Brain teaser thread

[Japher]

JohnM

Your right, I see my mistake. I counted spammer/flammer twice when I also counted flammer/spammer...doh!

For yours though how about:

Apostrophe
Catastrophe

[Japher]

oops, never mind. I read the question again

[Ignorance]

Headache.

Uhmm.
heartache?
hemotrophe?

[Immortal Wombat]

heterotrophe should be valid

[JohnM2433]

Yes, "headache" was one of the ones I was looking for.

Those other words are certainly good answers, but I was looking for a different one. Hint: it's surprisingly short.

[Ignorance]

Guess "he" fits the requirements too.
Can't come up with anything else.

[SnowFire]

Hmm... maybe I'm getting a wee bit impatient, but being that I just noticed this thread now and Ignorance hasn't provided a new puzzle yet... a classic game theory one.

MLeonard, Ming, and MtG, deciding to settle their differences once and for all, decide to have a truel (as opposed to a duel). MLenard is the worst shot and hits his target only 1/3 of the time, at least at the distance to be used for this truel (the men will stand in an equilateral triangle in an open field, with some witnesses & spectators from the OTF nearby). Ming still has his skills from the wild advertising world sharp, and hits his target 2/3 of the time. MtG has secretly refitted his duelling revolver into a tiny homing-missile launcher with a laser sight and GPS guidance to target, and like the FBI will always get his man. Sportingly, our three antagonists decide to let MLeonard shoot first, then Ming, then MtG, and go around in the circle so forth with any surviving members.

Where would you advise MLeonard to shoot on the first "turn" so to speak? Assuming you want him to win, of course.

[Urban Ranger]

Yes. He should shoot at MtG the first turn.

This gives him a 1/3 chance of hitting, and 2/3 chance of missing. If he hits, Ming will target him with a 2/3 chance of hitting. If he misses, Ming will target MtG because Ming knows MtG will try to take him out first as Matt is such a lousy shot.

So, first turn:

Matt hits MtG, Ming misses Matt : 1/9 (A)
Matt hits MtG, Ming hits Matt: 2/9 (B)
Matt misses MtG, Ming hits MtG: 4/9 (C)
Matt misses MtG, Ming misses MtG: 2/9 (D)

This gives Matt only a 2/9 chance of being eliminated the first round, but a 7/9 chance of surviving.

In cases (A) and (C), Matt gets a second chance of shooting at Ming, improving his odds to 2/3.

In case (D), it's just like everybody passes and back to square 1.

In case (B), oh well, you can't win them all.

[SnowFire]

Urban Ranger: Incorrect. You didn't think it'd be that easy, now did you?

[Urban Ranger]

Well then, it is the other choice.

Quick and dirty way of answering yes/no questions

[Skanky Burns]

Case (D) is wrong, as MtG would fire at (and hit) Ming.
Matt would then have a 1/3 chance of hitting MtG before he too would be taken out in round 2.

[SnowFire]

Well, a bit more specific than not shooting. I said there were spectators around. Do you want the cops called and all of you arrested for participating in a duel? Scare 'em off first by shooting at them so they know not to report it.

Let's look at it this way: Either wasting your shot -or- shooting at MtG and missing make no change whatsoever on the outcome of the duel. If that happens, MtG & Ming duke it out. Only one of them will be left for sure, given MtG's sure-shot. Then MLeonard will functionally have the first shot in a duel, with a 1/3 chance of winning outright (and if Ming won, then a slight chance of still winning if Ming misses and his second or third shot connects).

HOWEVER, if Matt hits MtG with his first shot... then what's Ming going to but start gunning at him! Now Ming has the first shot, and Matt will be gunned down quite likely. This is bad. You want to go first, not second. (you can see that shooting and hitting Ming is even more insane, it signs your death warrant).

If you want to look at it in terms of probabilities...

Miss/Intentionally Miss MtG: 33% chance of duel with MtG (Ming missed and MtG killed him), in which case there is 33% chance of success (hit with your first shot), for a total of 1/9 = .11 of a win; 66% chance of a duel with Ming, in which case there is a (1/3 + (2/3)(1/3)(1/3) + (2/3)(1/3)(2/3)(1/3)(1/3) + ... ) = 43% chance of victory, for a total of .28 of a win. Add the two together and you get 39% chance of victory.

