PROBABILITY AND STATISTICS PLEASE HELP | Mathematics | Page 1

novel tendon Jul 17, 2025, 9:20 PM

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Twenty-five ants are placed randomly on a meter-long rod; the thirteenth ant
from the west end of the rod is our friend, Ant Alice. Each ant is facing east
or west with equal probability. They proceed to march forward (that is, in
whatever direction they are facing) at 1 cm/sec; whenever two ants collide,
they reverse directions. What is the expected number of collisions that Alice herself has?

meager pilotBOT Jul 17, 2025, 9:20 PM

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novel tendon Twenty-five ants are placed randomly on a meter-long rod; the thirteenth ant fro...

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white galleon Jul 17, 2025, 9:50 PM

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novel tendon Twenty-five ants are placed randomly on a meter-long rod; the thirteenth ant fro...

What happens when an ant reaches the end of the rod?

novel tendon Jul 17, 2025, 9:58 PM

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white galleon What happens when an ant reaches the end of the rod?

it falls off

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and doesnt participate again

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in any collision

white galleon Jul 17, 2025, 10:01 PM

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Which is logically equivalent to just having an infinitely long rod.

novel tendon Jul 17, 2025, 10:05 PM

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well yea

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it is , ig

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though i do not see how that helps us whatsoever

white galleon Jul 17, 2025, 10:19 PM

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novel tendon though i do not see how that helps us whatsoever

Filtering irrelevant information out of the question is helpful.

novel tendon Jul 17, 2025, 10:22 PM

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white galleon Filtering irrelevant information out of the question is helpful.

Any idea on how to actually solve the problem though?

white galleon Jul 17, 2025, 10:26 PM

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novel tendon Any idea on how to actually solve the problem though?

Well, yes.

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I mean, how is expectation defined?

novel tendon Jul 17, 2025, 10:28 PM

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simulate the event

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millions of times

white galleon Jul 17, 2025, 10:28 PM

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No.

novel tendon Jul 17, 2025, 10:28 PM

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the average answer

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thats one way to look at it

white galleon Jul 17, 2025, 10:28 PM

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It's really not.

novel tendon Jul 17, 2025, 10:29 PM

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other way is sum X.p(x)

novel tendon Jul 17, 2025, 10:29 PM

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white galleon It's really not.

the intuitive meaning still holds true

novel tendon Jul 17, 2025, 10:29 PM

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novel tendon other way is sum X.p(x)

unfortunately though there really is no nice way to do it this way

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i have tried

white galleon Jul 17, 2025, 10:29 PM

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novel tendon unfortunately though there really is no nice way to do it this way

Isn't there?

novel tendon Jul 17, 2025, 10:29 PM

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nope

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try to find the probability that alice undergoes exactly k collisions

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i am not seeing any way to do this

white galleon Jul 17, 2025, 10:30 PM

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Well, for k = 0, it's trivially 1/2^12.

novel tendon Jul 17, 2025, 10:31 PM

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yes

white galleon Jul 17, 2025, 10:33 PM

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Look, what we can do to try to find a pattern is start with a smaller example.

novel tendon Jul 17, 2025, 10:33 PM

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yea i understand that but i am confident this problem is not really gonna be doable with a strategy like that

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like there was another variation of this problem

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that i was able to do

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what is total expected no of collissions between ants

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the way i did that was the following

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u assume each ant carries a flag

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whenever 2 ants collide

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you can say they just exchange flags

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and the flags sort of just pass thru any ant , like a ghost

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so using this , we can just analyze flags

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there are 25C2 flags

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pairs of flags*

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and for each pair , either both flags point in same direction east or west, or they point towards each other , or away from each other each being equally likely

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so probability for any 1 pair to have a collission is 1/4

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using linearity of expectation the total number of expected collissions comes out to be 25c2*1/4 which is 300/4=75

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i know we have to do something similar to this ghost flag process in this problem too but i am struggling to figure how exactly we do that

white galleon Jul 17, 2025, 10:37 PM

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How do you "know" that?

novel tendon Jul 17, 2025, 10:38 PM

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problem is part of 3 problems each of which utilizes this fact

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if you want , feel free to try and find a pattern for general k , i am confident you wont be able to

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bcuz idts there exists any pattern

white galleon Jul 17, 2025, 10:41 PM

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You do realize there are literally infinitely many possible functions from N to N, right? And yet you're confident that literally none of them count what we want to know here?

novel tendon Jul 17, 2025, 10:41 PM

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no of course there may exist some function

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but it being simple enough that you just notice it from a pattern?

