Stand-up Maths
Check out Jane Street’s icosahedron puzzle:
https://www.janestreet.com/IMO2022/
2022 International Mathematical Olympiad!
https://www.imo2022.org/
If you want those d60 and d120 we sell them on Maths Gear or you can go direct to The Dice Lab.
https://mathsgear.co.uk/collections/dice
https://www.mathartfun.com/thedicelab.com/d120.html
Here is my terrible python code.
https://github.com/standupmaths/higher_of_two_rolls
Thanks to Gilad Levy for sending in the question. This is the best collection of maths about the problem we found at the time: https://rpg.stackexchange.com/questions/14690/how-does-rolling-two-d20-and-taking-the-higher-affect-the-average-outcome
But much of that looks at getting a certain value OR GREATER whereas I focused on specific values and the average value.
I’ve also noticed that Chalkdust just beat me with a similar article. Nice to see a different way (induction) to derive the same probability of getting specific value equation! https://chalkdustmagazine.com/features/slaying-the-dragon/
Cheers to my Patreons for buying me all those dice. If you think I still don’t have enough dice, get involved here: https://www.patreon.com/standupmaths
CORRECTIONS
– Yes, on the bar chart axis it goes 2%, 4%, 2%, 4% instead of 2%, 4%, 6%, 8%. First spotted and pointed out by Deadeaded. It was just because I was copying and pasting in photoshop and forgot to edit it. Not because I was making that chart in a bar.
– At 23:07 I have (n=1) in the graphics which should be (n-1). Or maybe I put in two – and you should pick the highest. (Pointed out by Leick Robinson.)
– Marco Davi correctly noticed that the fifth Rhombic Dodecahedral Number is 369, not the 269 you see at 21:46.
– Let me know if you spot any other mistakes!
Filming and editing by Alex Genn-Bash
Dice gluing by Alex Genn-Bash
Putting 1/n in front of everything by 1/nMatt 1/nParker
Music by Howard Carter
Design by Simon Wright and Adam Robinson
But what about disadvantage? A lot of people are asking about rolling m n-sided dice and picking the lowest value.
Nicely it is a completely symmetric situation! For a d20 (or any dice) you can reverse the numbers on the bar chart to go 20 to 1 instead of 1 to 20 and it all works out. Which is to say: you swap the values of x for 21-x. And that feeds through everything else. The average value for disadvantage is (n+1) subtract the average roll for advantage in the identical situation. So for a d20 it's 21 – 13.825 = 7.175
The final conclusion also holds but now the expected average for disadvantage is 1/(m+1) × n + 0.5
Oh, and for those asking about the d60 and d120 dice: we sell them on Maths Gear or you can go direct to The Dice Lab.
https://mathsgear.co.uk/collections/dice
https://www.mathartfun.com/thedicelab.com/d120.html
anyone know what brand of jumbo dice are on the table at the intro? i would like a set of very large dice.
Correction, one is called die, multiple are called dice.
the golden ratio!!!
What happens if you roll N dice?
You can also get that formula of 3x(x-1)+1 by seeing it as the volume of x cube minus of x-1 cube. So: x³-(x-1)³ = x³ – (x³ – 3x² + 3x -1) = 3x² – 3x +1 = 3x(x-1)+1
Love the glued together dice. Can't say I agree with the solution to the sponsors problem being to just assume that you aren't unlucky. We are looking for the minimum, not the assumed minimum. I don't care how many times you walk around randomly. If you can't say for certain then wtf is the point.
Let's not ignore that rolling 1 die averages out at 1/2 of the highest die number, so that pattern is just, baked right in
Hey there! I think I have a correction. We missed something here by using geometry, by using the graph it combines two possibilities in the center line of the graph. Lets say you wanted a 3. There's 6 ways to roll a 3 not 5. (1,3) (2,3) (3,3) (3,3) (3,2) (3,1). On the graph however it would combine the two chances of rolling a 3,3 or a 3,3.
So the equation is just 2x/n^2 meaning you start with a .5% chance of rolling a 1 then go up by .5% each time. Rolling a 2 would be a 1% chance, rolling a 3 would be a 1.5% chance… rolling a 20 would be a 10% chance now instead of the 9.75% proposed which fits because doubling the dice just doubles chance, this is what sparked this as I was confused how .25% was stolen from me when I go to roll a 20. I would love to know if this effects the average value, I don’t think it does but I don’t know summation or calc in general so I can't be sure. Please let me know, I’m dying to be checked.
How about the odds for the system where you roll x d6's, and take the highest one – however for every six after the first you add one to your score.
So, with 5d6:
11111 = 1
12345 = 5
12346 = 6
12366 = 7
12666 = 8
16666 = 9
66666 = 10
I presume you would calculate 'highest value' and 'number of extra sixes' as two seperate functions and add them, or is there a quicker way?
incredible
"Look at what happens if you roll 2 distinguishable D6, so I've now got a blue one and a green one."
Me, who has a color deficiency and can't differentiate between blue and green very well: "Well darn."
I think I've figured out how it works after seeing the bar graph in the intro by thinking of just a d2 (aka a coin). There are only 4 possible rolls when flipping a d2 with advantage: 11, 12, 21, and 22. If younalways take the higher number, that's a 25% chance for a 1 and a 75% chance for a 2. Extrapolate that to a d20 and it looks the same with the decimal moved, 0.25% for a one and 0.75% for a two. It's all in the possible conbinations for any given die! Can't wait to watch the rest of the video to see if i puzzled it out correctly!
Edit: hey look! It's the thing I thought of almost immediately after the intro but with a d6! It illustrates it much better than a d2, but I only wanted to think about 4 sets of possibilities in my head lol.
In rolling up a DND character, often you'll roll 4d6 drop the lowest. It'd be very interesting to see what the average there would be and the maths behind it!
I'm a little confused here. You've got the average for 3 d20s as 15.4875, but three-quarters of 20 is a mere 15? or is it three quarters of the range? Maybe I'm not so hot at stats.
DICE = PLURAL
DIE = SINGULAR
I love everything about this video, EXCEPT that you keep referring to a single die as a dice
I wonder how doing roll n dice and choose m best results affects the math. For example in D&D it's very common to roll 4d6 keep the highest 3.
This video is incredible. I'm so happy I found this. I was actually trying to work out this exact stuff a few months ago. I was able to find the formula for two n-sided dice (using the spreadsheet method, naturally) but got hung up when I got to three dice, and generalizing it to any number of dice with any number of sides seemed beyond my ability. And as far as I could see via Google, there was no indication that anyone else had tried to do it. It never occurred to me to try a physical representation, but I would have gotten a lot further if it had. 😅
I know this is an old video but doesn't x^3-((x-1)^3) give the same results for the centred hexagonal numbers and is a tidier equation
Thank the algorithm gods for delivering this video onto me… I teach high school computer science, and do a lot of simulation work with python and dice. This is a fantastic video… Thanks!