CaptainEd

a set of samples into two sets of different sizes, each of which is balanced? I don't think that's what you were saying, and I don't feel confident about the physics. But If we superimposed and rotated a 2-sample (12,6) with a 3-sample(1,5,9) to create (12 1 5 6 9), we would naturally get 5 and (via complementation) 7.

 

Let P(n) be the probability of going bust, starting with N dice, where a die is said to go bust if all of its descendants go bust. The descendants of a die are those dice that are put into the gambler's stake after a successful wager, and (recursively) the descendants of those dice.

 

I believe P(2) = P(1)^2, because in order to go bust with both dice, each die needs to go bust, and they are independent.
Similarly, P(3) = P(1)^3.

 

So, what is P(1)?

P(1) = 1/4 (1 + P(1) + P(2) + P(3))
= 1/4 (1 + P(1) + P(1)^2 + P(1)^3)

 

so 3P(1) = 1 + P(1)^2 + P(1)^3

P(1)^3 + P(1)^2 -3P(1) + 1 = 0

The roots of this equation are 1, sqrt(2)-1, and sqrt(2)+1.

In the interval of interest, we have two roots: sqrt(2)-1 and 1.

I'm inclined to rule out 1 (without good justification--this is infinity we're talking about...), yielding a probability of (eventually) going bust of 0.414.

     I'm not happy with my pictures, they do not capture my tictactoe feeling
The only 3-chip combinations that add to 15:

 
The A and B below indicate that, at this stage of the game, both A and B have this possibility available to them. As a player chooses one of the chips, the other player loses that possibility.

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
A A A A A A A A
B B B B B B B B

 
Note that there are four combinations with 5, three combinations with each of 2,4,6,8, and two combinations with each of 1,3,7,9

 

(1) 5, A chooses 5, because it has the most combinations

 

1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
A A A A A A A A
B   B   B   B 

 
(2) B chooses 6, (it knocks out 3 of A's combinations--I believe 4 would have been equivalent)

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
  A   A A A A 
B   B   B   B

 
(3) A chooses 4, because it knocks out half of B's possibilities, while advancing two of A's own possibilities

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
  A   A A A A 
B   B      

 
At this point, A has 54 and B has 6.
B had better not choose 7, appealing though it be, as A is forced to choose 2, which creates a fork, with A threatening both 258 and 249. I believe the choice of 2 has the similar failure.
B's better choice is 8.

 
(4) B chooses 8, threatening 168

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
  A     A A   
B   B      

 
(5) A has no immediate win, so is forced to choose 1, threatening 159

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
  A     A A   
    B      

 
A has 145, B has 68

 
(6) B is forced to choose 9

 
1 1 2 2 2 3 3 4
6 5 6 5 4 5 4 5
8 9 7 8 9 7 8 6
          A   
    B      

 
A has 145, B has 689

 
(7) A chooses 7, threatening 357 and blocking B's last combination (267)
(8) B chooses 3, blocking A's last combination (357)

 
-----
After the first three moves (A5, B6, A4), I think it is clear that both players can force a draw (or can mess up and lose)

 
One could question whether one of the three first moves could be replaced by more powerful ones. But each move seems to have a good rationale...

2 / 7

z=0
y=1
w=1/5
t=10
ni=9
o=1
en=1
u=1
sev=7
fr=4
five=5
six=6
eigh=4/5
elve=6
fife=3/2
hir=3/10
hdrd=100
hsand=100
h*sand=100
mill=1000000/9
b=1000m
three=3
hree=3/10
ir=ree

 

unanswered=
nansered=
ansred=
h*sand*re/h=
100*re/h

The conditions for trianglehood are that

for sticks of lengths a, b, and c, a triangle results if

(a < b+c) AND (b < a+c) AND (c < a+b)

but you can rework that into

(2a < a+b+c) AND (2b < a+b+c) AND (2c < a+b+c)

let s be the sum a+b+c, then

(a < s/2) AND (b < s/2) AND (c < s/2)

 

In the case at hand, we have four sticks:

First, we pick p between 0 and 1.
Then, we pick q between 0 and p.
FInally, we pick r between 0 and (1-p).

Thus we have four sticks, of lengths q, (p-q), r, (1-p-r).

 

When we discard a stick of length x, the sum "s" is (1-x), so the triangle conditions become:

Discard       Condition (each remaining stick must satisfy)
   q          < (1-q)/2
  p-q         < (1-(p-q))/2
   r          < (1-r)/2
 1-p-r        < (p+r)/2

As Bonanova said, I don't get the "Aha!" that causes this to look like the case of "Break once, break again."

OK, I see, all these solutions (bodhri's solution, the pipet solution, the lever solution, and this solution) rely on the same principle at the highest level: construct one reading that adds together different fractions of the different glasses, so that the observed fraction of X identifies the glass. Makes sense to me...

on the laboratory bench next to the digital scale, place a knife edge as a fulcrum, and balance a plank of length 2 meters on the knife edge, with one end touching the digital scale.
At this point, the plank is adding exactly no weight to the scale.

Now,
* place G1 on the plank directly over the scale,
* place G2 on the plank at a distance of 1/2 from the fulcrum
* place G3 on the plank at a distance of 1/3 from the fulcrum
* place G4 on the plank at a distance of 1/4 from the fulcrum

* read the scale

* Without the poison, the scale would read G1 + G2*1/2 + G3*1/3 + G4*1/4 will be (A+A/2+A/3+A/4) =

(12A + 6A + 4A + 3A)/12 = (25A/12)

However, the poison of weight X will either appear as an addition of X (if in G1), X/2 (if in G2), X/3 (if in G3) or X/4 (if in G4).
(Firefox doesn't seem to play well with this spoiler editor.)

I think I've calculated the second case properly, though it would be gilding the lily to call my calculus proficiency merely "rusty".
Rather than randomly generating a place to break the stick, I'll integrate over all break points.
However, we don't want to break the short stick, as it will never lead to a triangle.
So, let p be the length of the long half of the stick. That is, 1/2 < p < 1.
Then break p at q.

The possible range of q is 0 < q < p, but the admissible range of q is (p-1/2) < q < 1/2, because
if q > 1/2, then q is too long, and
if q < p-1/2, then p-q is too long.
So the admissible fraction is of length (1/2 - (p-1/2)) / p.

So we want integral (1/2 - (p - 1/2))/p from p = 1/2 to 1.
= integral (1-p)/p from p = 1/2 to 1.
= integral (1/p) - integral (1), from p=1/2 to 1
= [ integral (1/p) from p=1/2 to 1 ] - 1/2
= ln(1) - ln(1/2) - 1/2
= 0 +.69315 -.5 = .19315

I sort of expected it to be twice this, so I may have either missed a factor of two, or done something totally irrelevant. Please let me know.

But .19315 fits in with my crude probabilistic simulations, and is close (but not close enough, I fear) to brifri238's results.

when we used octal arithmetic,
6x4 = 24(10) = 30(8)
7x4 = 28(10) = 34(8)
and 5x4x7 = 140(10) = 214(8)