How to roll an imaginary 6-sided die in your head

A few minutes ago I got a copy of Mindhacker: 60 Tips, Tricks, and Games to Take Your Mind to the Next Level, a new book (released today), by Ron Hale-Evans and Marty Hale-Evans. I started flipping through it and it looks fun! I'll review it soon, but here's one cool thing from it I came across:

Hack 36: Role the Mental Dice
Quick! You need a way to generate random numbers using just your brain, because you're playing in the pool, driving, falling asleep, or doing something else that makes it difficult to roll physical dice. Maybe you're playing games (board, role-playing, or purely mental) or breaking out of a rut by making decisions randomly. Don't panic – you can use simple math on random words from your surroundings to quickly create random numbers, just about anytime and anywhere.

For each of these procedures, you'll need to be able to come up with a short random word, either spontaneously or from the environment. To get one, check nearby signs or reading material, ask a friend or passerby, or do anything similar that seems good to you. You'll also need to know or work out the number value of each letter, corresponding to its place in the alphabet: A = 1, B = 2, C = 3, and so on through Z equals 26.

To emulate a 6 sided die, follow this procedure:

1. find a short word, such as “cat.”

2. Sum the numerical values of the letters in the word: C = 3, A = 1, T = 20, and then 3+1+20 = 24.

3. Calculate the number modulo 09 (see the section “How It Works” for the procedure).

If the remainder is a 0, 7, or 8, which will happen one third of the time, discard it and try a new word. Otherwise, the remainder will be a pseudo random number from 1 to 6. Congratulations! This is your 6 sided die roll. (The results for the word “cat” is 6.)

Mindhacker: 60 Tips, Tricks, and Games to Take Your Mind to the Next Level

Mark Frauenfelder is the founder of Boing Boing and the editor-in-chief of MAKE and Cool Tools. Twitter: @frauenfelder. Come and hear Mark speak at the ALA conference in Chicago on July 1.

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http://www.facebook.com/cStyle Mathew Lehmann

Why don’t you just do mod 6 and then add 1 to the result?
- Peter James
  
  Yeah, that was pretty much my first thought as well. It’s not like mentally doing mod 9 is much different than doing mod 6. That seems pretty strange. It’s like “instead of a coin flip, roll a 6 sided die. If it’s 1 it’s heads, 2 it’s tails, 3-6 re-roll” instead of just doing even/odds…
- http://www.facebook.com/people/Jim-Smith/714041653 Jim Smith
  
  Not random enough I guess
http://www.facebook.com/frank.farmer Frank Farmer

“modulo 09″?

Does not parse.
Mark Alawo

hmmm 8 = 1? right..
BarBarSeven

C = 3, 8 = 1, T = 20

8 = 1? Huzzah, wuzzah?
- rrh
  
  Well, how do you spell c8t?
http://twitter.com/cjporkchop cjporkchop

And for those of us who have forgotten advanced math skillz because we haven’t used any since college, and therefore don’t even remember what “mod” means…

1) Ask a random person what month they were born in.
2) January =1, Feb=2, March=3, … , July=7-6-=1, August=8-6=2, Sept=9-6=3, etc.

Not completely random, but close enough, amiright?
- http://www.facebook.com/cStyle Mathew Lehmann
  
  I have an even better idea, just look at the second hand of a watch!
Geoff Peck

“Alright, we need a random number, 1-6, hang on a sec…” 45 seconds of mental math later… “Why did I need a random number again?”

I get that just choosing a number isn’t very random because our brains like patterns and such, but how is “choose a word, use tedious algorithm that always produces the same result for a given word,” really that different. Sure instead of 6 options you have thousands, but it’s still subject to the patterns inherent in our brains. Soon enough you’ll learn to choose “cat” when you want a 6, consciously or not.

The trick needs to have an independent variable that changes the output… like include the time or something.
Josh Burns

8 has always been equal to 1. ALSO THERE ARE *FIVE* LIGHTS.
http://www.facebook.com/sylvar Ben Ostrowsky

It’s not bad; it’s only off by roughly 1%. Doing this to /usr/share/dict/words on my iMac gave me totals of 26230, 26157, 26136, 25936, 26107, 25998 for 1-6 respectively. Of course, once you remember a few words and how they come out… %)
endymion

Seriously… Why is it mod 9 instead of mod 6?

