• DEMOCRATs: Will respond to question #1 with the truth, question #2 with a lie, question #3 with the truth, etc., also toggling each time.

Senators are busy people. They'll allow you to ask any two questions you want (questions don't have to be true/false), but only two. Any question can be addressed to any senator. You can ask two different senators questions, or you can ask the same senator two questions.

You can only ask one senator at a time (no 'group' questions), and he/she will give you just one answer per question.

What you want to know with 100% certainty after your two questions is: Which senator is the INDEPENDENT?

The main complication: Although you know that four of the senators are either REPUBLICANs or DEMOCRATs, you don't know which ones are which, or even how many of each there are. There might be four DEMs, or four REPs, or two and two, or any other mixture. You just don't know.

What questions do you ask to find out what you want to know?

(Note: I'm using a 'senator' analogy with INDEPENDENTs, REPUBLICANs, DEMOCRATs for conceptual clarity. The correct solution has nothing to do with party ideology, politics, the current government, etc.)

A reminder to use '(Hint)' and '(Spoiler)' where applicable, and thanks to all posters who have been most diligent so far.

[Virgil Showlion]

beagleowner
Message #21 - 04/21/10 04:47 AM



Just a guess, I have no clue how to figure this one out. Ask one question.



Name me two senators not a in this room but in your party that you would like to see replaced with a new senator. (I would think only the independant would answer.)

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Virgil Syonid
Message #22 - 04/21/10 07:28 AM

Beagle,

Calling the actors 'senators' was a conceptual simplification. It's a logic problem; in the original problem the INDEPENDENT is called a 'truthteller', the REPUBLICAN is called a 'type A toggler' and the DEMOCRAT is called a 'type B toggler'. From the standpoint of the problem, you can consider the actors to be omniscient 'things' that answer truthfully/untruthfully with the indicated true/false patterns. Using senators just happens to be a convenient analog.

I've shown this problem to others before, and I find their eyes glaze over if I start talking about 'type A toggler' and 'type B toggler', etc.

Also, I've added some clarification: each question only goes to one senator. No 'group' questions.

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Virgil Syonid
Message #23 - 04/22/10 02:00 AM

(Solution to Problem 2)

Pick any senator, say senator 'A', and ask: Are you an independent?

If A happens to be the INDEPENDENT, he/she will answer 'YES'.
If A happens to be a REPUBLICAN, he/she will lie for question #1 and answer 'YES'.
If A happens to be a DEMOCRAT, he/she will tell the truth for question #1 and answer 'NO'.


If senator A replies 'NO' (meaning that A is a DEMOCRAT), make your second question (also to senator A): Who in this room IS NOT an independent senator?
Senator A, a DEMOCRAT, will lie for question #2, thereby naming a person in the room who IS an independent senator. Namely, the INDEPENDENT.


If instead senator A replies 'YES' (meaning that A is either the INDEPENDENT or a REPUBLIC), he/she will answer question #2 truthfully (either way).
Hence, ask senator A: Who in this room IS an independent senator? Senator A will then name the INDEPENDENT.

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Virgil Syonid
Message #24 - 04/22/10 02:02 AM

Problem 3 - Crotchety Ol' Joe

Crotchety Ol' Joe lives by himself in an ol' house he built with his own two hands.

And Ol' Joe's house has a cellar. The cellar is lit by a single filament light bulb hanging down from a string in the middle of the room.

Now Joe wasn't the greatest electrician, and he installed the only switch that turns the cellar light bulb on/off on the main level of his house. Also, the switch is on a panel that has two other switches that Ol' Joe left unwired. They don't do anything.

Because Ol' Joe is so old, he's run into a bit of a problem. Joe remembers that he left the cellar light off after he went down last, but (Joe's ol' memory not being what it used to be) forgot which of the three switches is the 'real' one that controls the light.

To make matters worse, Joe can't see whether there's light in the cellar unless he actually goes all the way downstairs and inside the cellar. And Ol' Joe (his knees not being what they once were) only wants to go down into the cellar once to figure out which switch is the 'real' one.

But Ol' Joe has some wits left in him yet. He thinks it over, twiddles with the three light switches, then heads on down into the cellar. By the time he's coming back upstairs, he knows for sure which of the three switches is the 'real' one.

How did Ol' Joe manage to do it?

(Note: Ol' Joe don't have no wife, neighbors, friends, cats, dogs, smart talkin' parrots, mirrors, or astral projectin' knowhow.)

A reminder to use '(Hint)' and '(Spoiler)' where applicable, and thanks to all posters who have been most diligent so far.

[Virgil Showlion]

beagleowner
Message #25 - 04/22/10 04:44 AM

(Solution to Problem 2)

Clear as mud, there, thanks.



I hate word games. They make no sense whatsoever. If one is lying or telling the truth and you don't know which is doing what, how the he11 are you supposed to follow that round about logic that goes nowhere except to fry one's brain.