Hit MtG: Well, chance of hitting first after Ming misses, plus the chance of hitting second, etc. We get
(1/3)(1/3) + (1/3)(2/3)(1/3)(1/3) + (1/3)(2/3)(1/3)(2/3)(1/3)(1/3) + .... = 14% chance of victory.

It's obvious that when you miss, you win more often. So why not do it intentionally?

[Ramo]

Bah, why do I always get here when the latest puzzle is already solved?

[Zero-Tau]

Now I'm here, I'll just add that JohnM2433 has indeed given the right answer to my riddle.

[SnowFire]

Okay, UR, you naughty boy. I wouldn't have given away the answer if you hadn't come up with your own bollocks calculations that needed refuting. What's this edit and this magical new "it must be the other one" thing?

Just so y'all know, UR originally said "Oh, I considered the shoot nowhere possibility, it just doesn't work as well!" and had some random probabilities from nowhere showing that somehow MLeonard dies more often from that. Which is why I responded.

[Urban Ranger]

Okay, you caught me

Since you have answered your own question, time for a new one.

[Zero-Tau]

Here's a quick one:

Divide 12 into two equally big parts, of which one equals 7.

[Lemmy]

12 decimal equals 14 octal, 14 decimal divided by 2 is 7.

[Zero-Tau]

Nice try, but that wasn't what I was thinking of. You don't need to change the base to solve it.

[Urban Ranger]

Seems like some kind of word play...

[JohnM2433]

Originally posted by Ignorance
Guess "he" fits the requirements too.
Can't come up with anything else.

That's it! Thanks to Douglas Hofstadter for that problem, BTW: It originally appeared in Godel, Escher, Bach: an Eternal Golden Braid.

Originally posted by Zero-Tau
Now I'm here, I'll just add that JohnM2433 has indeed given the right answer to my riddle.

Thanks. At first I didn't know what you were talking about; I forgot who posted what, so I thought that you were referring to the "dwarf" problem posted by Stefu near the beginning of the thread, which I supplied a more general answer to. Looking back to it, I saw that my answer should have referred to gnomes, not dwarves. Imagine my embarassment.

[Ignorance]

12 is XII in roman numerals.

\/ I I
/\ I I

Split in two

\/ I I
------
/\ I I

VII is 7

[Zero-Tau]

Right, Ignorance. Your turn.

[Ignorance]

It starts snowing before 4 pm on a cold day. At 4 pm snow plow starts plowing.
It travels 2 km in the first hour, and 1 km on second hour, slowed down by the increasing snow.
Plows speed is inversely propotional to the height of snow. Snow falls at constant rate.
When did snow start falling?

[Lemmy]

If the plow travels at half the speed after an hour, then the amount of snow must have doubled.
If the snow falls at a constant rate, then it must've started falling at 3 pm: after 1 hour there is (duh) 1 hour worth of snow, then after 2 hours, there is 2 hours worth of snow, which is twice as high since it is falling constant......

So the answer is 3 pm

[Ignorance]

It's not that simple.
Snow does not come hourly, it falls all the time.
Assuming it started at 3pm, there would be 1.5 hours worth of snow on the first hour to plow on average, and 2.5 on second. Which does not add up.

[Dauphin]

Then 3:30pm.

[Az]

It smells like diff. and Integ. Calculus.

[Ramo]

It starts snowing before 4 pm on a cold day. At 4 pm snow plow starts plowing.
It travels 2 km in the first hour, and 1 km on second hour, slowed down by the increasing snow.
Plows speed is inversely propotional to the height of snow. Snow falls at constant rate.
When did snow start falling?

The distance travelled by the plow in any time is the integral of the plow's velocity over its time.

And the plow speed is inversely proportional to the snow's high, or v ~ 1/h, or v = c₁/h, c₁ belonging to the set of real numbers

And the the rate of increase of the snow height is constant, or h = c₂t, c₂ belonging to the set of real numbers.

Putting all that together,
[integral]dt{t=4hr..5hr}c₁/(c₂t) = 2 km
[integral]dt{t=5hr..6hr}c₁/(c₂t) = 1 km

c₁/c₂ln(5/4) = 2
c₁/c₂ln(6/5) = 1

Solve that (which I can't be arsed to do ).

And once you get the constants, you'd plug those back into the original equations, and get the answer.

[Dauphin]

I'm more interesting in knowing how fast the snowplough can go when it hasn't snowed.