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#1 bushindo

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Posted 20 December 2011 - 07:55 PM

Suppose that there is a thin, circular ring of wire with circumference of 10 meters, and that we have some ants with the following properties,

1) All the ants when placed on the wire will travel at a fixed, constant speed of 1 meter/minute in the direction that they are heading.
2) When any two ants collide on the wire, each ant will instantaneously turn around and travel in the opposite direction at the same speed.

Suppose that we simultaneously place 14 ants at random locations on the ring. The orientation in which each ant is headed (clockwise or counter-clockwise) is determined by a fair, random coin flip. 13 of the ants are colored black, and 1 of the ants is colored red. We let the ants run around the ring as specified by the conditions above. After precisely 10 minutes, what is the exact probability that the red ant will end up at the spot it started in the beginning of the game?

Extra bonus:
Spoiler for Prove that

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#2 rjsghk107

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Posted 20 December 2011 - 10:49 PM

Spoiler for Hint

Edited by rjsghk107, 20 December 2011 - 10:56 PM.

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#3 superprismatic

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Posted 20 December 2011 - 10:50 PM

Spoiler for the exact probability

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#4 exagorazo

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Posted 21 December 2011 - 01:57 AM

Spoiler for smaller cases


Spoiler for hypothesis


Spoiler for If hypothesis is true...

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#5 CaptainEd

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Posted 21 December 2011 - 06:53 AM

I'm with exagorazo--if there were three ants, the R going CW and the two B going CCW, I think the R does not get home. I realize the OP has an even number of ants, but the Superprismatic argument does not seem to depend on the parity of the number of ants. So we wonder (exagorazo and I) whether the argument is really as strong as he suggests.

Edited by CaptainEd, 21 December 2011 - 06:53 AM.

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#6 thoughtfulfellow

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Posted 21 December 2011 - 07:06 AM

Spoiler for Since

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#7 thoughtfulfellow

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Posted 21 December 2011 - 07:12 AM

I'm with exagorazo--if there were three ants, the R going CW and the two B going CCW, I think the R does not get home. I realize the OP has an even number of ants, but the Superprismatic argument does not seem to depend on the parity of the number of ants. So we wonder (exagorazo and I) whether the argument is really as strong as he suggests.

If the ants are evenly spaced, I could not find and scenario of direction of movement or number of ants that would interfere with all ants reaching original positions in 1o minutes. The problem comes to how much variation from evenly spaced is allowed before a shift occurs. In my example above, I made the red ant having a small gap on one sife and large gap on the other but note it makes no differece which of the positions is the red ant for all must shift.
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#8 CaptainEd

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Posted 21 December 2011 - 07:41 AM

What fun! I love the argument that the ants' positions move independent of the actual ant identities.
BUT...
Let's refer to a specific point O as an admittedly arbitrary Origin.
The locations have increasing labels, going clockwise.
let's refer to times with distance labels. That is, if an ant is at location x, facing O, we could say that, after time x, the ant has reached O.
And let "e" be "epsilon", a tiny amount.

Now, arrange the ants as follows:
R is at O - e facing CW. In other words, after e time units, R will be at O.
the Bs are located at O+e, O+3e, ..., O+25e, all facing CCW. We can think of them as 13 blind mice.
GO!
At time e, R and B1 meet at O, and reverse direction. Now R is the "image" of B1. After about 13e, R will be leading the other 12 blind mice CCW, with B13 acting as the image of R, moving CW at location 13e.
Since the ring is 10 meters in circumference, halfway is 5.
At time e + 5, R and B13 will meet at location O+5. R will bounce, and start moving CW again, but will hit B2 at e + 5 + e, and bounce again. R will arrive back at location O+5 at time 5 + 3e, moving CCW. To get back home (O - e), R has to travel more than 5, but he has less than 5 units of time remaining.
So, here is a layout that appears to fail--R does not get back home, someone else gets to R's home.
If this is right, then I think exagorazo's insight comes into play: the pattern of CW and CCW ants may affect whether R acts as his own image or not.
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#9 14.swapnil.14

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Posted 21 December 2011 - 11:43 AM

Spoiler for I think.......

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#10 bushindo

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Posted 21 December 2011 - 06:21 PM

Spoiler for smaller cases


Spoiler for hypothesis


Spoiler for If hypothesis is true...


This is indeed the correct probability. Congrats! (14.swapnil.14 made the same conjecture but made some mistake in the calculations). However, the solution to this puzzle is slightly incomplete (for me) because the probability computation relies on the unproven fact that

Spoiler for unproven fact


There is a neat proof for this fact, and I think the fine denizens of the den would appreciate the challenge of working why the above statement is true. Instead of posting that fact as a new puzzle, I think I'll declare this puzzle to be half-solved, and revise the OP to include this new challenge.
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