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Candy Cane Colleenz
Message #26 - 04/22/10 02:32 PM



SPOLIER (I think)...















Flip the first switch and wait a few minutes. Flip the second switch and go to the cellar right away. If the light is off it is switch #3. If the light is on, touch the bulb. If the bulb is cool it is switch #2. If the bulb is warm it is switch #1.

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Gameshow
Message #27 - 04/22/10 07:23 PM

Now Joe wasn't the greatest electrician, and he installed the only switch that turns the cellar light bulb on/off on the main level of his house. Also, the switch is on a panel that has two other switches that Ol' Joe left unwired. They don't do anything.

According to the OP's statement.
If the other two switches "don't do anything", why not just flip all 3 switches on.
One of them is going to light up the cellar, and his old knees doesn't have to worry about coming back up the stairs.

-Ben

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Virgil Syonid
Message #28 - 04/22/10 09:33 PM

If the other two switches "don't do anything", why not just flip all 3 switches on.

*lol*

OK. Let's say that the two other switches are wired up to give Dr. Pig an electric shock.

[Virgil Showlion]

SlowbutSure
Message #29 - 04/22/10 09:40 PM

Let's say that the two other switches are wired up to give Dr. Pig an electric shock.

Then I would definitely turn on all three switches. Love ya Doc.

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Scottish_Lassie
Message #30 - 04/22/10 11:09 PM



My Answer to Problem 3

None of the switches will work, because at the start of the problem it says the lightbulb is hanging down from a string. Not an electrical wire.

DUH

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IHearYou2
Message #31 - 04/22/10 11:13 PM



(Spoiler)



The question that needs to be asked first is what position are all the switches in, when he first fiddles with them. If one of them is up then he has a method to figure out with certainty which switch is the connected one. If all three are down then he has no method mathematically of figuring out with certitude which switch is the connected one with one trip into the cellar.

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Virgil Syonid
Message #33 - 04/23/10 12:39 AM

Good creative answers for problem 3.

The 'correct' answer is given by colleenz in #26, although all others showed some excellent out-of-the-box thinking.

[Virgil Showlion]

Virgil Syonid
Message #34 - 04/23/10 12:40 AM

Problem 4 - The Evil Beer Queen

Queen Beerella III is evil, queeny, and loves beer. In fact, she just received 1,000 kegs of her favorite beer for a royal beer ball she's hosting in one month (31 days).

But Beerella's rule hasn't always been the most popular. In fact, her chief jailer recently rounded up 1,000 dissidents and is awaiting Beerella's order to execute them. And worse, Beerella's informants tell her that before the dissidents were captured they managed to poison the beer in (exactly) one of the 1,000 kegs Beerella received for the beer ball.

This poison is incredibly devious. If a person drinks even a drop of poisoned beer, he/she will feel perfectly fine for exactly 30 days and then BAM! drop dead. Drinking any greater amount of the poisoned beer will have exactly the same effect.

Naturally, Beerella thinks: I'll assign a keg to each one of 1,000 dissidents, feed each dissident a drop of beer from their keg, and see which one drops dead after 30 days to determine which keg is the poisoned one.

That's fine, her jailer tells her. But he reminds her that every dissident she brings to the castle for her test will have to be individually guarded at a cost of 1 gold per dissident. Hence, 1,000 dissidents will cost 1,000 gold.

You, being Beerella's most trusted adviser, tell her you can determine (with 100% certainty) which keg is the poisoned keg using far fewer dissidents (and hence, costing much less gold).

What is the smallest number of dissidents you require to determine the poisoned keg with certainty? How do you do so?

(Note: None of the dissidents will tell you which keg is poisoned. You have to find out by feeding them beer.)
(Note: Postponing the beer ball, buying new beer, or poisoning guests are all unacceptable!)
(Note: Beerella (and therefore you) don't care how many dissidents die during the test. The survivors will be executed anyway.)

A reminder to use '(Hint)' and '(Spoiler)' where applicable, and thanks to all posters who have been most diligent so far.

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SlowbutSure
Message #35 - 04/23/10 02:05 AM

((shudder)) I don't even want to think about a world with an evil Beer queen.

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beagleowner
Message #36 - 04/23/10 03:44 AM

This sounds like a problem for Beerwench, obviously she is related.

[Virgil Showlion]

Virgil Syonid
Message #37 - 04/24/10 05:09 AM

(Solution to Problem 4)

It is possible to determine which keg is poisoned using only 10 dissidents.

This is accomplished by assigning a unique 'pattern' of dissidents to each keg, and seeing which pattern dies off. The most natural choices are bit patterns for binary numbers.

As a smaller example where 3 dissidents are used on 8 kegs, we would have:

K1 K2 K3 K4 K5 K6 K7 K8
D1 * * * *

D2 * * * *

D3 * * * *

Where a '*' indicates that a particular dissident (D1, D2 or D3) is made to drink a sample from a particular keg (K1, K2, K3, etc.).