Is this mentioned in the book? If not, this book will be far more annoying to me than enlightening. :)
- Camilo Polymeris
  
  Randomness, perhaps?
  I simulated the frequency of the numbers you get using modulo 6, and it turns out to be:
  0: 12%
  1: 21%
  2: 23%
  3: 20%
  4: 7%
  5: 16%
  
  https://docs.google.com/spreadsheet/ccc?key=0AltDNAydfoc5dF8yU1lCMnU3NVVDNWpTWUFIVjVBSEE&hl=en_US
  
  This is using data from wikipedia, which (I think) includes all english words, not just short ones.
  - Camilo Polymeris
    
    Nah.
    
    Did it for modulo 9, and it is only slightly better:
    
    1: 14.647%
    2: 11.32%
    3: 9.565%
    4: 7.637%
    5: 21.811%
    6: 9.885%
    
    It is probably what @boingboing-5f94b59347f7087ae8bced48a87eca60:disqus said.
    - IceCream
      
      Dude, you cannot do this using letter frequencies! You need to use words, from a dictionary, like Ben Ostrowsky did above.
      
      And like knappa said, it is not mod 6 because there is no simple way to calculate the remainder when you divide by 6. Doing it that way would no longer involve “simple math” you can do easily in your head.
      - Camilo Polymeris
        
        Oh, I see, you are right. You have to sum before taking the modulo. And I agree that probably @boingboing-5f94b59347f7087ae8bced48a87eca60:disqus ‘s argument is correct — mod 9 is easier than mod 6. But why not use mod 10, then?
      - IceCream
        
        I think your suggestion about mod 10 is really good. I can’t see any reason why it wouldn’t also provide a uniform distribution, and it definitely wins hands-down for ease of calculation! The only drawback is discarding 4/10 tries instead of 3/10, but that’s trifling.
        
        Is there something we’re missing? If not, you should drop MindHacks a line…
herrnichte

So this is suggesting that Gematria is not the hand of GOD speaking to us, but rather it’s some sort of randomizing proceedure?? ooo, i’m telling Madonna!

by-the-bye, i think where-ever it says “8 = 1″ up there they meant it to be “a = 1″ (’8′ and ‘a’ aren’t very close on my keyboard, but i suppose keyboards vary)
http://twitter.com/_BluShine Mark

I agree with Geoff, you’ll eventually choose words that you’ve chosen before, even if only subconsciously. And asking for a random word from a stranger only works if you’re walking down the street or something.

A better algorithm could use a random word from a friend (whom you don’t explain the algorithm to), and the current time/date. Asking the friend prevents you from subconsciously choosing specific words. Taking into account the date/time means that you only have to worry about them saying “cat” twice in the same day/minute, but if you use the algorithm a few times daily it’s fine if they say “cat” every day.
KanedaJones

if it is driving you are distracted by while playing DnD or whatevehs might I suggest picking the first number under six you see on a street sign??

and ya I dunno why people gotta act confused when it sez we are using the word CAT and the things around it are C= and T=

MAI GAWD 8=1 WHAT COULD IT MEAN?
knappa

I think the reason they want to add mod 9 is that 10 mod 9 is 1. This means that a number mod 9 is equal to the sum of its digits. It makes arithmetic very easy. Watch: 26+123=2+6+1+2+3 mod 9 so if we do these additions we get 2+6+1+2+3 = 8+1+2+3 = 9+2+3 = 11+2+3 = 1+1+2+3 = 2+2+3 = 4+3 = 7 mod 9.

Otherwise, I would just do it mod 6 as well.
- Camilo Polymeris
  
  OTOH, modulo 10 would be even easier!
  - knappa
    
    Yeah, that’s true. On the other hand, 8=1 so maybe we should be working mod 7. :)
  - Camilo Polymeris
    
    Also, how do you generate D8, D9, D12, D15, etc? Just discarding other values or is there some other way?
    - clockworksaint
      
      > Also, how do you generate D8, D9, D12, D15, etc? Just discarding other values or is there some other way?
      
