(inspired to post by Anne’s 30-Day Blog Challenge)
So I was playing Scrabble last night (I lost – it’s one of those board games I’m not the best at) when we talked about how, when you are playing with good competent players, the board often winds up with knots of small words close together.
Kinda like this one.
So we talked about how we could promote long and fun words instead of those same short words all the time, and thought you could have a variant where you get bonus points based on how long your word is, regardless of which letters you use or where you place it.
Such a bonus somewhat already exists – you get a 50 point bonus if you use all 7 of your tiles. So we thought we could add other bonuses for other lengths. We agreed we should keep the 50 point bonus for 7, and that you shouldn’t get a bonus for only using 1 letter. As well, we thought a 2 tile go should get 1 point as a bonus. So I said I could definitely model it from there.
I tried to feed those data points into Wolfram Alpha for a fit but they provided linear, logarithmic, and period fits, all of which were terrible. I then forced them on a quadratic fit (after all, 3 points make a parabola), which was alright, although maybe too many points for a 6 tile play. Then I did an exponential one (though I had to use (1,0.1) since Wolfram didn’t like using (1,0) in an exponential fit, as if we couldn’t shift the curve down.) Then I just fed them into Desmos and rounded.
Below are the graphs and the tables for each fit. What do you think of this variant? Which point spread would be better? Of course, we’d have to play it to see….
James Cleveland-Tran
The Roots of the Equation 
Comments on: "Scrabble Variant" (4)
Great idea.
Here’s my thoughts… There’s no such thing as a 1-letter word, so that isn’t in the analysis. You don’t want 2-letter words, that’s the point of this, so no bonus there. 3-letter words aren’t much good either. I’d give them a 0 too. Then I’d round the others to end in 5 or 0, to help people keep it in their brains. So, using what you have, I’d use either:
4-letter gets 5, 5-letter gets 10, 6-letter gets 20 or 25, 7-letter gets 50,
or
4-letter gets 10, 5-letter gets 20, 6-letter gets 30, 7-letter gets 50.
I’d test-play both these version to see which was more fun.
Well, remember, using 1 tile isn’t the same as a 1 letter word. Using 1 tile means you’ve made a 2 letter word. Using 2 tiles is iffy – sometimes it’s a 2 letter word (because you played them beside something) and sometimes it’s a 3 letter word.
Though the bingo bonus is based on the number of tiles used (all 7), you could use that to make a 7 letter word or an 8-letter word. Maybe this bonus should be based on the length of the longest word you make when you play – so as to further emphasize acrostic-style playing instead of crossword-style play.
I like Sue’s suggestions (1st one is better; 5, 10, 25, 50 is still quadratic), but would do it for number of tiles played as opposed to length of word formed. You could consider a penalty, too: only 1 or 2 tiles played means no bonus scores from the board, or some such. (Probably still worth it as you’re denying your opponent the bonus, but at least you’re not double rewarded.)
If it gets around all that nonsense of people making two-letter words based on a list of archaic words from a Scrabble dictionary, then I’m all for your version. (As you can see, I don’t want to put the effort into being a “good” Scrabble player.) Or you could scrap Scrabble altogether and play Quiddler, which I find to be much more fun.