In this example, if D2 and D3 (but not D1) dropped dead after 30 days, Beerella could be certain that K7 was the poisoned keg. If D1 dropped dead (but not D2 or D3), K2 is the poisoned keg, etc., etc.

For the larger problem, nine dissidents would work for up to 512 (= 2^9) kegs. As this is not enough, we move to 10 dissidents, which works for up to 1024 (= 2^10) kegs. This is more than enough to assign a unique death pattern to each of the 1,000 kegs.

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Virgil Syonid
Message #38 - 04/24/10 05:12 AM

Problem 5 - Rockin' the Boat

You are sitting in a boat in the middle of a lake. You have a rock with you in the boat. You throw the rock overboard into the lake and it sinks to the bottom. By doing this, will the level of the water in the lake:

a) go up slightly
b) go down slightly
c) stay the same

Why?

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genuine ga peach
Message #39 - 04/24/10 04:02 PM



my guess:

















stay the same...the displacement will not change, only the location of the rock

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Virgil Syonid
Message #40 - 04/25/10 07:42 PM

(Solution to Problem 5)

The correct answer is b) go down slightly.

To understand this (somewhat unintuitive) result requires looking at the principle of buoyancy. Buoyancy tells us that when the rock is in the boat, the level of the water displaces the mass of the rock.

Hence, suppose the density of the rock is D, the volume of the rock is V, the density of the water in the lake is W, and the surface area of the lake is S.

The mass of the rock, M, is given by: M = D x V

If H is the rise in the water level due to mass displacement, and M_disp is the mass of displaced water, we have: M_disp = H x S x W.

Since buoyancy tells us that M = M_disp, we have:

D x V = H x S x W

And hence,

H = D x V / (S x W)

When the rock is underwater, it displaces its volume in water. If H_vol is the rise in the water level due to volume displacement, we have: V = H_vol x S

And hence,

H_vol = V / S

To calculate the change in the level of the water (from the mass displacement of before to the volume displacement of after), we look at:

H_change = H_vol - H

Substituting in, we find:

H_change = V / S - D x V / (S x W)
= V/S x (1 - D/W)

We know that V/S is positive (since volume and surface area are both positive). Hence, to determine the sign of H_change, the question is:

1 - D/W > 0?

Or equivalently:

1 > D/W?

So if the rock is more dense than the water, H_change will be negative (water will go down). Otherwise, H_change is positive and the water level will go up. But we know that the rock is more dense than the water since the rock sinks in the lake.

Hence, D > W, and H_change is negative. The water level goes down when the rock is in the lake.

[Virgil Showlion]

Virgil Syonid
Message #41 - 04/25/10 07:44 PM

Problem 6 - The Mutilated Chessboard

Take a standard 8x8 chessboard. Remove the tiles of two opposing corners (e.g. NW and SE, or NE and SW) so that you are left with a 'mutilated' 62-tile board.

Now suppose you have 31 dominoes, and the dominoes are sized so that one domino fits over exactly two tiles on the board. Can you come up with a layout for the 31 dominoes to completely cover the 62 tiles on the board?

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Candy Cane Colleenz
Message #42 - 04/26/10 01:33 PM



Spoiler alert..































No

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IowaSwirl
Message #43 - 04/26/10 02:06 PM

Can you come up with a layout for the 31 dominoes to completely cover the 62 tiles on the board?

Can I use a chainsaw?

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whiskmav
Message #44 - 04/26/10 02:36 PM

Since you have taken away like colored squares (ie black and black or white and white) the answer is no.

[Virgil Showlion]

Virgil Syonid
Message #45 - 04/27/10 12:19 AM

(Solution to Problem 6)

whiskmav is correct in #44. Since either 'mutilation' will result in an unequal number of black tiles and white tiles remaining, and since each domino must cover one black tile and one white tile, there is no way to cover all of the tiles.

And no, you can't take a chainsaw to it.

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Virgil Syonid
Message #46 - 04/27/10 12:20 AM

Problem 7 - The Monty Hall Problem

You are a contestant on a game show where the host presents you with three doors. Behind one of the doors is a brand new car (the prize); behind the other two doors are goats (the booby prizes). You do not know which of the three doors conceals the car and which two doors conceal the goats.

The host asks you to pick a door at random, and you do. He then goes to one of the two doors that you didn't pick and opens it up to reveal a goat. (He can do this because he knows what's behind each of the doors.)

Then he asks you: do you want to claim the prize behind the door you initially picked, or do you want to claim the prize behind the remaining door (that is, the door you didn't pick and that he didn't open).

Assuming you want to win the car, what do you do? Does it matter whether you switch doors or not?

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IowaSwirl
Message #47 - 04/27/10 03:34 AM



SPOILER ALERT!















It very much matters. 2 out of 3 times, switching will win.



