Araver, I guess that means we don't have it right. So I've got a question for you
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#327866 YSan's centrifuge
Posted by CaptainEd on 31 January 2013  04:14 PM
#327827 Colorful foreheads
Posted by CaptainEd on 30 January 2013  08:22 PM
denoting a distribution with
AABBCC: the two colors on A's forehead, followed by those of B, followed by those of C
for example:
rbbbrr means A is wearing rb, B is wearing bb, C is wearing rr.
A(1) didn't see ..bbbb, or would have announced rr so rrbbbb is out
B(2) didn't see bb..bb or would have announced rr, so bbrrbb is out
B(2) didn't see rr..rr or would have announced bb, so rrbbrr is out
C(3) didn't see rrrr, or would have announce bb, so rrrrbb is out
C(3) didn't see bbbb, or would have announced rr, so bbbbrr is out
C(3) didn't see rrbb, or would have announced rb, so rrbbrb is out
C(3) didn't see bbrr, or would have announced rb, so bbrrrb is out
A(4) didn't see ..rrrb, so rbrrrb is out
A(4) didn't see ..rrbb, so rbbbrr is out
A(4) didn't see ..bbrr, so rbbbrr is out
A(4) didn't see ..bbbr, so brbbbr is out
Remaining cases are
rr rb rb
rb rb rr
rr rb bb
rb rb rb
bb rb rr
bb rb rb
rb rb bb
 1
#327813 Staying alive with an eightsided die
Posted by CaptainEd on 30 January 2013  06:35 PM
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 justificationthis is infinity we're talking about...), yielding a probability of (eventually) going bust of 0.414.
 1
#327802 First to Fifteen
Posted by CaptainEd on 30 January 2013  05:06 PM
The only 3chip 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 combinationsI 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...
 1
#327575 East meets West
Posted by CaptainEd on 23 January 2013  06:54 PM
This gives 6 faces, four LFs and two Es.
I believe that you are saying for each item S in {5, 6, 7, 8, 9, 20,11,12,13,14,15,16}
there are distinct x and y items from {LF1, LF2, LF3, LF4, E1, E2}, where X+Y=S, and that all X+Y are in that set.
If each individual die obeys the principle that opposite faces total 7, then if a Long Face has total X, its opposite LF has total 14X.
What LF combinations could total 20?
12,8,6,2this results in a sum of 10, which is not shown as possible
11,9,5,3
10,10,4,4this does not permit enough crosssums to make 12 different sums.
So the LFs must be 11, 9, 5, 3.
What are the Es?
First, which sums are due to LF + LF?
20, 16, 14, 12, 8 (need to add 5, 6, 7, 9, 11, 13, 15)
Then the Es must be
4 (yielding 15, 13, 9, 7) 2 (yielding 13, 11, 7, 5)
The end points are 4 and 2, so the glued faces (7X) are 3 and 5.
What an interesting puzzle! I thought it would require computer work or tedium!
 1
#327260 Hitting 137
Posted by CaptainEd on 14 January 2013  09:32 PM
 1
#326802 unanswered
Posted by CaptainEd on 20 December 2012  06:54 PM
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
 1
#326787 Triangular sticks
Posted by CaptainEd on 19 December 2012  07:01 PM
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 (1p).
Thus we have four sticks, of lengths q, (pq), r, (1pr).
When we discard a stick of length x, the sum "s" is (1x), so the triangle conditions become:
Discard Condition (each remaining stick must satisfy)
q < (1q)/2
pq < (1(pq))/2
r < (1r)/2
1pr < (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."
 1
#326771 Lethal Solution
Posted by CaptainEd on 18 December 2012  07:17 PM
 1
#326745 Lethal Solution
Posted by CaptainEd on 17 December 2012  07:53 PM
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.)
 2
#326550 Triangular sticks
Posted by CaptainEd on 20 November 2012  05:38 PM
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 (p1/2) < q < 1/2, because
if q > 1/2, then q is too long, and
if q < p1/2, then pq is too long.
So the admissible fraction is of length (1/2  (p1/2)) / p.
So we want integral (1/2  (p  1/2))/p from p = 1/2 to 1.
= integral (1p)/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.
 1
#319882 Some Like It Hot
Posted by CaptainEd on 22 June 2012  10:18 PM
Your 13th test distinguishes the last pair, so no 14th test, agreed.
My 13th test distinguishes my only pair, so no 14th test. Unfortunately, my 12th test touches 3, so it takes two more to find which one for sure.
This is why your solution ending in 3 pairs is better than mine, ending in a trio and a pair.
Good job, Ysan. Hontoo ni utsukushii desu!
 1
#315243 Weird Multiplication
Posted by CaptainEd on 27 January 2012  11:45 PM
when we used octal arithmetic,
6x4 = 24(10) = 30(8)
7x4 = 28(10) = 34(8)
and 5x4x7 = 140(10) = 214(8)
 1
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