Extra footage from this interview: kzfaq.info/get/bejne/p79xis16u7-biqs.html

Accidental Derangement would make a great name for a band...

The correct title for this video is “Parker permutations”

I definitely find these videos interesting, but the main reason I watch them is to see the mathematicians get really, really excited. It's so fun.

I always knew he was a tiny person

Need much more 'surprise e' in my life

All the other professors (and hosts) are really really great (seriously they're awesome) but Dr. James Grime is my absolute favorite.

e is like a Spanish Inquisition

We did this in class today. If only this video released yesterday...

I pondered this during my years of watching auto racing, it seemed that every race, at least one of the cars finished in the same position it started in. Never worked out the probability though, but it didn't come as a surprise that it's genuinely over 50%.

Only watching because of James Grime

The man, the myth, the legend is back. Best videos when you're in them

James Grimes + Mathematics has just taken the No 1 spot in my all time list of best things I have ever watched.

Thank you very much I finally understand fixpoint, I was scared that my time was wasted because I didn't saw the term fixpoint in the title or description of the video. It wasn't clickbait after all thank you haha. :)

My guess for the thing in the beginning: 1: Rephrase question - if you can choose freely between all the cards not used, then put it down on a random position (where there hasn't yet been a chance), what chance is it that you'll get one in the right position 2: For any N cards you have a 1/N chance of getting it correct the first draw, other wise go to step 3 with x being 1 3: If it's not correct, choose the card that should have been in that slot, and put it in a random position, this gives you a 1/(N-x) chance of being back at step 1 with the new N being (x+1) less than the original 4: If you didn't go back to step 1, repeat 3 with x one higher than last time, until you run out of cards. We can split this into two states, having n cards and a chance to win, or having n cards and a chance to close a loop: Win: 1: 1/1 chance to win 2: 1/2 chance to win 3: 1/3 chance to win, 2/3 chance to go to Close:2 - 1/3 + 2/3 * 1/2 = 2/3 4: 1/4 chance to win, 3/4 chance to go to Close:3 - 1/4 + 3/4 * 1/2 = 5/8 5: 1/5 chance to win, 4/5 chance to go to Close:4 - 1/5 + 4/5 * 13/24 = 19/30 (....) Close: 2: 1/2 chance to go to Win:1, 1/2 chance to lose 3: 1/3 chance to go to Win:2, 2/3 chance to go to Close:2 - 1/3 * 1/2 + 2/3 * 1/2 = 1/2 4: 1/4 chance to go to Win:3, 3/4 chance to go to Close:3 - 1/4 * 2/3 + 3/4 * 1/2 = 13/24 5: 1/5 chance to go to Win:4, 4/5 chance to go to Close:4 - 1/5 * 5/8 + 4/5 * 13/24 = 67/120 (...) Since both Close:2 and Win:2 have a 50% chance of winning, and all streaks of loss will eventually go to one of those two, we know for certain that there's at least 50% chance of winning with any amount of cards larger than 0. However, continuing like this is boring, and prone to error (I probably did something wrong in all that, actually), so let's look for abstractions. We could consider a game where you start with a stick, and a number N. There's a 1/N chance of you picking up a stick, and an (N-1)/N chance of losing a stick, if you have one. If you manage to get two sticks, you win. This means that you have to get sticks twice in a row, except on the start, meaning we can abstract having 0 sticks to: 1/(N*(N-1)) chance of winning => 1/(N^2 -N) (N-1)/N chance of subtracting one from N => 1-1/N (N-2)/(N*(N-1)) chance of subtracting two from N => 1/(N-1) - 2/(N^2 - N) Then we just start from 10 and add downwards to get the chance of having a specific amount on N: 10: 1/1 9: 9/10 (from 10) 8: 4/45 (from 10) + 9/10 * 8/9 (from 9) = 8/9 7: 9/10 * 7/72 + 8/9 * 7/8 = 623/720 6: 8/9 * 3/28 + 623/720 * 6/7 = 703/840 5: 623/720 * 5/42 + 703/840 * 5/6 = 4841/6048 4: 703/840 * 2/15 + 4841/6048 * 4/5 = 28423/37800 3: 4841/6048 * 3/20 + 28423/37800 * 3/4 = 137897/201600 2: 28423/37800 * 1/6 + 137897/201600 * 2/3 = 527383/907200 1: 137897/201600 * 1/6 + 527383/907200 * 1/2 = 1468457/3628800 Since 1 is the only loss-state, we know that there's a 1-1468457/3628800 = 2160343/3628800 ≃ 0.595332616843 chance of winning, unless I did something wrong. I have no idea what a generic method would be though. EDIT: Okay, turns out I was wrong. EDIT 2: NVM, I was right, I just forgot that I was supposed to start with 1 stick, not 0: 2160343/3628800 * 10/11 + 1/11 = 2523223/3991680 = 0.6321205607664 (though that's for 11 cards)

Yay! James Grime!!!

