"Knowledge Base" February 16, 2024

[WC50]

Sensing the x,y path but not sure what to use as the origin, I plugged the grid into Excel and used OFFSET formulas to test an array of starting points, wondering if any English word would emerge. Only one did: NERDS. An admittedly nerdy way to get to the answer without spotting the hex theme.

[Joepickett]

To.continue with the nerd theme, you have to enter 5.3 years once a week at 400 to 1 odds to get to 50% chance of winning. So yeah our chances are low.

I don't understand how the odds would change over time. What would change to give you a 50% chance of winning?

@Joepickett is saying that if the odds of winning the mug each week was 1 in 400, then...

The chance of not winning the mug each week is 399/400, or 99.75%

The chance of not winning a mug at that consistent chance for N weeks is 99.75% to the Nth power.

So in order to have a 50% chance of winning the mug, you would need to submit a correct answer for 277 straight weeks.

99.75% ^ 277 = approximately 49.99%, so you'd have a 50.01% chance of winning at some point during that period.

277 / 52 = 5.33 years.

Unfortunately, the odds are usually not that good, so it would take even longer than that to have had a 50% chance.

Yeah, what he said.

Log(1-<percent chance to win you want>)/log(399/400)

[Conrad]

We have been running these contests weekly for more than 8 years, and I think this week's has a fair claim to be the most difficult one yet. For all that, we got a weekly reminder of just how clever our top solvers are: we had 537 entries and 73% were correct, which in this case was quite impressive.

One thing I have noticed while solving metas, and especially since I've been writing up the WSJ meta for Crossword Fiend: a lot of people react negatively when they miss a meta. Or they mash the 1 feedback button on Crossword Fiend. Especially for the WSJ. This is one of those metas.

This meta was a bit inelegant due to OVERIT (IMO), but it was otherwise amazing. It happened to be locked into my wheelhouse, but maybe not yours. That happens to all of us: sports metas are a common example. But the WSJ meta is always fair. HEX and ROW were symmetric, and showed the path.

I was terrible at metas years ago, but I improved. I never blamed the metas that I missed, or mashed the 1 button in frustration: I tried to find a lesson learned, and applied that to future metas.

[Beleaguered Castle]

Sensing the x,y path but not sure what to use as the origin, I plugged the grid into Excel and used OFFSET formulas to test an array of starting points, wondering if any English word would emerge. Only one did: NERDS. An admittedly nerdy way to get to the answer without spotting the hex theme.

Same concept as my template!
(But my method is analog.)

[Nlobb]

Count me among those who never heard of a hexadecimal !! It wasn’t in the curriculum in my rural Montana high school back in the 1960s!!
I do think the chance of winning the mug starts over with each contest. If it was cumulative some of us who have been doing these for awhile would have a better chance of winning, but I don’t think so… maybe I am wrong…

[MajordomoTom]

a series of independent random draws, each from a different sized pool

the probabilities do "reset" - i.e., if the draws are truly random, from the submitted pool of answers each week, you have the same ability to win each time

probability - and the "5+ years" math (above) - that's only to mathematically have at least a 50% chance of having one (at least) one mug. It doesn't mean that you'll ever actually definitely win one - that's not how these coins flip.

[rjy]

We have been running these contests weekly for more than 8 years, and I think this week's has a fair claim to be the most difficult one yet. For all that, we got a weekly reminder of just how clever our top solvers are: we had 537 entries and 73% were correct, which in this case was quite impressive.

One thing I have noticed while solving metas, and especially since I've been writing up the WSJ meta for Crossword Fiend: a lot of people react negatively when they miss a meta. Or they mash the 1 feedback button on Crossword Fiend. Especially for the WSJ. This is one of those metas.

This meta was a bit inelegant due to OVERIT (IMO), but it was otherwise amazing. It happened to be locked into my wheelhouse, but maybe not yours. That happens to all of us: sports metas are a common example. But the WSJ meta is always fair. HEX and ROW were symmetric, and showed the path.

I was terrible at metas years ago, but I improved. I never blamed the metas that I missed, or mashed the 1 button in frustration: I tried to find a lesson learned, and applied that to future metas.

Hear hear!

I am so glad you wrote that. We should all hope that constructors are imaginative enough to be innovative within this crazy, constrained art form, but that will be at the risk of occasional metas that will be tough to get. It’s the only way this will stay interesting and challenging! (…that said, it can be fair to quibble with red herrings or ineffective sign posts)

On to next week!

[MajordomoTom]

no one gets 100% of these - I've been doing them roughly 4 years and have gotten pretty good, but not 100% ... that's why they're "puzzles", if we could solve all of them, wouldn't that be boring?

[WC50]

Sensing the x,y path but not sure what to use as the origin, I plugged the grid into Excel and used OFFSET formulas to test an array of starting points, wondering if any English word would emerge. Only one did: NERDS. An admittedly nerdy way to get to the answer without spotting the hex theme.

Same concept as my template!
(But my method is analog.)

Ha. I did see that. And your method works during a power outage!

[ship4u]

To.continue with the nerd theme, you have to enter 5.3 years once a week at 400 to 1 odds to get to 50% chance of winning. So yeah our chances are low.

I don't understand how the odds would change over time. What would change to give you a 50% chance of winning?

@Joepickett is saying that if the odds of winning the mug each week was 1 in 400, then...

The chance of not winning the mug each week is 399/400, or 99.75%

The chance of not winning a mug at that consistent chance for N weeks is 99.75% to the Nth power.

So in order to have a 50% chance of winning the mug, you would need to submit a correct answer for 277 straight weeks.

