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CaptainEd added an answer to a question Random bullets
Rank order of velocities does not always determine outcome, which is why I can't compute the probabilities in a closed form

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CaptainEd added an answer to a question Random bullets
I agree with bubbled about Jasen's statement:
Bubbled, I agree that you have to recognize that bullets disappear, but I disagree that you have to simulate each second; merely calculate collision timestamps

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CaptainEd added an answer to a question Random bullets
Error! I left out a case: corrected cases are:
20000 samples of this yields 0.371. This is close enough to bonanova's statement of 3/8 being spot on, that I'm looking forward EAGERLY to the simple analytical explanation :)

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CaptainEd added an answer to a question Random bullets
I've run 20000 samples, with p(zilch4) of 0.319

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CaptainEd added an answer to a question Random bullets
My Monte Carlo for zilch4 yields 0.357.

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CaptainEd added an answer to a question Random bullets
I'm surprised that the choice of annihilation is so simple as these programs suggest. Perhaps there's a constraint I haven't comprehended yet. Suppose the bullets fired on one second intervals are b0, b1, b2, and b3, And suppose their velocities are v0, v1, v2, v3. Even the 4bullet case seems to have several outcomes, one of which is that b2 catches b1 before b1 catches b0. Nowhere in these code bodies is there a threebullet comparison. In fact, even that comparison isn't quite enough, as it's possible that b3 could catch b2 before b1 catches b0.
Am I missing something in the code? or am i missing something in the problem?

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CaptainEd added an answer to a question Count the Flags
I assume that
( 1 ) as the number of flags is bounded (between 100 and 1000), we can require enough memory to store a record for each found flag, and can declare how long a record is
( 2 ) a memory cell has finite precision, and that the dimensions of the maze could be arbitrarily greater than the largest value of a cell. So, although we could store some information about each found flag, we could NOT store coordinates. In fact we cannot even store coordinates of one spot (after all, we may have walked a googolplex number of steps in a straight line).
( 3 ) we are NOT permitted to leave anything in a square except a flag, and are not permitted to distinguish the orientation of the flag.
Are these all true assumptions about the rules?

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CaptainEd added an answer to a question Partitioning a Triangle into similar pieces
Bonanova showed that right triangles could be partitioned into similar but noncongruent triangles. All I showed is that no other triangles can be partitioned into similar triangles.

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CaptainEd added an answer to a question Base 10 and base 3 equivalents

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CaptainEd added an answer to a question Partitioning a Triangle into similar pieces
Neither. It doesn't prove that the condition can be met for right triangles, only that IF the condition can be met at all, it can only be for right triangles.

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CaptainEd added an answer to a question Partitioning a Triangle into similar pieces
Here's my attempt at a proof that only right triangles are solutions to the OP.
oops, sorry, don't know how to attach the files INSIDE the spoiler...
.

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CaptainEd added an answer to a question Find the radius
I no longer remember whether I ever knew this, but...
So, I've caught up with Rob

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CaptainEd added an answer to a question The mean of four numbers
If we interpret the clause as "...d has to decrease by two to keep the same median of (a,b,c,d)", I get an unequivocal answer (edited from a moment ago)

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CaptainEd added an answer to a question The mean of four numbers
BMAD, I'm also having difficulty interpreting the words "...d has to decrease by two to keep the same median". The only median mentioned so far is the median of a and b. As Rob and Araver have said, the value of d has no effect on the median of a and b. So, should we interpret the fragment above as: "...d has to decrease by two to keep the same median of (a,b,c,d)"?

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CaptainEd added an answer to a question Double up!
Araver, I'm assuming that the shortest path is what counts, and that a stakes assignment with two or three equal numbers counts as zero. I think the coinflipping is window dressing...Of course, Bonanova's opinion is more reliable :)
Here's my answer at 8 steps:

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