      From a D6 and a D20 you have prime factors of 2, 3 and 5, so without discarding values you can simulate a die with any number of sides that’s a product of those numbers:
      
      D8 = (D6 mod 2) * 5 + (D20 mod 4) + 1
      D9 = (D6 mod 3) * 3 + (D6 mod 3) + 1 # You need to roll the D6 twice
      D12 = (D20 mod 2) * 6 + D6 + 1
      D15 = (D6 mod 3) * 5 + (D20 mod 5) + 1
      D18 = (D6 mod 3) * 6 + D6 + 1 # You need to roll the D6 twice
- Ron Hale-Evans
  
  Yes! It turns out mod 11 is pretty easy too, so it’s easy to emulate a d10. This is in the book.
Kapuhi

I’ll bet there’s a free app that does this for you.
http://www.facebook.com/john.olmstead John Olmstead

Umm…
http://pulse.yahoo.com/_O73GS2LP4ARSDL3DDRC4OZXMQI Keith

A buck says Mark dictated this into some magical dictation software. (8 instead of A)

Also, I bet prisoners know all sorts of stuff like this… at least the lifers, the thinking lifers.
- Daniele A. Gewurz
  
  And the same (dictation software), I believe, holds for “role” in the place of “roll”.
moonglum

Or you could do what I do and *always* carry both a smart phone and a d6 + d20 in my pocket (and a friggin towl on my shoulder– yes I am that hard core). With d6 + d20 you can generate d2, d3, d4, d5, d6 (duh), d8, d9, d10, d12, d15, d16, d18, d20 (duh), …, d100, d1000, …

The trick is memorizing your d1000 critical hit table.
- Camilo Polymeris
  
  Or use a coin. Flip it e.g. 3 times if you need a D6:
  HHT = 001 base 2 = 1
  HTH = 010 base 2 = 2
  HTT = 011 base 2 = 3
  … and so on.
  
  Discard 0s and 7s.
  - knappa
    
    I think that part of the point was that you would be doing this entirely in your head, without coins and such. Otherwise, roll some D6 dice.
    
    The discarding of answers bothers me though. It seems like it can only add bias once you know which words “work”. Here’s a possible solution: Compute the same sum of digits then take the reminder mod 3. Now compute the sum mod 2 – which is also easy. Multiply the two results and add 1. That way the result will always be from 1 to 6.
    - Camilo Polymeris
      
      “I think that part of the point was that you would be doing this entirely
      in your head, without coins and such. Otherwise, roll some D6 dice.”
      
      Yes. That was in response to someone who suggested carrying dice around.
      
      “Compute the same sum of digits then take the reminder mod 3. Now compute
      the sum mod 2 – which is also easy. Multiply the two results and add 1.
      That way the result will always be from 1 to 6.”
      
      I think the result would be either 1, 2 or 3, with 1 far more likely, unless I misunderstand you.
      
      An option would be to take modulo 3 of the sum (which is easy), then add 1 if sum is odd or 4 if sum is even.
      This is actually equivalent to taking modulo 6, plus 1, I think.
      - knappa
        
        Oops, of course you are right. My mind was doing the chinese remainder theorem but my hand was typing BS. I should make that procedure: multiply the result of mod 3 addition by 2 is the sum is even.
- bibulb
  
  Ah, so you play Rolemaster.
Ron Hale-Evans

Yes, 8=1 is a typo in the transcription above. There is also more to this hack in the book, and of course many more hacks. :)
dorkhero

*cough* *cough*
Ugh! Math fumes!

…and that boys and girls is why the well prepared gaming geek always carries a full set of polyhedrals with them. You never know when you will be called upon to game.
pjcamp

8=1?

And I always thought it equaled 0. That’s pretty random.
http://www.facebook.com/people/James-Redekop/666775545 James Redekop

Quick, think of a three-letter word, such as “cat”!

Chances are, you thought of “cat”.
jimbuck

I always ask my kids “what do you think of when I say the word zebras? One will say stripes, one might say Africa. Then they’ll ask me what I think of and I’ll say zebras.
Mark_Frauenfelder

Sorry – A = 1 not 8 = 1. I was using Dragon Dictate and I should have reviewed the results!
jimkirk

Um, if you’re driving, why would you need to be rolling dice in the first place?
Please just concentrate on driving…

“But officer, it wasn’t my fault, I just rolled Yahtzee when that light turned red.”
invisiblemonkey

Even easier, pick a number between one and six, turn to the person next to you, and say “Pick a number between one and six”, then mod 6 that.

The word thing would be easy to game, since you could fairly easily remember a short list of short words that calculate to 6, or 1, or whatever number you “need” at the time.

Or, you know, I have a free app on my phone called Dice Bag that has a virtual 4, 6, 8, 10, 12 and 20-sided die.
Jonathan Waltz

Obviously this method is flawed as I immediately thought of the word “cat” before reading that far in the message. Everyone will be rolling 6′s.
futnuh

I’ve got a random number from one to six memorized: 4. I’m saving it for an emergency, though I’ve also memorized 2 as a backup.

