When you have:

Total Boards = (20^0 * 6^16) + (20^1 * 6^15) + … + (20^16 + 6^0)

It should be:

Total Boards = (20 choose 0)(20^0 * 6^16) + (20 choose 1)(20^1 * 6^15) + (20 choose 2)(20^2 * 6^14) … + (20 choose 20)(20^16 + 6^0)

since you need to CHOOSE which of the “n” spots you put your “n” consonants. The total is actually a lot higher.

Also, the eight-fold symmetry comes from reflection, since reflected boards have the same score.

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You could divide the alphabet into 20 consonants and 6 vowels, and for a 4×4 board the total number of possibilities would be the sum of every combination of consonants and vowels, namely:

Total Boards = (20^0 * 6^16) + (20^1 * 6^15) + … + (20^16 + 6^0)

Of course, in English every word must have at least one vowel, so the last term drops out since it represents the all-consonant boards. Therefore, the total is:

Total Boards = 2.8 x 10^20 ~ e^47

At 10,000 boards/second, that would still take almost 900 million years to evaluate. However, at this point one could employ a couple heuristics. For example, by examining the frequency distribution of letters in the English language, it is reasonable to assume that the letters J, X, Q, and Z will not be part of the best Boggle board. If those letters are excluded, then the number of consonants falls to 16, and the new number of boards decreases by a factor of 25, to 1.1 x 10^19 boards (35 million years).

From here, one could go a step further, and remove B, V, and K from the running, to decrease the total to 5.7 x 10^17 boards (1.8 million years). Or, one could examine the letter frequencies again and surmise that the best Boggle board will in all likelihood contain an E. If one E is guaranteed, the total boards changes to:

Total Boards = (20^0 * 6^15) + (20^1 * 6^14) + … + (20^14 + 6^1)

Or, 5.1 x 10^19 (162 million years)

With one E guaranteed and J, X, Q, Z removed, you would have 6.9 x 10^17 boards (2.2 million years), and with one E guaranteed and J, X, Q, Z, B, V and K removed, you would have 4.4 x 10^16 boards (139,000 years).

Of course, rotating a Boggle board 90 degrees will produce a different board with the same score, so by symmetry you could divide all of these totals by 4 (I saw in your posts that you found eight-fold symmetry – how is that?).

Removing or guaranteeing certain letters is pretty unscientific, but taking about a Z or putting in an E is also pretty reasonable. I guess it depends on how thorough one wants to be. Also, even with the most ‘flexibility’ employed (one E, no J, X, Q, Z, B, V or K, and four-fold symmetry), the solution would still take about 35,000 years to solve. But I imagine that’s a bit more palatable than 2 billion years.

– Nathan