Grant Sanderson
Region: US
Friday 16 January 2026 18:15:41 GMT
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Salim.slm163 :
50%. The last number is either six or not❤️🩹
2026-01-16 20:26:44
18300
machh :
probably involves π for some reason
2026-01-16 19:07:58
28884
JLW 1808 :
can confirm based on an extremely large sample size it is 1/11 for all numbers other than 12 of course
2026-01-16 23:35:36
4023
Rocket in the rough :
1/11 just because i wrote a Python script with 1 milion simulations and this is the result
2026-01-16 20:02:17
5649
Fighterjet991 :
How would you even account for the possibility that it gets stuck bouncing off of 2 numbers for the rest of eternity?
2026-01-16 19:37:18
381
Cheddar Goblin :
1/12
2026-01-16 18:42:48
3509
razeth :
Because it's completely random my guess is 1/11
2026-01-16 18:58:33
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Eddi :
50/50 either it happens or it dont
2026-01-17 16:56:32
237
Wut :
Why would a ladybug land on a clock?
2026-01-16 22:53:56
840
Austin :
I monte Carlo’d it with a million simulations and got a range of 9.06-9.14% for 1-11 so it seems like the answer is 1/11 for any given number.
2026-01-17 00:17:18
97
Bendix :
solution: its 1/11 but its not easy at all. We actually want to avoid to calculate the exact probability right away. we first make some observations: before the ladybug reaches the 6 it must past either the 5 or the 7. with a 100% certainty it does. We want to start at the point where the ladybug is for its first time on the position 5 or 7. WLOG we assume that the ladybug is at position 5. For 6 beeing the last new number the ladybug visits it must go all the was around to position 7. since by our assumption its hasn't been there. we call the probability for this p. [1]
2026-01-18 15:27:18
21
Peter Piper :
I can't do the maths but I can write computer simulations. But I'm in bed doom scrolling, not at my computer.
2026-01-16 22:11:08
95
Mohamed :
the answer is 1/11, all the numbers are equal in chance to be the last number and we have 12 numbers in a clock but the number 12 is eliminated since we start on it, and so 1/11
2026-01-17 13:27:40
84
cello :
50/50 it either happens or nor
2026-01-17 07:38:03
12
Thavud :
1/65536
2026-01-17 13:07:48
8
Medd🇲🇦 :
Every move is independent of the one before it, classic random walk. The answer is much more straightforward than one would assume. Because of that independence and the equal probabilities of going either C or CC, reaching any number last would simply be logically equivalent to picking a random number from the set of naturals 1 through 12, 12 excluded; so 1/11!
2026-02-03 21:18:30
30
Cool Ranch Luke :
Well the last number is definitely not going to be a twelve, so chances are 1/11?
2026-01-16 23:56:34
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ally :) :
I know a Markov chain when I see one 🤨
2026-01-16 19:18:44
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Pedro :
Yeah dawg you’ll have to figure this one out yourself
2026-01-17 00:44:33
30
H :
I figured it was 1/11 and then looked at the comments and was right. This doesn’t seem very hard
2026-01-17 06:49:21
5
Just Sayori🎀✨ :
50% either the last number is 6 or it’s not
2026-02-21 08:10:06
6
deu5 5um :
50/50, either it happens or it doesn't
2026-01-17 23:53:54
5
Jasiek Krężel :
1/11th. Which ever number you choose (apart from 12), you can assume that at some point you hit one of its neighbour and then you have to “climb” all the around and hit it from the other side without “falling” back from the side you started on. It doesn’t matter which number you choose so the probability that you do that for any of the numbers is equal, which means it must be 1/11th for all numbers. To be more precise, you can draw a Markov Chain with absorbing states and realise they all have the same properties no matter which number you choose (apart from 12)
2026-01-16 21:36:51
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M.s :
Why am I seeing Markov chains on my fyp????
2026-01-17 00:43:41
160
aviiii_aa :
how have stochastic processes followed me to my fyp 🥲
2026-01-16 19:44:44
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