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i dont think so

white galleon Jul 17, 2025, 10:46 PM

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Well, let's think about the other ants in this line.

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What's the expected number of collisions for an ant on the end?

novel tendon Jul 17, 2025, 10:47 PM

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for an ant on the end

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i will have

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with 1/2probability

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0 collisions

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with 1/2 probability 1 collission

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so 1/2 a collision

white galleon Jul 17, 2025, 10:51 PM

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novel tendon with 1/2 probability 1 collission

Show your work there.

novel tendon Jul 17, 2025, 10:51 PM

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well its trivial isnt it

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oh wait

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it is possible that all ants go to the right

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so its actually 1* (1/2-1/2^24) + 1/2^240 + 1/20

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so its 1/2 - 1/2^24

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except for the case where all ants are going away from our ant there will always be 1 collision for this corner ant

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bcuz the moment it collides it will again go off to infinity with no ant being able to catch up due to all of them moving at same speed

white galleon Jul 17, 2025, 10:54 PM

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Right.

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So now, what about the ant second from the end?

novel tendon Jul 17, 2025, 10:56 PM

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with 1/4 probablity it goes to infinity straight away

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and suddenly it becomes very complex

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and i am not really sure how to end up with a closed form expression now

white galleon Jul 17, 2025, 10:58 PM

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Well, I don't see how it's that complex.

novel tendon Jul 17, 2025, 10:58 PM

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it may undergo as many as 3 collisions

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but calculating probability for each does not seem simple

white galleon Jul 17, 2025, 10:58 PM

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The probability of zero collisions is 1/2^2 + 1/2^23.

novel tendon Jul 17, 2025, 10:59 PM

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yea

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for 1 collision ig it wud be 1/4*(1-1/2^23) ig?

white galleon Jul 17, 2025, 11:00 PM

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Hmm. Show the work?

novel tendon Jul 17, 2025, 11:01 PM

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this only happens

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again assume its the left end we are talking about WLOG

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then the ant even more left to it has to go away from it

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and this ant we are looking at has to go towards right

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and from the remaining 23 ants

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at least 1 has to be looking left

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probability for which is 1- 1/2^23

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complement of probability that all go right

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so 1/4* that

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but you do see how this will get more and more complicated with no visible pattern as we go further?

white galleon Jul 17, 2025, 11:03 PM

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I agree that it will get more complicated. I disagree that there will be no visible pattern.

novel tendon Jul 17, 2025, 11:05 PM

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for 2 collisions ig it is again 1/4*(1-1/2^23) , as this time both left most ants have to move towards each other , then after collission we need 1 more collision with one of the ants from the 23 ones , so at least one must move left from those 23 which is same as before

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for 3 collisions we can just look at the remaining probability that is 1- ( 2* (1/4)(1-1/2^23) + 1/2^2+1/2^23))= 1-(3/8 + 3/2^24)

white galleon Jul 17, 2025, 11:07 PM

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Maybe it would be more useful to go back to considering the probability of n collisions of the center ant.

novel tendon Jul 17, 2025, 11:07 PM

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hmm

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for 1 collision it wud be 1/2^12 (1-1/2^12)

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or is it wait

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yea

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it is

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all ants that are on that side facing away from alice must keep going away from her with prob 1/2^12

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and all ants that are on the side that alice is looking at

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we must need at least 1 to move towards alice

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prob of that is 1-1/2^12

white galleon Jul 17, 2025, 11:10 PM

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Right.

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2 collisions?

novel tendon Jul 17, 2025, 11:13 PM

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its much more complicated for 2

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cuz we need to ensure that alice collides with only 1 ant from the 12 she is facing and then these 12 go on to infinity without collissions

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then we multiply this by (1-1/2^12) for 1 collision with 1 of the ants from direction she is not facing

novel tendon Jul 17, 2025, 11:14 PM

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novel tendon cuz we need to ensure that alice collides with only 1 ant from the 12 she is fac...

oh ig this can be done in the way we choose any 1 ant from these 12 to go towards alice rest all look right

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so thats 12C1/2^12

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so 12C1/2^12(1-1/2^12)

white galleon Jul 17, 2025, 11:16 PM

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And of course, 1 = 12C0, right?

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Can you sense the pattern?