Best explanation so far on derangements. Thank you for making such a wonderful video.

What do you mean old-fashioned? Don't you check your hat at the theatre? What do you do with it? How do the people behind you see the stage?

Just watched an ad on the derangements video! Glad we got that sorted. Tims rejoyce.

"We're gonna get deranged, let's do that." Okay?

You guys never fail to amaze me. Keep it up!

Fascinating video! Thank you! Please do more probability videos.

A new video!!! and it's from my boy grime!!!! IM SO HAPPY

I loved this explanation for Derangements by Jamie!

YAY new video from Dr. James Grime

I absolutely love these videos thanks

I'm a simple man, I see a video with Dr. Grimes, I drop everything to watch it.

That was soooo amazing!!!!!

Dr Grimes is BACK ! Yesssssss

actually finished discrete mathematics this semester and we solved this in one of the lectures. suddenly i feel so competent, knowing what you're talking about

Awesome explanation Thank you Dr. James

Love the videos with Dr Grime.

So I watched 8 minutes for seeing how he calculates the number of derangements, because that is the hard part in this, but then he explains everything else than that and just throws !n=n!/e at us with no explanation

I JUST WATCH THESE VIDEOS TO LOOK SMART

I am reminded of "manotazo", where players take turns laying down cards and have to slap down the pile when the card layer down coincides with name of the card. The names are spoken in order during turns from ace to King and then ace to King again. Last hand on top of the pile takes the pile. You notice derangements quite often during play

I thought he would mention this. When you divide n! by e, you get a number between two integers. The closer one is the number of derangements. The other one is the number of permutations with exactly one fixed point. For example 4!/e is about 8.83. 9 derangements and 8 permutations with one fixed point. 5!/e gives you 44.145. 44 derangements and 45 permutations with one fixed point. For n even, the number of derangements is larger. For n odd, it is smaller.

James Grime 😍😍😍

You explained so well

I might be wrong, but if instead of a discrete deck of cards you had a "continuous" deck with infinite cards (each having a number between 0 and 1, for example) then you'd always get a fixed point :D Edit: As Donald pointed out, that's wrong since the 'shuffle' function isn't continuous.

Simple and clever. I like this concept.

When the number of cards is big, the probability of getting k follows Poisson distribution with lambda equals to 1, which yields 1/e for 0 or 1 and 1/(2e) for two.

6:31 #ParkerSquare

Can’t unsee tiny James Grime

Please also make a video about sterling function in permutation and combination. Or various ways we distribute objects to people. It would be a great help. Thanks

He's a tiny smart hell of a mathematician and I love it ❤

2:04 Right before he said "What of it's 2 cards? I had it in my mind... Cool!

you are amazing man ...

I’m imagining James with those cards now not at a 6’4” guy but as a tiny leprechaun holding a normal deck.

( 1- (1/n) ) ^ n = 1/e, as n --> infinity

1:40 Thx for the example. Was so hard to figure out :D

my favorite professor

The probability of getting a derrangement with n cards is (n-1/n)^(n-1) which approaches to 1/e. The probability for n=10 is actually 38. 7%. NOT 36. 8% which is 1/e

"It looks like you've got the world's smallest hands." Perhaps James could be President of the United States.

This video should have also covered arrangements, which show the number of ways you can arrange n distinct objects, and are essentially the inverse of derangements. The way to work out the number of arrangements is the same as derangements except instead of alternating between adding and subtracting, you just add. For example, for five objects: Number of derangements = 5! - 5(4!) + 10(3!) - 10(2!) + 5(1!) - 0! = 120 - 120 + 60 - 20 + 5 - 1 = 44 Number of arrangements = 5! + 5(4!) + 10(3!) + 10(2!) + 5(1!) + 0! = 120 + 120 + 60 + 20 + 5 + 1 = 326 The interesting thing about this is that the ratio between number of permutations (120) and arrangements (326) is the same as the ratio between the number of derangements and permutations. They both converge to e (~2.718).