The title of this post is somewhat misleading. I envisioned a trick that would actually let me visualize a spinning die, whose sequence of rotations were being governed by an external stimulus (like passing license plates or street signs).
E T

A = 1 not 8 equals 1
http://www.facebook.com/profile.php?id=595537411 Marty Hale-Evans

Everyone who’s commenting on the problems in “just thinking of a three-letter word” and whether you’d be unconsciously or consciously gaming the system, please read more carefully. If you look again, the excerpt Mark quoted explicitly says to “come up with” a three-letter word, and suggests ways to do that from an outside source. Certainly each person’s brain tends to come up with similar things in patterns. In the complete version of the hack that appears in the book, we discuss some of the problems with finding a good three-letter sequence; for example, we suggest ways to find sequences that aren’t complete English words, because the patterns of English tend to favor certain letter combinations. We also do discuss the mathematical underpinnings for the techniques in a bit more detail.

There may be other ways to do this task, for sure. It’s great fun to see this group riff on it and devise new approaches. And you could surely do it with a phone app or a die – but what if you don’t have one with you, or your phone is out of power, or you’re in the pool, say? What if you don’t need to roll dice intensively enough to game, but you want to make a quick random decision about pizza toppings? We’ve found it to be useful the way we wrote it for quick, light applications, both as a tool and as a mental exercise. If it doesn’t appeal to you, though, we have 59 more for you to try….
- http://twitter.com/MartianEmpress Rezeya Montecore
  
  It’s cool. I think you’ve just wandered into a community with a LOT of gamers in it. We’ve put way too much thought into problems like this already. If you’re coming at this from a cog psych angle instead, we’ve probably got a slight advantage over you. :)
  
  Your method, let’s face it, is not an especially good number-generation algorithm, and I think most of us have seized onto that. But I stopped and looked at it again, in context, and I think it’s still a pretty good example for a mindhacks book.
  
  Some people need to start from the bottom up; this very useful habit of “I can take this random bit of data from my environment, apply a filter to it, and use it as an arbiter” is one that not everybody has… but I think a lot of geeks come by it naturally and we take it for granted. Even if it’s not great at what it supposedly does on the surface, i.e. simulate die rolls, it’s a good exercise to put a newbie through, because it gives them a grounding in useful mental tools like digital roots (which are useful in all sorts of math shortcuts) and automatism (like, in the Surrealist sense of letting a process do the creative work for you).
  
  Anyhow, I ramble. This was interesting. n.n
  - http://www.facebook.com/profile.php?id=595537411 Marty Hale-Evans
    
    Thanks, Reyeza. FWIW, we’re gamers too. We did put some thought into it, but yes, we’re writing for a more general audience who doesn’t come to this stuff as naturally, and it’s true that one of the reasons we couch it in terms of virtual dice is as a practical motivator for people to learn how to do the math techniques and get more comfortable with this kind of mental technique.
    
    As I said above, we like this hack, but we certainly don’t claim that everything we’ve written is the only way to do anything. We’re sharing stuff that works for us, and may be useful to someone else. We want people to hack the hacks; we LOVE it when people take our ideas and run further with them. Continuing discussion and capturing new ideas that grow from previous ones is of the main reasons for the Mentat Wiki, which is older than either of our books. (We’d encourage you to add your table there, by the way – we think it’s ingenious.)
    
    Mainly, I hope people will wait to see the more comprehensive review and not judge the book by whether they like one hack. It was our experience with Mind Performance Hacks (our earlier book) that not every hack appealed to every reader, but almost every reader found a number of useful hacks in the collection. We expect the same to happen with Mindhacker.
    - SamSam
      
      Hi Marty, can you explain why you picked mod 9?
      
      The “it’s easier” argument posited above doesn’t hold any water, as mod 10 is even easier.
      
      I’m assuming that the distribution table for mod 9 across common English words looks better than it does for mod 6 or mod 10, but I’d love to get a verification on that.
      