Let's say Door 1 is the winner.



You pick #1. Switch = lose, keep = win.

You pick #2. Switch = win, keep = lose.

You pick #3. Switch = win, keep = lose.

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whiskmav
Message #48 - 04/27/10 01:47 PM

But at that point you only have 2 doors to choose from. You have only a 50/50 chance.

[Virgil Showlion]

IowaSwirl
Message #49 - 04/27/10 03:14 PM

But at that point you only have 2 doors to choose from. You have only a 50/50 chance

No, you have the choice of taking the door you originally chose (with a 1/3 shot of being correct) or the two doors you didn't choose (even though one was revealed as being incorrect).



I mapped out above how it plays out.

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whiskmav
Message #50 - 04/27/10 03:17 PM

So you're saying that the door with a goat behind it (it's already been revealed) is a viable choice for you?

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IowaSwirl
Message #51 - 04/27/10 03:21 PM



Keep Winner 1 2 3 You Pick 1 Win Lose Lose 2 Lose Win Lose 3 Lose Lose Win Total wins = 3 Switch Winner 1 2 3 You Pick 1 Lose Win Win 2 Win Lose Win 3 Win Win lose Total wins = 6

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IowaSwirl
Message #52 - 04/27/10 03:22 PM

So you're saying that the door with a goat behind it (it's already been revealed) is a viable choice for you?

I'm saying it is part of the choice to switch.



See the post directly above this one proving how it plays out. If you still think it is incorrect, show me where my data is wrong.

[Virgil Showlion]

IHearYou2
Message #53 - 04/27/10 04:04 PM

No this is a common misconception it still is 50/50 whether you stay with your door picked or switch to another. People think that the odds get greater for them but it is a constant. The only difference is it isn't a 33% chance of winning. The same is with a flip of a coin you can flip a coin 20 times and get heads every time. The 21st flip you have no better chance of getting tails then on the other 20 it is still 50/50.

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IowaSwirl
Message #54 - 04/27/10 04:14 PM

No this is a common misconception it still is 50/50 whether you stay with your door picked or switch to another.

Actually, this is the common misconception. You would be correct if Monty was opening a door at random, but he isn't - he is opening a door that he knows loses.



There have been math papers written about it, there is a wikipedia article about it, and there's even a simulator showing how the odds work.

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whiskmav
Message #55 - 04/27/10 04:23 PM

If the host knows where the car is then switch. If the host does not know, its anyone's guess.

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IHearYou2
Message #56 - 04/27/10 07:09 PM

I don't understand the logic. One door has been eliminated that has all that has occurred, you don't have any reason to think that the other door is the one the car is behind then before. All that has increased is your probabilities of winning but there is no change in preferred door just like with the flip of a coin.

[Virgil Showlion]

IowaSwirl
Message #57 - 04/27/10 07:33 PM

just like with the flip of a coin.

This is not at all like the flip of a coin.



Picture it this way instead. Let's say instead of 3 doors, it was 3 cards set next to each other, one of which was the Ace of Spades which you need to find. You pick the first card. The dealer stacks the second and third cards. Do you switch from your one card to the pile?



Let's take this to a little higher level. You have a full deck of cards. You pick one, the dealer stacks the other 51. Do you switch from your one card to the pile?



Take this back to Monty Hall - there's a donkey behind two doors, and a car behind one. You pick a door. Monty then offers you either your door or both other doors (which is no different than him revealing a loser, or stacking other cards). Do you switch?

---

Saturn8
Message #58 - 04/27/10 07:51 PM



If you look at the problem from this point of view, it might make more sense.

Before you chose any door, you had a 33% chance of being right, yes? Door 1 = 33%, door 2= 33%, door 3 = 33%. So chances are, you picked the wrong door. You have a greater chance of being wrong.

So you pick door #1. There is now a 66% chance the car is behind door #2, or door #3. Regardless of whether Monte opens one of those doors, there is still a 66% chance the car is behind door #2 or door #3. When Monte opens door #2 and reveals the goat, there is still a 66% chance the car is behind door #2 or door #3...the fact that you now know it is NOT behind door #2, just shifts all that probability to door #3!

Lesson: always switch

---

Virgil Syonid
Message #59 - 04/28/10 12:04 AM

(Solution to Problem 7)

You should in fact switch doors.

IowaSwirl gives a good explanation in #51.

To summarize. You have a 1/3 chance of initially picking the door with the car (case A), and a 2/3 chance of picking a door with a goat (case B). In case A, switching loses. In case B, switching wins. But since case B is twice as likely as case A, switching gives you twice the odds of winning.

As others have mentioned, the problem doesn't work if the host doesn't know which door the car is behind.

Good participation in this one.

---

Virgil Syonid
Message #60 - 04/28/10 12:05 AM

Problem 8 - The Missing Dollar

Three men visit a hotel for an overnight stay. As they check in, the concierge tells them that the room fee is $10 per person. They each pay $10 ($30 in total) and go to their rooms.