99.75% ^ 277 = approximately 49.99%, so you'd have a 50.01% chance of winning at some point during that period.

277 / 52 = 5.33 years.

Unfortunately, the odds are usually not that good, so it would take even longer than that to have had a 50% chance.

I humbly admit that you guys are much better at numbers than I am. However, I recall from my statistics class of many years ago, an example that went something like this: If you flip a fair coin 10 times and it comes up heads each time, what are the odds that on the eleventh flip, it will come up tails. Answer: 50-50.

It seems to me, Dr. Watson, that what is not being taken into account in the "Nth power" argument, is that each week is a fresh game, meaning that each week, all of the entries for that week are put into a "hat" and one entry is randomly drawn. Whether a person has entered once previously, or on 277 previous occassions, would have no bearing on the current pool and random drawing.

[woozy]

99.75% ^ 277 = approximately 49.99%, so you'd have a 50.01% chance of winning at some point during that period.

The discouraging aspect of this is that probabilities are always 50% for five years from now. Never five years from then. I you play for three and a half years and never win it you still have to play for five more years (8 1/2) for the chances to be 50%. It's a never jam today situation.

[ship4u]

My one diversion was the clue for 71A, "Far From the Maddening Crowd" setting.

I spent the next two hours speed-reading the book online

Whilst the puzzle was certainly "Maddening," the novel wasn't. No doubt the product of speed reading.....

[DBMiller]

I don't understand how the odds would change over time. What would change to give you a 50% chance of winning?

@Joepickett is saying that if the odds of winning the mug each week was 1 in 400, then...

The chance of not winning the mug each week is 399/400, or 99.75%

The chance of not winning a mug at that consistent chance for N weeks is 99.75% to the Nth power.

So in order to have a 50% chance of winning the mug, you would need to submit a correct answer for 277 straight weeks.

99.75% ^ 277 = approximately 49.99%, so you'd have a 50.01% chance of winning at some point during that period.

277 / 52 = 5.33 years.

Unfortunately, the odds are usually not that good, so it would take even longer than that to have had a 50% chance.

I humbly admit that you guys are much better at numbers than I am. However, I recall from my statistics class of many years ago, an example that went something like this: If you flip a fair coin 10 times and it comes up heads each time, what are the odds that on the eleventh flip, it will come up tails. Answer: 50-50.

It seems to me, Dr. Watson, that what is not being taken into account in the "Nth power" argument, is that each week is a fresh game, meaning that each week, all of the entries for that week are put into a "hat" and one entry is randomly drawn. Whether a person has entered once previously, or on 277 previous occassions, would have no bearing on the current pool and random drawing.

Ah... Your example talks about the 11th flip, and ONLY the 11th flip. Each case is indeed individual. So yes, no matter what happened for the past 10 flips (Or the past 399 weeks), the chance this time is the same as the last. This is the Gambler's Fallacy - You can't expect your true chances each week to change based on past performance. The micro-mug calculator is one such usage. You can add up all the misses you like - You might end up with a value greater than one. Does it mean you won the mug? No. Does it mean you are due? No. You are no closer this week than last week. Gambler's Fallacy.

To make an apples to apples comparison would be:

Suppose you [flip a fair coin 10 / will submit the correct WSJ meta answer each week where 400 people do the same for 277] times. What are the chances that [at least one flip will come up heads / you will win the mug]?

For the coin flip case, the chance is pretty good that at least one flip will be heads, because it's a 50/50 shot on each flip. So the chance of getting tails every single time is .5^10 or .00098, or less than 0.1% - So in this case, there's a 99.9% chance that you will flip heads at least once in 10 tries. Go ahead, try and flip a coin 10 times in a row and get tails every time. We'll wait...

But with only the 0.25% of winning the mug, it takes quite a number of iterations to increase the .25% chance of winning to just over 50%. But again, this is saying that the chance is just about 50/50 if you knew you were going to submit the correct answer for THE NEXT 277 weeks and that there were only going to be 400 correct answers each week.

Oh and BTW, I did HORRIBLY in statistics class in college. Hated it. But the college had no computer degree. Only Math with a concentration of Computer Sciences. But I do like trying to figure these things out. Sometimes I get 'em right.

Then we get into things like the Birthday Paradox. That really makes your brain hurt.

[woozy]

Monroe's What If #2 had a question about driving to the edge of the universe and he basically did a large number riff and noted you had a 99.86% chance of dying in a car accident before you were a thousandth of the way there.

Same idea.

[clonefitz]

The probability of "NEVER" winning a mug decreases with each weekly attempt, therefore the probability of "NOT NEVER" winning a mug increases with each weekly attempt.

[BethA]

I meant it when I said that I enjoy metas with an additional aspect or subject beyond wordplay. Now a little worried that somehow there will be one based on probabilities! Aaaah…

[ricky]

20240216-WSJCC-GRA-JR-FYC.png

Wow, this looks pretty solid! Nice job! For some reason this got me thinking on a tangent: I wonder if there has ever been a puzzle that was not intentionally created as a meta, but ended up having a solid mechanism resulting in an unintended "answer".

Early in my blog's existence, a Daily Crossword Links email linked to one of my puzzles but incorrectly labeled it as a meta. This resulted in my blog's first mention on the Fill Me In podcast being Ryan lamenting that he was unable to get anywhere with solving the meta puzzle that was not actually a meta puzzle.

[Susan Goldberg]

Yikes. If this had been the first meta I ever attempted, I am not sure I would ever have come back! Fortunately, I was in Spain last week, so did not get to this one until after the contest was over!