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Let's think about 3 collisions now.

novel tendon Jul 17, 2025, 11:18 PM

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for 3 we need 2 collissions in direction she is facing so 12c2/2*12 and 1 collission from direction she isnt facing so thats 12C1/2^12

white galleon Jul 17, 2025, 11:19 PM

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Well, remember.

novel tendon Jul 17, 2025, 11:19 PM

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wait is it gonna be 12c1/2^12

white galleon Jul 17, 2025, 11:19 PM

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Hmm.

novel tendon Jul 17, 2025, 11:19 PM

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yea it is

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we need exactly 1 ant

white galleon Jul 17, 2025, 11:20 PM

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I guess at 3 collisions we do start caring about every ant?

novel tendon Jul 17, 2025, 11:20 PM

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wdym by that

white galleon Jul 17, 2025, 11:21 PM

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Well, because for 0 and 1, we only care about the ants in the direction Alice ends up going, and the one ant she collides with.

novel tendon Jul 17, 2025, 11:21 PM

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yea

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so for k=4 it shud be

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12c2/2^12 * 12c2/2^12

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for k=5 maybe 12c3/2^12 * 12c2/2^12?

white galleon Jul 17, 2025, 11:23 PM

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novel tendon for k=5 maybe 12c3/2^12 * 12c2/2^12?

Or should we double that?

novel tendon Jul 17, 2025, 11:23 PM

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so it becomes sum of (2r)(12Cr)^2/2^24 from r=2 to 12 and (2r-1)12Cr*12Cr-1/2^24 from r=2 to r=12

white galleon Jul 17, 2025, 11:23 PM

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Wait, no.

novel tendon Jul 17, 2025, 11:23 PM

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white galleon Or should we double that?

no

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we are assuming a direction WLOG

white galleon Jul 17, 2025, 11:23 PM

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Yeah, I figured it out.

novel tendon Jul 17, 2025, 11:23 PM

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novel tendon so it becomes sum of (2r)(12Cr)^2/2^24 from r=2 to 12 and (2r-1)12Cr*12Cr-1/2^...

does this seem right

white galleon Jul 17, 2025, 11:23 PM

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No, that's not what confused me.

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I forgot that the turns have to be in order.

novel tendon Jul 17, 2025, 11:23 PM

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yea

novel tendon Jul 17, 2025, 11:24 PM

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novel tendon so it becomes sum of (2r)(12Cr)^2/2^24 from r=2 to 12 and (2r-1)12Cr*12Cr-1/2^...

and to this we add the case for 0 and 1 and 2 collissions

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oh wait we have to find expected lmao

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so we multiply by the number of collissions for each term

white galleon Jul 17, 2025, 11:25 PM

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That would make sense to me.

novel tendon Jul 17, 2025, 11:25 PM

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ive edited it

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and i think this seems right now

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but unfortunately this sum is not something i know how to evaluate

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maybe wolfram alpha will help here

white galleon Jul 17, 2025, 11:28 PM

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I mean, maybe it's telescoping?

novel tendon Jul 17, 2025, 11:28 PM

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uh idts

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ill ask gpt to tell me prompt to put in wolfram

white galleon Jul 17, 2025, 11:29 PM

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Yeah, don't.

novel tendon Jul 17, 2025, 11:29 PM

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why though

white galleon Jul 17, 2025, 11:29 PM

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Don't ever use chatbots for math.

novel tendon Jul 17, 2025, 11:30 PM

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anyways wolfram is giving a much smaller answer

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the actual ans is around 10.7

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so even if we add the terms for 0 1 and 2 to this

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we wont be close to the right answer

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that means we messed up somewhere

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man thats disappointing af

white galleon Jul 17, 2025, 11:32 PM

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Is dividing by 2^24 accurate?

novel tendon Jul 17, 2025, 11:32 PM

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yea cuz we are essentially fixing directions

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for all 24 ants

novel tendon Jul 17, 2025, 11:33 PM

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white galleon Is dividing by 2^24 accurate?

the order of our answer is correct so dividing by 2^24 isnt the issue i think

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can someone please take a look at the calculations we have done and see if we went wrong somewhere??

novel tendon Jul 18, 2025, 10:13 AM

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Lmao I just realized

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@white galleon we have been stupid af xd

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In the case of 4 collisions say

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After all 4 collisions happens we just need all ants in the direction Alice was facing to be now all going away from her but not for the ants that Alice is facing away from

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We had written 12c2*12c2 /2^24

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Rather it would be 12c2/2^12 for ants on right side ie . Direction Alice is facing

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And for direction she isnt facing we just need at least 2 to move right but if it's more than that it still works

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SO it's 1- probability 0 ant goes right -probability 1 ant goes right

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Lmfao we gotta do it this way for all the cases

lost nightBOT Jul 19, 2025, 11:00 AM

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@novel tendon

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lost nightBOT Jul 19, 2025, 6:25 PM

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@white galleon The user still needs help with this help request.

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