But then the concierge realizes that the hotel was having a special and that the room fee should only have been $9 per person. He summons the bellboy, gives him $5 (in ones) and tells him to reimburse the three men.

The bellboy visits the three men and gives them back a dollar each. Since the bellboy has $2 left, he pockets it for his trouble.

But now the concierge is confused. Each of the three men effectively paid $9, making $27, and the bellboy kept $2, making $29. But $30 was transacted originally. Where did the extra dollar go?

[Virgil Showlion]

weltschmerz
Message #61 - 04/28/10 01:01 AM

Do you have any brainteasers that don't involve math? I really stink at math.

---

weltschmerz
Message #62 - 04/28/10 01:04 AM

Three men visit a hotel for an overnight stay. As they check in, the concierge tells them that the room fee is $10 per person. They each pay $10 ($30 in total) and go to their rooms
But then the concierge realizes that the hotel was having a special and that the room fee should only have been $9 per person. He summons the bellboy, gives him $5 (in ones) and tells him to reimburse the three men.

Why would the concierge give the bellboy $5 in ones, instead of $3?

---

Virgil Syonid
Message #63 - 04/28/10 01:16 AM

Why would the concierge give the bellboy $5 in ones, instead of $3?

Um...

Yeah.

Suppose the concierge gives the bellboy a fiver. The bellboy trades the fiver for five ones from the men. The rest is the same.

(Good to see that people are thinking about this harder than I was. )

---

beagleowner
Message #64 - 04/28/10 03:22 AM

But now the concierge is confused. Each of the three men effectively paid $9, making $27, and the bellboy kept $2, making $29. But $30 was transacted originally. Where did the extra dollar go?

Spoiler:

















simple math. the 27 + 2 = 29 doesn't matter. What matters is that 3 men paid $10 each to get the room. They found out that the special was $27, divide that by 3, you get 9. Each man gets one dollar back. The busboy or whatever had a $5. Each of the men get returned the $1 that is owed back to them. Remember? 3 guys one dollar each, and busboy kept 2. 3 + 2 = 5 simple as that.

OR $30 - $27 = 3. Busboy had $5. He kept $2 and returned 1 each to the 3 men, the math still works. 3+2=5, The 27 + 2 is just in the word puzzle to distract you.

[Virgil Showlion]

Candy Cane Colleenz
Message #65 - 04/28/10 04:00 PM



Spolier Alert...









Has the concierge been drinking? (must be an EE poster )



If he had $30, and needed to refund $1 to each guest, why did he give the bellboy a fiver? Each man did pay $9 for a total of $27. Of that, the concierge kept $25 ($30 minus the $5 he gave the bell boy) and the bellboy kepy $2. There is no missing $1.

---

Gameshow
Message #66 - 04/28/10 04:28 PM

Beagle, I think your logic is correct, but your starting point is a bit off.

Room special is $25, not $27.

<SPOILER ALERT>

The cost basis has changed. The room no longer costs $30, it cost $25.
Therefore, each guy paying $9 = $27 results in $2 more than the room cost. Which is what the bellman pocketed. No missing dollar.

-Ben

---

PaladinHG
Message #67 - 04/28/10 04:51 PM



Sorry, Virgil, but I think you got the riddle wrong. Here it is as I remember it:



Three guests check into a hotel room. The clerk says the bill is $30, so each guest pays $10. Later the clerk realizes the bill should only be $25. To rectify this, he gives the bellhop $5 to return to the guests. On the way to the room, the bellhop realizes that he cannot divide the money equally. As the guests didn·t know the total of the revised bill, the bellhop decides to just give each guest $1 and keep $2 for himself.

Now that each of the guests have been given $1 back, each has paid $9, bringing the total paid to $27. The bellhop has $2. If the guests originally handed over $30, what happened to the remaining $1?

---

beagleowner
Message #68 - 04/28/10 05:40 PM



paladin, my math still works. It does not matter if the bill was reduced to $25. What matters is the $5 difference.

Each of the three paid $10 which comes to $30. Bellhop, steals $2 because he can't split it evenly. He gives the other $3 to the men who paid for the room. Thus equalling $5 difference. Math still works, I'm sticking to my original thing. the other numbers don't matter. 3 + 2 = 5, there's no missing dollar.



If you want to go further, 25 + 3 +2 = 30. Either way the math works. The 27 + 2 is just in there to confuse you. It doesn't matter.

[Virgil Showlion]

PaladinHG
Message #69 - 04/28/10 05:51 PM

paladin, my math still works. It does not matter if the bill was reduced to $25. What matters is the $5 difference.

I understand that. But it just makes more sense when you have the riddle right.

---

beagleowner
Message #70 - 04/28/10 06:25 PM

oh, ok.

---

SantaSpydah
Message #71 - 04/28/10 06:27 PM



Spoiler Alert!!!!!

