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JohnM2433
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Milwaukee, Wisconsin
Jul 2002
time: 21:23
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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.
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SnowFire
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New York City, NY
Jan 1970
time: 00:23
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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.
Last edited by SnowFire on 26-10-2002 at 09:00
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Urban Ranger
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Apolyton Duke of Off-Topic
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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. 
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SnowFire
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New York City, NY
Jan 1970
time: 00:23
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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?
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Zero-Tau
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Elsewhere
Aug 2002
time: 06:23
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Now I'm here, I'll just add that JohnM2433 has indeed given the right answer to my riddle.
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Zero-Tau
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Elsewhere
Aug 2002
time: 06:23
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Here's a quick one:
Divide 12 into two equally big parts, of which one equals 7.
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Zero-Tau
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Elsewhere
Aug 2002
time: 06:23
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Nice try, but that wasn't what I was thinking of. You don't need to change the base to solve it.
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Zero-Tau
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Elsewhere
Aug 2002
time: 06:23
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Right, Ignorance. Your turn.
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Ramo
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Austin, Texas, USA
Oct 1999
time: 23:23
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quote:
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.
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26-10-2002 03:07
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