      Maybe Ben Ostrowsky can run his program again to check.
      - http://www.facebook.com/profile.php?id=595537411 Marty Hale-Evans
        
        We suggested mod 9 because we thought mod 10 is a little too subjectively predictable and therefore easier to game, consciously or not, because you’re simply looking at the last digit of the result. Performing a slightly more difficult calculation means that it’s harder for you to predict the answer from the start; mod 9 is easier than mod 6 albeit not as easy as mod 10, so we think it hits a sweet spot between too easy to guess and too hard to bother with. Also, as Rezeya mentioned, knowing how to calculate mod 9 quickly is also useful for other math hacks, for example being able to calculate a fast checksum for basic arithmetic operations, as we describe in Mind Performance Hacks.
        
        One good thing about using mod 10 is that it effectively gives you a d10, not just a d6, which enables you to generate random digits, which enables you to emulate any size of die, such as percentile dice, d30, and so on, provided you are willing to discard enough rolls. In the book we do this with mod 11 (which also has a shortcut and also lets you do nice checksums, etc.), Obviously mod 10 is no more objectively predictable than mod 9, so if you’re willing to sacrifice “subjective unpredictability” and therefore a certain level of trust in the randomness of the answer, by all means, use mod 10.
http://www.kmoser.com kmoser

I have a simpler method: just think of a random number between 1 and 6. It should be random enough for most purposes.
- Klaus Æ. Mogensen
  
  Not very random, actually. You are less likely to choose 1 or 6 than numbers in between. I read once that when asked to think of a random number between 1 and 10, more than half chose 3 or 7.
silkox

Then there’s the whole role-should-be-roll misspelling problem.
- Ron Hale-Evans
  
  I believe role-vs.-roll is Dragon Dictate again. It’s spelled correctly in the book.
http://twitter.com/MartianEmpress Rezeya Montecore

Better yet, why not just pick a single letter and compare it to the following table?

1: A, B, C, D
2: E, F, G
3: H, I, J, K, L
4: M, N, O
5: P, R, S
6: T, U, V, W, X, Y, Z

If I did my math right and “Letter Frequency” at Wikipedia is accurate, each of those groups should comprise ~16.67% of letter frequency. It could probably stand to be refined much further, but it apparently gives more balanced results than this silly digital root method.
Joe in Australia

Here’s how to do mod-6 easily in your head.

First, is your number odd or even? If it’s odd, then your number modulo 6 must *also* be an odd number (i.e., 1, 3, or 5). If it’s even then your number modulo 6 must *also* be an even number (i.e., 0, 2, or 4).

Next, add up the digits of your number, and add up the digits of that sum, and add up the digits of *that* sum, until you have a single digit number. Divide that final amount by three and look at the remainder.

If it’s zero then your number modulo 6 was 0 or 3, depending on whether the original number was odd or even.

If it’s 1 then your number modulo 6 was 1 or 4, depending on whether the original number was odd or even.

If it’s 2 then your number modulo 6 was 2 or 5, depending on whether the original number was odd or even.

Easy-peasy, and you don’t need to discard a third of your answers.
librtee_dot_com

This reminds me of an old mental game I used to play, I came up with as a kid.

1) Look at any random word. Let’s look at “Unicorn”

2) Break it into it’s numbers – 21 14 9 3 15 18 14

3) Try to make an equation with it. After some thought, I get (21-14) + 9 +3 -15 = 18 – 14

You can use all sorts of things, such as ^, sqrt, etc. Also, I like to play that the alphabet loops around, so Z+A=A. Then, Z*anything always = Z, M^ anything = z or m, etc.

A fun little diversion if you’re sitting around bored. Certainly has filled many idle hours in my life :)
http://twitter.com/MartianEmpress Rezeya Montecore

Here we go. PIck a random wikipedia page (or what’s handy). The entry’s first letter (excluding A, An, The, etc.) is the die roll. If you get anything but an L, O, or T, consult the following table:
(1)ABCD (2)EFG (3)HIJK (4)MN (5)PQRS (6)UVWXYZ

For an L, O, or T, pick another random page (because just using the next letter would screw up the distribution this relies on) and use the first letter to consult this table:
[L] (3)A-N (4)O-Z
[O] (4)A-R (5)S-Z
[T] (5)A (6)B-Z

And there you go. I’ll leave it to the math people here to figure out exactly how close I got to 1/6 probability per result, but I think I got as close as anyone will ever need. :)

LOT/DGKNSZ/NRA. That’s a reasonable thing to memorize, right? Hell, put it on a business card and pull it out when you’re gaming.
Mark Leone

Look at your watch. The current number of seconds is a pretty good random number, provided you sample it at random times.
Glen Able

If you’re doing something by yourself, presumably (hopefully!) you’re not motivated to cheat and “roll” a 6 every time. So, just choose a word you can’t remember having chosen before and follow the procedure.