---

Virgil Syonid
Message #72 - 04/28/10 11:44 PM

(Solution to Problem 8)

Several ways of looking at the problem are given in problems #64 - #68. As most point out, the problem is to realize that only $27 is transacted: 3 x $9 (men) = $25 (concierge) + $2 (bellboy). The number $30 is one of the intermediate steps and ultimately has no significance.

Paladin points out in #67 that there are ways to make the riddle more elegant (from a delivery standpoint). Although I acknowledge this, I will defend the 'rightness' of #60 in the sense that the problem was presented to me in this way.

[Virgil Showlion]

Virgil Syonid
Message #73 - 04/28/10 11:45 PM

Problem 9 - Tipsy Turvy

Suppose you have eight marbles identical in every way except that one of them is slightly heavier than the other seven. You are given an old balancing scale (the kind with the two dishes and a needle that tells you which side is heavier) and asked to determine which marble--exactly--is the heavier one. If you were allowed three weightings, finding the heavier marble would be a breeze. But suppose you can only perform two weightings using the scale.

What do you do?

Do you have any brainteasers that don't involve math? I really stink at math.

This one doesn't involve math.

---

Scottish_Lassie
Message #74 - 04/29/10 12:17 AM



SOLUTION TO PROBLEM #9??

Place 4 marbles in each of the two scale trays. When one side of the scale gets lower than the other, the heaviest marble will roll to the outer edge of the lower tray due to gravitational pull.

---

Virgil Syonid
Message #75 - 04/29/10 12:56 AM

the heaviest marble will roll to the outer edge of the lower tray due to gravitational pull

*lol*

I never thought of a density-based solution before.

Technically, if you tilted the scale and shook it a lot, due to density laws the heavy marble would have an (ever so slightly) greater-than 1/8 chance of ending up on the bottom. (Say a 12.50001% chance rather than a 12.5% chance. It would depend on the weight difference and the 'mobility' of the marbles.) If you could somehow keep track of which marble was which, you could discover the heavy one through a process called 'tomography'. Basically, run a huge number of experiments, shaking the scale for each 'experiment' and recording which marble ends up at the bottom.

The tally would eventually show one of the marbles was present at the bottom more often. But for the results to be statistically significant you'd have to run millions of experiments, and even then you would only get 'very, very close' to discovering the heavy marble--there would always be a finite margin of error.

So I'll give you partial credit.

---

GeenaMercile
Message #76 - 04/29/10 01:02 AM

Spolier Alert























You split the marbles up into 3 groups, Group A has 3, Group B has 3 and Group C has 2. Put group A on one side and B on the other.

outcome 1 from weigh one-- they equal, use your second weigh with 1 marble from group c on one side the other on the other side, the one that goes down is the one you want.

Outcome 2 from weigh 1 is that A goes down. You know that the heavier marble is in group A now. Split group A up with one marble on one side, a second marble on the other side and one marble off. If one side goes down then that is the one you want. If they balance the one off to the side is the one you want.

Outcome 3 from weigh 1, B goes down. Do the same with group B as you did with A in outcome 2.


ETA- the spoiler at the start...

[Virgil Showlion]

Virgil Syonid
Message #77 - 04/30/10 12:57 AM

(Solution to Problem 9)

GeenaMercile gives the solution in #76, saving me from having to type it.

---

Virgil Syonid
Message #78 - 04/30/10 12:57 AM

Problem 10 - Population Control

In a third-world backwater country called Misogyna, the government decides they want to increase the ratio of males to females. After consulting with the 'scientists' at the OPT, they decide to implement an ingenious, diabolical solution: each family (i.e. couple) is allowed to have as many boys as they want, but only one girl. That is, as long as a couple's children are all male, they can continue to have new children, but as soon as the couple has one girl they cannot have any more children.

Will their diabolical plan have the intended effect?

---

IowaSwirl
Message #79 - 04/30/10 02:35 AM

Will their diabolical plan have the intended effect?

That depends if I can round or not - I realize that the chances aren't 50/50 for boy/girl.



If I could, let's say there are 100 couples having children. 50 will have a girl and be done, 50 will have a boy and continue. Of the 50 with boys, 25 will have a girl and be done, 25 will have a boy. Of the 25 with boy-boy, 12 will have a girl and 13 will have a boy. Of the 13 boy-boy-boy, 7 will have a girl and 6 will have a boy. Yes, I rounded.





This will, of course, continue, but at this point you'll have:



50 + 25 + 12 + 7 = 94 girls

50 + 25 + 13 + 6 = 94 boys



If you factor in that something like 105 boys are actually born for every girl, it would - but not to a major degree, especially considering infant/childhood mortality.

---

beagleowner
Message #80 - 04/30/10 06:57 AM

wouldn't you need to know the boy to girl birth ratio?

