Wikipedia:Reference desk/Archives/Mathematics/2022 May 10
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May 10[edit]
curious about calculating a wordle average[edit]
Hi friends. Some of the wordle offshoots give you an average, such as globle, which tells you the average number of guesses you take. So I was curious about how to calculate the average for wordle.
I think I can figure it out in the case of winning every game; so if I played 20 games, and won 4 times on the second row, 6 times on the third row, 6 times on the fourth row and 4 times on the fifth row the average number of guesses to win would be (4*2 + 6*3 + 6*4 + 4*5) / 20 = 3.5.
But suppose I'd actually played 25 times and lost the other five times. Would there be any calculation that could allow for that? Because a person who played 20 times and won every time would be a better player than someone who only won 20/25, but their averages could be the same by this method. Thanks, 70.67.193.176 (talk) 22:43, 10 May 2022 (UTC)
- Suppose some player never succeeds in finding the word within the five guesses allowed. (So they fail unfailingly.) What is the average score for this loser? Clearly, it is undefined in the globle sense: 0/0 = ?. What can we do if we need a number? One possible approach is to assign them a fictitious "average" that is definitely greater than 5, so that in a ranking of players based on their average scores such losers end up at the bottom. This "losers' average" could then also be used as the score for a loss by more successful players. Choosing the actual value is necessarily somewhat arbitrary. I think you have to be somewhat of a dolt not to get a given word in 7 guesses (if the game would allow one to keep playing), so you could use 7 as the score for a lost game. --Lambiam 06:46, 11 May 2022 (UTC)
- That is a version of additive smoothing which is sometimes used in these situations. 2601:648:8202:350:0:0:0:738F (talk) 07:23, 11 May 2022 (UTC)
- As a side note, it is important to note that the word "average" is a very mathematically imprecise term, we have several dozen common methods of calculating an average. The most common is perhaps the arithmetic mean, but if the answer doesn't match, then perhaps one of the other various methods of calculating such an average are used. The summary in the table at the average article lists 14 ones, but that is hardly comprehensive. --Jayron32 14:01, 11 May 2022 (UTC)
- My Wordle scores per number of tries (0-6) run 0, 1, 16, 34, 10, 4. I have (so far) never failed, so all higher values are 0. The values form a rough curve, which might be matched to a probability distribution (Poisson springs to mind, but I never was any good at probability or statistics). Therefore, I suggest deriving a curve from the observed values, and taking the mean (or whatever average might be appropriate) from the curve.
- Subsidiary question: how many attempts does it take for that method to suggest/prove with a given confidence that I am honest about never failing (or otherwise)? -- Verbarson talkedits 14:47, 11 May 2022 (UTC)
- The 34 hump is much too pronounced to make the Poisson distribution's probability mass function a plausible candidate. Given a model and a good fit, one can estimate the probability of losing a game, in which the random variable denotes the number of tries needed. Under the null hypothesis that the score keeper is not cheating, the -value, in this case the probability to observe zero losses in games played, equals --Lambiam 06:55, 12 May 2022 (UTC)
- As a side note, it is important to note that the word "average" is a very mathematically imprecise term, we have several dozen common methods of calculating an average. The most common is perhaps the arithmetic mean, but if the answer doesn't match, then perhaps one of the other various methods of calculating such an average are used. The summary in the table at the average article lists 14 ones, but that is hardly comprehensive. --Jayron32 14:01, 11 May 2022 (UTC)
- That is a version of additive smoothing which is sometimes used in these situations. 2601:648:8202:350:0:0:0:738F (talk) 07:23, 11 May 2022 (UTC)
Thank you everyone who replied. The only answer I understood how to apply was the first one...it let me discover, as I had guessed, one language is easier for me to wordle in than the others. By the method of counting a loss as a 7, anyway! As a followup, was just wondering if any of the other options are accessible to a non-mathematician.
2601:648:8202:350:0:0:0:738F, is there a simple formula for calculating an additive smoothing that could be explained to say, someone with maths skills of a 10-year-old?
Jayron32, thanks for the term of what I was trying! Is there one of those other 14 you’d recommend? And if yes could it be explained how to calculate it?
Verbarson, is it possible you could show me how to derive a curve and take a mean from it in a formula a 10-year-old, say, could apply? 70.67.193.176 (talk) 15:41, 12 May 2022 (UTC)
- I'm afraid that I am outside of my area of (very limited) expertise here as far as recommending a type of average. There are certain reasons why one might choose a particular average over another; for example in large data sets with a very wide distribution (i.e. "long tails"), then median is often a better average because outliers can dramatically throw off the arithmetic mean greatly, but median tends to have it's own problems as well. Then you have geometric mean, which is the multiplicative analogue of the arithmetic mean, which tends to be used in areas like measuring average growth rather than average value (i.e. it tracks better for exponential functions than for polynomial functions). As far as which average the program in question uses to calculate your average score, I have no clue. The best thing I can offer is just try each one individually and compare results. The formulae for calculating them are listed in the article average. --Jayron32 15:58, 12 May 2022 (UTC)
- No worries at all Jayron32, I understand. I think with the wordle setup there won't be any outliers because there aren't that many possible scores. Alas, none of the formulae in the average article make sense to my very-beginner-level math brain. But thank you again, I can use the simple method of counting a loss as a 7. And I'm glad the mathematicians had a good discussion from the question! 70.67.193.176 (talk) 17:24, 13 May 2022 (UTC)
- IP, my mathematical skills do not extend to selecting a suitable probability model and curve-matching it at any age level, let alone for a ten-year old. I'm just aware that such things may be possible. Happy Wordling. -- Verbarson talkedits 17:38, 13 May 2022 (UTC)
- No worries! I probably wasn't clear that I hoped for an answer I could execute. Appreciated your reply. 70.67.193.176 (talk) 20:45, 13 May 2022 (UTC)
- IP, my mathematical skills do not extend to selecting a suitable probability model and curve-matching it at any age level, let alone for a ten-year old. I'm just aware that such things may be possible. Happy Wordling. -- Verbarson talkedits 17:38, 13 May 2022 (UTC)
- No worries at all Jayron32, I understand. I think with the wordle setup there won't be any outliers because there aren't that many possible scores. Alas, none of the formulae in the average article make sense to my very-beginner-level math brain. But thank you again, I can use the simple method of counting a loss as a 7. And I'm glad the mathematicians had a good discussion from the question! 70.67.193.176 (talk) 17:24, 13 May 2022 (UTC)