If you’re playing something with competitors, invisiblemonkey seems to be on the right track. But you and a competitor would need to choose a number from 1-6 at the same time – e.g. with that gesture-based countdown used for “rock paper scissors”. Add your number and theirs, subtract 6 if necessary. The nice thing about this is that there’s a component of skill, looking for patterns in the other players’ choices.
Max

Regardless of doing mod calculations in my head, I don’t know what number letter T is. If the example had said it was 17 or 23 I would have believed it quite happily, so now, I’m sat counting through the alphabet for 3 different letters, adding them all, mod-ing them and then, sometimes having to throw the result away. Alternative way to find a random number, just ask a random stranger or toss a coin a couple of times.

And really, if you’re driving and need a random number, you’re probably not concentrating enough that you can use your excess speed above the limit as a seed… A few quick mods or halvings should get you into the 1-6 range.
Klaus Æ. Mogensen

I would look at my watch and choose the last digit. Or start the stopwatch, stop it after an interval, and use the last digit in the 1/100 seconds.
L_Mariachi

Why discard 7-9? Just convert the numerification of your seed word to base 6 and then take the result mod 6. Or am I missing something?
http://www.facebook.com/daen.de.leon Daen de Leon

If you only have a coin, I have a way of getting random and unbiased numbers from 1-6 … but you do need to remember three numbers (1,2 and 4), and be able to add.

Toss a coin 3 times – assume heads = 1, tails = 0 … coin tosses go from b2 = 2^2 = 4, through b1 = 2^1 = 2, to b0 = 2^0 = 1. Sum the bits (4+2+1) – in the event you get 000 (0) or 111 (7), do it again.

Or, to save an extra three coin tosses and the chance that you get *another* 000 or 111, toss the coin again – if the coin is heads, flip the middle bit (so if it’s 000, it becomes 010, which is 2 – if it’s 111, it become 101, which is 5). If it’s tails, toss the coin again – if it comes down heads, flip b2 (so 000 becomes 100, which is 4, and 111 becomes 011, which is 3). if it comes down tails, flip b0 (so 000 becomes 001, and 111 becomes 110, which is 6) …

This bit-flipping works because there is at minimum 1 and maximum 2 bits set in the binary domain representing the numbers 1-6, so doing a single bit NOT on 000 or 111 as I describe will correct the bitcount, without bias (I think).

Maximum of 5 coin tosses, no modulus arithmetic required. There’s probably a more elegant way to do this in fewer coin tosses, but it’s 6:21am, and I’ve got pneumonia so :p …
SamSam

All of those suggestions above that recommend picking a random letter, and then grouping the letters to make the distribution even: how are you going to pick a random letter?

You can look around for the first sign you see and pick its first letter, but that distribution is not the same as the distribution of all letters in English. Picking the middle letter won’t help either, and you can’t pick a “random” letter any more than you can pick a random number.

BUT: this can be remedied. Use the first letter frequencies table. Do the same grouping so you end up with six even groups of letters.

Now you can just look around, note the first word you see, and use its first letter. Of course… the first-letter-frequency of words on road signs might not be exactly the same…
Vnend

It is nice when the comments are as much fun as the article.
Hodge

>You’ll also need to know or work out the number value of each letter…..

Oh yeah ? That’s sooo easy !!

An easy method: Car number plates – just take the last number on the plate; discard if it is a 0, 7, 8, or 9 and take the next plate.
TFox

I used to have a book, published by the Rand Corporation, with a million random digits. These were very high quality random numbers, generated by a hardware process (radioactive decay, perhaps). It was intended for situations where an ordinary dice roll wasn’t random enough, and the algorithmic pseudorandom number generators weren’t as good as they are now. Of course, you had to be careful to not reuse them, or to end up always picking numbers from the middle of the book.
- Hodge
  
  Hard to memorise a million random digits !
- http://www.facebook.com/daen.de.leon Daen de Leon
  
  You can also toss a coin 20 times, and use heads = 1, tails = 0. Write that down in order, you’ve got a 20 bit high quality random binary number between 0 and 1048575.
Quibbler

Using about 17000 dictionary words of between 3 and 5 letters long the frequency is

1 16.87%
2 16.83%
3 16.66%
4 16.64%
5 16.65%
6 16.35%
It doesn’t get much better than that. Using either MOD 6 +1 or MOD 10 does not make much difference to the distribution.
http://nefariousnewt.blogspot.com NefariousNewt

You mean you folks don’t carry dice around with you? Oh…
snakedart

I could carve a six-sided die in the time it would take to run this “simple” algorithm. And doing it would be much more fun.
TimothyWells

Flip a coin 6 times, heads equals 1 tails equals 0? I’m not sure how the probability there compares to an actual dice roll, but it seems faster, and coins are more abundant than dice.