[Virgil Showlion]

Virgil Syonid
Message #81 - 04/30/10 05:22 PM

wouldn't you need to know the boy to girl birth ratio?

A solid 'proof' should not assume either of the following:

• a 50/50 birth ratio
  
• a couple will continue to have children as long as they can

The answer to the problem remains the same regardless of whether these assumptions hold or not. I suppose one unstated assumption of the problem is that the 'current' male/female ratio in Misogyna reflects their natural male/female birth ratio. Hence if no special measures are taken to proscribe births, the ratio will just stay the same.

Amazingly, the correct answer still holds even if you assume that gender of siblings is correlated (e.g. if your first child is a boy, your second one is more likely to be a boy, etc.) However, a) gender correlation between siblings in humans is as close to zero as you can realistically get, and b) including correlations complicates the problem beyond what most EE'ers would find interesting, methinks.

---

IHearYou2
Message #82 - 04/30/10 06:09 PM

You needed to tell us that unstated assumption or there could be multiple answers. Theoretically everyone could have three boys and then one girl which would allow thier nefarious plan to succeed. Assuming current female/male ratio into the future then no the plot would not succeed based off of a variation of what Iowa proposed above.

---

Virgil Syonid
Message #83 - 04/30/10 06:23 PM

Theoretically everyone could have three boys and then one girl which would allow thier nefarious plan to succeed.

A probability law called the 'law of large numbers' essentially states that this is impossible if a population is large.

More specifically, the law states that a population size exists at which you can guarantee the plan will fail 'almost certainly'. Consider the plan to have succeeded if the expected ratio of males:females changes (rather than the ratio changing due to random happenstance).

---

Candy Cane Colleenz
Message #84 - 04/30/10 07:42 PM

Shhh - nobody tell China

[Virgil Showlion]

whiskmav
Message #85 - 04/30/10 07:53 PM

Will their diabolical plan have the intended effect?

Off the cuff....no. You might end up with a decrease.



As an aside, I was reading somewhere, when the birth ratio favors males, there is more violence in the world.

---

Virgil Syonid
Message #86 - 04/30/10 08:17 PM

when the birth ratio favors males, there is more violence in the world

It affects the love:war making ratio.

---

PaladinHG
Message #87 - 04/30/10 08:29 PM

They should just make the mothers eat more cereal. I read somewhere that eating a lot of high fiber cereals will increase your likelihood of conciving a boy.

---

Virgil Syonid
Message #88 - 04/30/10 08:43 PM

I read somewhere that eating a lot of high fiber cereals will increase your likelihood of conciving a boy.

The gender of a child is entirely determined by the father's genetic contributions.

I'm going to write you a prescription for staying away from ads put out by cereal factories in India.

[Virgil Showlion]

Virgil Syonid
Message #89 - 05/02/10 06:30 PM

(Solution to Problem 10)

IowaSwirl's worked example in #79 is close to a complete solution.

A probabilistic solution is as follows:

Let the probability of having a boy be b. The probability of having a girl is therefore 1 - b.

For any given couple, let k be the maximum number of children they intend on having (if they were allowed to have any number of children).

Probabilistically, after one child, a couple will have E boys and E[G] girls, where the expectation values are given:

k = 1:
E = b and
E[G] = 1 - b.

In the event that they want more than one child, the sequence extends to:

k = 2:
E = 2*b^2 + b(1 - b) = b(1 + b), and
E[G] = (1 - b) + b(1 - b) = (1 - b)(1 + b)

k = 3:
E = 3*b^3 + 2*b^2*(1 - b) + b(1 - b) = b^3 + b^2 + b = b(1 + b + b^2), and
E[G] = (1 - b) + b(1 - b) + b^2(1 - b) = (1 - b)(1 + b + b^2)

...

One could note by this point that E is given by a simple geometric distribution, and E[G] = k - E is an exponential distribution. A more useful observation is that E/E[G] is fixed for all k: E/E[G] = b/(1 - b).

Hence, regardless of the distribution for k, we have E[E/E[G]] = b/(1 - b).

And of course, the ratio b/(1 - b) is exactly what the Misogynians would get if they implemented no population controls at all. Hence, they've pulled a 'Canadian government special': spent buckets of money on regulations that serve no useful purpose.

---

Virgil Syonid
Message #90 - 05/02/10 06:33 PM

Problem 11 - The Mysterious Slide

Take a standard chessboard and make an off-diagonal cut as shown in the figure below. Then slide the 'top' part of the board down along the cut until the tiles line up on the next row (also as shown).

Take the triangular piece 'left over' in the top left and use it to fill the gap in the bottom right.



But the top (8x8) board in the figure has 64 tiles, and the bottom board (9x7) has 63 tiles. Where did the extra tile go?

---

SlowbutSure
Message #91 - 05/02/10 07:02 PM

Isn't this basically the same as the first problem?

---

Virgil Syonid
Message #92 - 05/02/10 07:36 PM

Isn't this basically the same as the first problem?

