Grant Sanderson
Region: US
Tuesday 16 June 2026 18:15:01 GMT
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zwenzwen :
its gonna be something including pi
2026-06-16 18:26:32
529
gautier :
expected value is linear so it's n(n-1)/2 * the probability that 2 chords intersect, and my napkin math says that this is 1/3
2026-06-16 18:40:50
67
Dash :
for 10 chords: 9.167. For 100 chords: 841.67
2026-06-16 22:10:58
5
unbebelievable :
Let's start with a single pair: For any pair of cords, I think it's got to be 50-50 whether they cross, because whether, say, the red cord crosses the yellow one reduces to whether the order of the points around the cricle is RYRY or RRYY, which are otherwise fully symmetric scenarios. So is it just a bunch of 50% cointosses for every pair of cords? that feels too easy.
2026-06-18 01:34:41
1
IBlowSpeed :
Take a fixed point on a circle, and chose a second to make a chord. Let the chord divide the circle into 2 arcs. The probability ,that some other chord intersects it, is equal to the sum of probabilities that the first end of the chord is on the first arc and the other is on the second, and that the first end of the chord is on the second arc and the other is on the first, which both are equal to x(1-x), with x equal to the length of the first arc. So the probability is 2x(1-x) and the average is equal to the integral of 2x(1-x) dx from 0 to 1, which is 1/3. Lets add each chord in separate steps. When adding the nth chord, the expected number of new intersections will be (n-1)/3. So the total number of expected intersections is sum from 1 to n-1 divided by 3, which is equal to n(n-1)/6. For n=10 it's equal to 15
2026-06-22 20:44:26
2
Lexinathan~ :
~165?
2026-06-16 18:34:35
2
Austin Nar :
for any 2 chords the chance is 1/3; imagine picking 4 points at random, and then deciding which two will be the first chord. no matter what you choose as the first point, there is exactly one choice that "crosses the middle" and cuts a line between the other two. hence 1/3 In general for n≥2 chords, when you add the nth chord it has a 1/3 chance of intersecting each of the other n-1 chords. so the total expected number is: sum from i = 2 to n of (i-1)/3; equivalently 1/3 × (sum from i = 1 to n-1 of i); equivalently 1/3 × (n-1)n/2 = n(n-1)/6
2026-06-16 20:08:50
21
polirt23 :
would love to solve this but i aint that good at math
2026-06-16 18:25:54
38
hiokseeya :
How does one even practice maths to build intuition to solve stuff like this. Genuinely curious.
2026-08-10 04:58:03
0
wheelercode :
Given 4 random points on the circle, there are three possible pairings into 2 chords, exactly one produces an intersection. So (10 choose 2) * 1/3 = 15. And (100 choose 2) * 1/3 = 1650. Simulation confirms.
2026-08-01 05:06:19
0
patrickbrauchtfennec :
mega great video bro
2026-07-13 20:55:46
0
Willstray :
15 and 1650?
2026-06-16 18:58:25
0
Dustin :
The surgeon was his mother.
2026-06-17 15:32:07
1
Peregrin Took :
For 10 it’s the average of the minimum 0 and the maximum 45, so the avg. intersections are 22.5. This is the bell curve number.
2026-07-25 19:14:59
0
Rieri_Shrimp :
Educated guess. Start with 2 chord solution then generalize. 2 chords you pick 4 points and leave them unpaired. each point has 2 "close" points and one far, only when paired with the far point do the chords intersect, 1/3 for any 2 chords to intersect. Generalize to N chords their intersection is all Independent so solution in (N choose 2)/3, 10 choose 2/3 =15 and I am inclined to believe that was a hint
2026-06-16 18:45:22
1
SoInnocent. :
prolly 3 and 31
2026-06-29 16:48:08
0
ValidatingUsername :
You Grant figure out the correct math words for non constructed infinitely long "non repeating" numbers where the construction or equation (pi) where there is no set digit for specific digit places, as in the construction has a pattern for the claimed digits...
2026-06-17 02:31:43
0
łaCarįngë :
very interesting. waiting for the answer
2026-06-16 18:27:17
0
drew :
a million
2026-06-17 12:20:42
0
Strega Nona :
i guess 40
2026-06-17 14:31:54
0
S G453 :
Is this Dunbar?
2026-07-06 23:03:55
0
ManuBanana :
I will say, at least 3
2026-08-15 05:03:38
0
legs.plus.brain :
like 16
2026-06-18 04:55:10
0
XtraPath :
Amazing 😍
2026-06-29 19:15:35
0
lesser_d :
I feel like e comes into play here
2026-06-16 20:16:13
0
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