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The Word Game (math version) - October 28, 2011
Rating: 4/5 (59 votes cast)
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In math terms, this is how you play the game:

Spiked Math Comic - The Word Game (math version)

The rules in another language:

Spiked Math Comic - The Word Game (math version)


And now the comic! Yay!



Spiked Math Comic - The Word Game (math version)


Alternate third panel.


It's similar (but not isomorphic) to the number game (Harold's game??) where people instead have to choose a positive integer with the winner being the person who wrote down the smallest positive integer not written by anyone else.

Also, I vaguely remember playing this game (or something similar perhaps?) as a board-game, but cannot remember what game it was!! Anyone know?

Some sample (mathematical) categories you can try out:
  • Mathematicians whose surname starts with ___
  • Mathematicians with Erdős number 1
  • Categories
  • Clopen sets
  • Functions that are continuous everywhere but differentiable nowhere
  • For n players: nth roots of 1
  • Fields medal winners
  • Misnamed Theorems
  • Transcendental numbers
  • Math Comics
You can also turn it into a mathematical drinking game where people who say the same answer have to take a drink! :D




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16 Comments

Are you thinking of Scattergories?

you mean 1821 not 1921 for Chebyshev and Cayley's year of birth :)

Yup 1821, fixed now!

Yes, Scattergories is like this, except that you write down as many items as you can think of in each category, and then you get credit for each item that no one else has (as long as your opponents agree that it belongs in the category).

Quite similar to Scattergories yup! Except with that game, all your answers must start with the letter on the die. The way I usually played is you can only write down one thing (though it could be double/triple worded like "Michael Moore", which is worth 2 points for two M's - assuming M is on the die and the category fits).

When you say, it will be a tie for two players, you assume that they won't make any mistake, and they will always both find an answer or both not find any answer.

It is a big assumption, but still is funny!

I think we can assume that hey always find an answer, the trick of the game is finding an answer that no one else says.

I got confused at line 5 because I hadn't realised C was a set of sets. While wondering how this could have been made clearer, I learnt that 'Often, families of sets are written with either a script or Fraktur font to easily distinguish them from other sets' (according to wikipedia) and I also learnt about classes, which is what C should probably be. It's a good day!

The C in the above description is a class. It is also essentially a joke on the math term 'category'. Which have a little more structure than simply sets or classes. In that case the C would be the category of small categories Ci.

The condition for getting one point looks bogus: |A_i(j) ∩ A_i|. A_i is the set of all A_i(j), and will normally have an empty intersection with each individual A_i(j). I think you mean something like |{ l | A_i(l) = A_i(j)}| = 1 here.

I agree that the types don't match in the "if" clause. An easier way to fix it is |\{A_i(j)\} \cap \mathcal{A}_i| = 1.

I somewhat agree with what you say but I don't think the intersection would be empty; moreover, the intersection between the set of all A_i(j) and an individual A_i(j) would be A_i(j). Therefore it would need to be |A_i(j) ∩ A_i\A_i(j)|.

Actually, I'm going to retract the last part of that. After looking at it closer I think the original one is correct. Because |A_i(j) ∩ A_i)| will equal 1 if your answer is unique because then the intersection will contain the single A_i(j) entry.

Mike, the problem is that A_i(j) is (presumably) a single string/thing, whereas A_i is a set of the same type of elements. You can't really intersect an element with a set, so it should be {A_i(j)}. Alternatively, if you think that A_i(j) is a set of one string/thing, then A_i is a set of sets, so again, the intersection won't work out.

If there are an infinite number of valid elements to choose from, the optimal strategy for 3 players is to choose the nth element from the beginning with probability 1/(2^n).

the numbers game would be really interesting with the nonnegative reals.

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