Today’s tape was soooo boring, I decided to go ahead and give the 10-balls riddle a try. Took me about 30 minutes to solve. Slipping there. I find exercising your mind with probability can help you with trading. Well, we never go and calculate exact probabilities when we trade a stock, but we do have a pseudo-probability-version floating somewhere in the subconscious of our minds before we enter and exit a trade. Kind of hard to explain.
This riddle was given to us by Joe Andra3’s hedge fund manager peer:
Riddle: You have 10 balls and they all look the same. One of the balls weighs either more or less than the other 9, but the other 9 are of equal weight. You have a scale and 3 chances to determine which ball is different from the rest. How so? The manager told me that this is a probability question. (Note: it’s one of those old fashion justice-scales, the ones where you weigh one item against the other)
I solved it using brute force conditional statements. Yay! Hat tip to Dr. Incognito. Every indented “if” is a new weigh, so I’m allowed 3 indentations. Oh yeah, and “!=” is another way to write “not equal to”
Find Q, where Q = odd ball
Given 10 balls: a,b,c,d,e,f,g,h,i,j
Create 4 groups:
group1 = abc
group2 = def
group3 = ghi
group4 = j
// Step 1: weigh abc vs def. Gives us two scenarios…
if (abc = def)
{
//it’s in ghi OR it’s j; weigh 2 balls from group ghi
if (g>i) //it’s either g or i
// weigh g or i with normal ball ‘a’
{ if g = a, then Q = i
if g != a, then Q = g
}
if (g=i) // it’s either h or j
{ if (g=h) then Q = j
}
if (g!=h) then Q = i
}
if (abc < def)
{
Q != j, ghi // rule out those hairy balls
// weigh abc vs normal group ghi
if (abc > ghi) // then Q is heavy, and in group abc
{
if (a > b) then Q = a // heaviest ball
if (a = b) then Q = c
if (a < b) then Q = b
if (abc < ghi) // then Q is light, and in group abc
if (a < b) then Q = a
if (a = b) then Q = c
if (a > b) then Q = b // lightest ball
}
… tricky! The trick was to find out if the odd-ball Q was either heavier or lighter.
Here’s Doc’s answer:
Separate them into 3 groups.
Weigh the first two groups. If they are equal its one in the last group.
Weigh two of the three balls in the last group. If they weigh the same its the last one. If they weigh different amounts, weigh one against the other one in that group.
——————————————————–
If the first two groups weigh different amounts, weigh one group against the last group. At this point you have one more try but you know whether the ball is heavier or lighter. Weigh two balls from the lighter or heavier group and you have your answer.
… have a good weekend. I still haven’t solved to 2358=36 impossible riddle.
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Shorted LULU today.
Maybe I looked at this too quickly, but I don’t think Doc’s answer works. It seems like he is assuming there are nine balls, not ten.
But if ball ten is the correct one, it still only takes 3 weighings…
Another way to solve is to split into two equal groups (5&5) Weigh the 5s to determine the heavier group. Then take the heavier 5 and split into (2,2,1) Weigh the 2s (2 v 2) if they are balanced then it must be the 1. Otherwise split the heavier 2 and weigh (1 v 1)
Tell me about the VIX today……if you like brain teasers try Professor Layton it has some real good ones like the above…they make cool games for Nintendo DS and the like……. I kinda liked the tape today?
My wife is hooked..good brain work out.
Docs solution still works because, this will detect ball 10 “j”…
if (abc = def)
{
//it’s in ghi OR it’s j; weigh 2 balls from group ghi
if (g>i) //it’s either g or i
// weigh g or i with normal ball ‘a’
{ if g = a, then Q = i
if g != a, then Q = g
}
if (g=i) // it’s either h or j
{ if (g=h) then Q = j //HERE IT IS ERIC
}
if (g!=h) then Q = i
}
hey gonzo, actually the tape was not “boring”, i meant choppy. ooops. i get little reads on choppy intraday tapes in medium term neutral zones,
@ckaltner
The odd ball can be either heavier or lighter. Splitting into two groups of 5 doesn’t help determine if the odd ball is heavier or lighter.
Eric is right, you’ll need to weigh it about 5 times to solve it that way.
Gio, you are, of course, right. The Doc’s solution does work. I guess that in my haste I was distracted by the Doc’s reference to “three” groups (of three balls each, plus the tenth ball ‘on the side,’ so to speak.”.
Doesn’t the third group of four balls have to be broken down into smaller groups? If you break the groups down to 2,2,3,3,where you weigh 2 balls versus 2 balls, you woukd be in trouble. If my mind is working correctly, you wouldn’t be able to determine which side of the scale had the abnormal ball without performing four weighings total.
I guess my point was that a solution couldn’t be had by breaking the challenge into only three disticnt groups. There must be four.
A small quible, but I think that’s just how different brains look at the same problem, and my brain is admittedly inferior.
Nevertheless, I really enjoyed this puzzle. Thanks for posting it.
Hi Gio,
OT, but, I know that you track the VIX so I thought you might find this of interest:
“The proposed pilot program will commence on the date upon which the SEC will approve the New Market Model and will continue for six months thereafter ending on April 30, 2009.” If the SLP is now over, should one expect GS’s principal volume trading to drop dramatically, if, as Canaday says, the volume is mostly SLP driven? Also, does that mean volatility in the market is about to spike as there are no entities (well, one entity) providing NBB and NBOs?…
http://zerohedge.blogspot.com/2009/05/observations-on-nyse-program-trading.html