Edit: nope, that won’t work. rolling a 0 is possible there.
- Uthor
  
  Flip it 5 times and add one? You’ll get a value between 0 and 5.
  
  Though I would think getting 0,0,0,0,0 or 1,1,1,1,1 to be much harder than getting, say, ”3″ (11000, 10100, 10010, 10001, 01100, 01010, 01001, 00110, 00101, 00011).
  - bwcbwc
    
    yeah. Flipping a coin multiple times and adding 1s for heads gives a bell-curve, not a linear distribution. Have to convert the coin flips to binary numbers, which leads back to doing 3 coin flips (values from 0 to 7) and discarding the 0s and 7s.
    - http://www.facebook.com/daen.de.leon Daen de Leon
      
      … which is what I suggested (also with a way of not discarding the 0 and 7, but tweaking them back into range with max two extra coin tosses).
    - Uthor
      
      D’oh. That should have been obvious to me after typing all those 1′s and 0′s.
http://www.facebook.com/profile.php?id=47204308 David Ritterskamp

Aside from all the math issues (and I am a mathematician, which is why the article caught my eye), isn’t the title redundant? Where else would you roll an imaginary die but in your head?
Uthor

Thanks! Next time I need a random number, I will think “cat” and get 6!
bwcbwc

Actually, my favorite mind-hack is if I can’t make a decision, I flip a coin. If my reaction to the coin-flip is “OK, let’s try for 2 out of 3″ I know which decision I want emotionally. Depending on circumstance I can say “go for it” or realize that I’m putting too much value on an option that I’m emotionally attached to and go for the other one.
Godfree

OK, if we’re allowing watches and coins as alternate methods, let’s use a pencil, too. 1) Procure pencil. 2) Number the sides 1-6, either mentally or physically. 3) Roll.
http://www.facebook.com/profile.php?id=595537411 Marty Hale-Evans

bwcbwc: Actually, that coin-flip emotional check hack is in the book, as part of another hack. I like it a lot myself.

Godfree: Love that idea, but I wouldn’t trust most pencils to be manufactured to high enough standards to be reliably random. I think the chances would be good that something would skew the results to favor some sides over others – internal weight distribution, surface imperfections that affect the rolling, warp, etc.
http://www.facebook.com/sylvar Ben Ostrowsky

I should point out that I didn’t filter out words based on length. If I think of it, I’ll find the program I used and pastebin it.
http://pulse.yahoo.com/_LRER3TP3HMXPAYE5VWM7XAO6HI jason

For anyone who does use this, an easy way to know the number of a certain letter in the alphabet is to remember the word EJOTY. This is the 5th, 10th, 15th, 20th, and 25th letters. That way you don’t have to count through the whole alphabet!
Te Ster

And here is my idea for coming up with a pseudo-random dice-ish number, provided there are at least two people playing a game with dices:

1. Countdown from 3, 2, 1..
2. Simulatenously, all players (or a sane subset, but ideally more than two) say or show (with their fingers) a number between 1 and 6.
3. Add up all numbers, modulo 6 the sum, add 1.
4. Profit. (You now have a pseudorandom number in the range of 1 to 6.)

I did this with 1 million rounds using pseudo-random numbers and got this distribution:

1: 16.65%
2: 16.62%
3: 16.72%
4: 16.65%
5: 16.71%
6: 16.66%

I think this would work ‘even better’ if we change the upper procedure to:

1. All players (or a sane subset, but ideally more than 2), shake their fists in a stone-sissors-paper way.
2. Simulatenously, players show between zero and five fingers.
3. Add all fingers, add number of shaken fists, modulo 6 the sum, add 1.
4. see above

This maps the numbers of a 6-side-dice to no to five fingers and makes the whole ‘rolling’ easier, I think.

Simpler math, and definitely no braincrunching letter-to-number mapping. And nothing more needed than a set of players. Doesn’t work when you play alone, but then you win anyway.