It has a similar construction, but the solution to problem 11 has very little in common with the solution to problem 1.

And besides, if two problems are the same, you should have no problem determining the solution to #11.

[Virgil Showlion]

SlowbutSure
Message #93 - 05/02/10 08:27 PM



Oh, I've solved it which is why I know it's similar to Problem 1. I just am too much of a slacker to show my work.







Hint to anybody to hasn't:











The 63 tiles are no longer perfect squares.

---

Virgil Syonid
Message #94 - 05/02/10 08:51 PM

SPOILER: Click to show

Hint to anybody to hasn't:

Actually in part (b) of the figure above, all of the tiles are still perfect squares. Something else is wrong.

---

GeenaMercile
Message #95 - 05/02/10 10:50 PM

But the tiles in part B that were cut and moved down wouldn't be perfect squares. If you look at the way the line cuts the squares in part A, it doesn't cut the squares into two equal halves. Lets look at the whites the top half of the whites are continuously getting smaller as you move down the cut, so if slide the top 1/2 down, when you match the 1st top white 1/2 to the second whites bottom, that top is going to be to big to make a a square, so the area of the 64th square isn't lost, it is just in the none square shapes.

---

GeenaMercile
Message #96 - 05/02/10 10:52 PM

Or is the answer that the 2nd picture doesn't accurately display the steps that are said to be taken place with the first, as the second picture does show squares along the cut line and that there shouldn't be any squares along the cut line.

[Virgil Showlion]

Virgil Syonid
Message #97 - 05/03/10 12:16 AM

SPOILER: Click to show

Geena, I'll post the official answer tomorrow.

You're right to suppose that the second figure doesn't accurately reflect what the board would look like after the proposed slide. If you pull it into MSPaint and slide it manually, you can see what's going on.

---

whiskmav
Message #98 - 05/03/10 03:54 AM

Okay, I am in no condition right now to be looking at this.

---

PaladinHG
Message #99 - 05/03/10 12:46 PM

null

---

Virgil Syonid
Message #100 - 05/04/10 01:58 AM

(Solution to Problem 11)

GeenaMercile gives a text-based solution in #95.

In short: the 'shifted' version of the board in the lower figure isn't the correct figure. The real shift looks like this:


where it's clear that the tiles aren't properly aligned.

Slowbutsure might also have been referring to this in #93.

I consider problems 1 and 11 to be distinct. Problem 1 is based on an optical illusion that looks wrong but is right (the shapes haven't been modified), while problem 11 is an optical illusion that looks right but is wrong.

[Virgil Showlion]

Virgil Syonid
Message #101 - 05/04/10 01:59 AM

Problem 12 - Don't Wine

From smart-kit.com:

You have a bucket of red wine and a bucket of white wine. You take a cup of red wine and pour it into the bucket of white wine. After thoroughly mixing, you then take a cup of this mixture and pour it back into the red wine bucket.

Is there more red wine in the white wine or is there more white wine in the red wine?

---

genuine ga peach
Message #102 - 05/04/10 02:06 AM



[images.google.com/imgres?imgurl=http://www.alastairbathgate.com/wp-content/uploads/2008/04/confessions-logo-small.jpg&imgrefurl=http://www.alastairbathgate.com/&usg=__mFbmtGLgq-esw5dqFyVT7-Tv9XE=&h=438&w=310&sz=26&hl=en&start=81&um=1&itbs=1&tbnid=xCZXsqEbI7cQOM:&tbnh=127&tbnw=90&prev=/images%3Fq%3Dwino%26start%3D63%26um%3D1%26hl%3Den%26sa%3DN%26rlz%3D1T4ACEW_en___US367%26ndsp%3D21%26tbs%3Disch:1] *hic*





i'm sorrryyy....wha wuzs the question?

---

SlowbutSure
Message #103 - 05/04/10 02:23 AM



A little bit for the bucket, a little bit for the mixer, a little bit for the bucket,...

[img src="http://www.superemoticons.com/emoticons/drinking/081-[Drinking]-[SuperEmoticons.com].gif"]



Easy! I have the most wine.

---

weltschmerz
Message #104 - 05/04/10 02:55 AM

More red wine in the white. What you're pouring into the red is already strongly diluted, no?

[Virgil Showlion]

genuine ga peach
Message #105 - 05/04/10 02:59 AM



one thing is certain: by the time the research for this one is complete, the socks will be OFF.

---

Virgil Syonid
Message #106 - 05/04/10 03:08 AM

I'm guessing wine is a big thing around here?

---

genuine ga peach
Message #107 - 05/04/10 03:13 AM

what makes you say that?

---

weltschmerz
Message #108 - 05/04/10 03:14 AM

I'm guessing wine is a big thing around here?

Not necessarily. Some of us don't drink. :::::sniffs haughtily::::







But I AM a pothead.