Matt Bayfield wrote:Would it be possible, in terms of comparing scores for different formats, to go through the entire database, and determine:

Lm = mean score for a Letters round, which I imagine is something below 5, bearing in mind that many contestants score zero because their opponent has a longer word

Nm = mean score for a Numbers round

Cm = mean score for a Conundrum (which must be below 5, since only one player can ever get the conundrum)

Then, for each format, calculate the "mean" score: e.g. 9R mean score would be S(9Rm) = 6Lm + 2Nm + Cm; old 15R mean score would be S(15Rom) = 11Lm + 3Nm + Cm, etc.

You could then even consider converting between format scores using e.g. S(15Ro)equivalent = S(9R)actual x S(15Rom) / S(9Rm). However, I have a feeling this might be a poor conversion at the very low and high ends of the score spectrum.

(Edit: Looks like Gevin has already suggested something similar, but I'm leaving this post here anyway.)

I'm going to answer this question and then also count it as an answer to Gevin's question.

The mean score for a player in a letters round, looking at all televised letters rounds where we know what the player scored, up to the end of the last CoC, is about 4.942.

For numbers, the mean score is 6.328.

For conundrums, the mean score is 3.647.

All these figures are rounded to three decimal places.

Our simulated mean score for a 9 rounder, therefore, is 6Lm + 2Nm + Cm = 45.956.

For 14 rounders it's 72.142.

For old 15 rounders it's 76.995.

For new 15 rounders it's 78.380.

The *actual* mean score for a player in a 9 rounder, looking at all televised 9 rounders, is 45.328. For 14 rounders it's 76.472, for old 15 rounders it's 77.526, and for new 15 rounders it's 78.433. These figures are reasonably close to what the formula predicts, but the actual mean score in 14 rounders is a bit higher, presumably because 14 rounders, being finals, featured stronger than average players.