# Convert Big Boggle to regular Boggle

I have a 5x5 Big Boggle set (in English). Is it possible to play regular 4x4 Boggle with this set by removing cubes? I'd like to remove 9 cubes and get a letter distribution that's close to the distribution in traditional 4x4 Boggle.

• Could you expand a bit on your definition of "close letter distribution"? What is "close enough" for you? – John Mar 17 '16 at 15:28
• @John "Close enough" would be a roughly similar distribution of vowels and consonants, while retaining most of the rare letters. If you want to get technical, let's find the 16/25 cubes that give the smallest Kullback-Leibler divergence from standard Boggle. – Nuclear Hoagie Mar 18 '16 at 16:50
• First time I've ever seen KL in boardgames. Well done sir. Well done. – John Mar 18 '16 at 18:04

## Summary

The best sixteen dice to pick in order to produce the lowest Kullback-Leibler divergence from the (new) standard Boggle distribution are:

``````AAEEEE, ADENNN, AEEGMU, AFIRSY, BJKQuXZ, CEILPT, CEIPST, DHHLOR, DHLNOR, DHLNOR, EIIITT, EMOTTT, ENSSSU, FIPRSY, GORRVW, NOOTUW
``````

The divergence for this set of dice is 0.0226, and it provides at least one of every letter. This dice set provides about 10% more words than official Boggle, and about 13% more points (from an average grid).

## Background

There are two different standards for 16 dice Boggle, and one for 25 dice.*

There are multiple ways to measure the 'closeness' of two sets of Boggle dice. A comment on the question suggests Kullback-Leibler divergence. Another simpler metric would be a count of the number of incorrect letters which I've referred to as the Manhattan distance.

The most interesting quality of a Boggle dice set is probably how easy it is to make words during a game, so I decided it was also worth investigating the average number of words in a grid, and the average score (if a player found them all).

* The BGG thread mentions another Big Boggle distribution that contains a AEMNNS die, but the set was incomplete, so I didn't include it here.

## Method

I wrote a Python script to search through all combinations of sixteen dice from the twenty five provided in Big Boggle. For each set of dice I measured the Manhattan distance and the Kullback-Leibler divergence from both the new and old standard Boggle dice.

Once I had the optimum set of dice for each metric, I used a Boggle solver to simulate 10,000 games of Boggle with them, and counted the number of words and the total score available in each grid.

## Results

``````Metric | Old/New | Best selection of dice                                                                                                                                          | Distance (using metric)
KL     | New     | 'AAEEEE', 'ADENNN', 'AEEGMU', 'AFIRSY', 'BJKQuXZ', 'CEILPT', 'CEIPST', 'DHHLOR', 'DHLNOR', 'DHLNOR', 'EIIITT', 'EMOTTT', 'ENSSSU', 'FIPRSY', 'GORRVW', 'NOOTUW' | 0.0226049400407
KL     | Old     | 'AAAFRS', 'AEEEEM', 'AEEGMU', 'AEGMNN', 'AFIRSY', 'BJKQuXZ', 'CEIILT', 'CEILPT', 'CEIPST', 'DDHNOT', 'DHHLOR', 'DHLNOR', 'ENSSSU', 'FIPRSY', 'GORRVW', 'NOOTUW' | 0.0403229165488
Manh.  | New     | 'AAAFRS', 'AAEEEE', 'AEEGMU', 'AFIRSY', 'BJKQuXZ', 'CEILPT', 'CEIPST', 'DDHNOT', 'DHHLOR', 'DHLNOR', 'DHLNOR', 'EIIITT', 'EMOTTT', 'ENSSSU', 'GORRVW', 'NOOTUW' | 12
Manh.  | Old     | 'AAAFRS', 'AAFIRS', 'AEEEEM', 'AEEGMU', 'AEGMNN', 'AFIRSY', 'BJKQuXZ', 'CEIILT', 'CEILPT', 'CEIPST', 'DDHNOT', 'DHHLOR', 'DHLNOR', 'FIPRSY', 'GORRVW', 'NOOTUW' | 16
``````

It turned out that the optimum dice sets all contained every letter of the alphabet, which seems like it should be a requirement for a game of Boggle.

Simulating 10,000 games with each set, and using a Boggle solver, I found the following results.

``````Dice set      | Average number of words in grid | Average maximum score in grid
Original New  | 87                              | 113
Original Old  | 83                              | 105
KL New        | 96                              | 128
KL Old        | 92                              | 122
Manhattan New | 99                              | 132
Manhattan Old | 92                              | 121
``````

## Conclusions

We used four different methods to pick 'the best' sixteen dice from Big Boggle, and it seems that any of the sets of dice are reasonably suitable for playing a game of standard Boggle with. The old standard Boggle dice are slightly harder to play with than the new dice, and it follows that it is also harder to play with the Big Boggle selections which were optimised towards the old Boggle standard.

All of the Big Boggle dice sets produced more words and higher scores than the original Boggle dice sets, and this may be due to the presence of dice like AAEEEE, AEEEEM and EIIITT in Big Boggle, which help to avoid consonant-only grids.

As a side note, Boggle is a pretty popular game, and is not too hard to pick up a second hand copy from a charity shop. If you really want to play standard Boggle then getting a copy might be easier than separating out the right set of dice each game, and you'll also get a four-by-four box to shake the dice in!

• This is just what I was looking for, great answer! – Nuclear Hoagie Mar 21 '16 at 8:32

You cannot "exactly" recreate 4x4 from 5x5. A list of the 4x4 and 5x5 dice are at this question. You can see that the Qu cube for the 4x4 is different than the one Qu cube for the 5x5.

I should think that the simplest solution would be to randomly select 16 of the 25 dice. You could do this by setting up the 5x5 game as normal but then removing one row and one column. You won't get exactly the same probabilities but it should be close enough to make little practical difference.

• I don't think this is actually a bad suggestion. I just ran 10000 simulations of grids made by picking 16 of the Big Boggle dice at random. On average there were 102 words per grid, with a total score of 140. This is a significantly higher number of words/points per grid than standard Boggle, but it suggests that games played in this way would definitely work. – tttppp Mar 23 '16 at 16:43