DR. JAMES!!!!!! YYYUUUUUSSSSSS!!!!!

I liked the video before fully watching it.

That's fantastic! ❤️

Excellent!

Interesting how both transcendental numbers can be expressed as a ratio of two quantities, tau : Circumference/radius e : n!/!n, n->infinity

I could honestly listen to James Grime talk all day long about math 😍😍😍🤓🤓🤓

A great video! It's a nice throwback to Dr. Grime's video on the topic of e, and, low and behold, there is some actual maths! Cannot wait for the Numberphile2 part, is !n = n! - sum(k=1 -> n)[nCk * !(n-k)]? Curious to see the link to e.

I'm a simple man. I hear or see James Grime on a Numberphile video, I hit "like"

I was just wondering when the next James Grimes video would happen, and then this happened.

I've been looking at what happens when you sum mutual digit derangements. For example 247, 742, 274 are derangements of each other. Any other permutation, such as 724 has at least one element, 7, in the same position. The sum always seems to be a multiple of 37. I don't think that's got anything to do with 1/e, but with what you get when you factorize repunits of varying size. 111 = 3*37, and 1111 = 11*101, so the sum of say 2473 and its derangements 3247, 7324, 4732 is 17776 = 11*101*16. For 5 digit derangement sets the sum is a multiple of 41 and 271.

Singingbanana!

great video

Dr Grime is the best.

The chance of getting at least one right is approaching 1/e for n going to infinity. The formula for calculating the chance for any number n can be derived from first principles..

1:16 "37.....Oh that number again" Veritasium sound in the background

I see what you did there! 1:46 #ParkerSquare

thank you

it started off so random but saying real life applications, even if not very popular situations, is still very interesting.

The probability that someone may get their letter may be a 63% chance, but the probability of the secretary being fired is damn near 100%

the cool feeling you get when you saw e coming right after seeing the percentages.

I see James, I like immediately!!

There's still coat checks in various places. Only time I've encountered one was at Beerfest, but I'm sure they'd have them for other conventions.

I love when the camera guy calls out general statements like at 2:04

Derangement judgment = " We thought you lost, you're actually a genius!"

finding out how likely it is for m out of n match up should be rather straightforward, as long as n-m>3. After all, if you have at least m-1 matching up, then the chance for m matching up should just be the same problem as 1 matching up out of the n-m+1 subset, so you can just concatenate to (1-1/e)^m. if you want to exclude any more after m to match up, you jsut multiply by 1/e again

e is that friend you don't like that always shows up when you're going out with your other friends.

Wow! James has a huge deck.

Just awesome explanation 🫡

This is awesome! Thanks to your video. The next type which I will pay my bills I will try this trick! Maybe I can turn the task easy this way! :/

He is just lovely.

I like this guy's 24/7 smile

Huh? You use the same print on the background as the 'square number glitch-y' in the Perfect Square video! _I love the little details by the way_

James is still my favorite

If the formula for the probability of a derangement is round(n! / e) / n!, then the formula still applies at n=2, since round(2/e)/2 = 1/2. since 2/e = 0.735758882 and at n=1, round(1/e)/1 = 0, you can't not get a derangement with 1 card, so the formula works for any number in the natural numbers. Also as n increases towards infinity the results approaches 1/e, despite never being exactly 1/e.

As soon as he said 37% I knew e was coming!

When he was saying that exactly numbers of matches were complicated to work out, you should have asked him the probability that you get exactly 9 matching.

Is this the problem with Secret Santa when someone gets their own name?

The n! over e result is pretty interesting, as math doesn't usually "round" in combinatorics...

I have used a whole deck of cards to play this game with rules where you lose if you call out the number of the next card (numbers 1...13 are repeated four times). Took me probably hundreds of games to finally win the game.

I felt so smart right until you said, "Divide by e," and then I felt super excited and confused. Surprise "e" indeed! I wish I could have seen that coming.

0:17 Aaaahahah! He was making the same joke as I was making it! Wonderful!

The moment he said "37%", I immediately thought "1/e", because that's how these things work.

Wow, this video is full of 'surprise-e-ing' information.

that cool. i suggest you a make on Bernoulli numbers

The number 37% made me think in conjunction with the structure of the game about the secretary choosing problem...

I thought this video would be about James Grime's wilder days!