c4countdown

Yep, already paid. Not sure about the hangover, though. It was Ryan, yes, but I'm not sure if either of us are interested now.

EGOMANIAC

I feel the numbers where you got 2 away but you could have got away as a max is up to opinion, as I know a few people don't think it should be classed as a max (I'm one of them).

I certainly wouldn't be a fan of ordering these in the way in which Countdown seed finalists. As in, most victories first, and then points. If so, that'd make Dylan 70th!

1) Conor Travers
2) Dylan Taylor
3) Jack Hurst
4) Edward McCullagh
5) Giles Hutchings

Anyway, we're posting all of this in the wrong thread! It's wasting space so if you'd like to keep debating, I'd suggest putting all future posts in the "Best ever contestants" thread.

From the neutral's point of view, there's no way he'd me remembered for being the best. From the person who knows much more about Countdown (Apterites etc.), they may see where I'm coming from and agree with me, perhaps. Just remember, if Dylan had been in any regular series before his time, he'd ha...

But yeah, I can totally understand why some of you don't feel Dylan should be #1. It's just like the seeding system with Countdown - more wins first, then points.

7/5 points is available as a max*

Andy was very unlucky that ODORIZER wasn't in. It used to be and is set to be added again next year. He only got 11 maxes in his final, Jack H got 14. Also, I don't class a numbers game where you get 7/5 points but a better 7/5 points is available, and I know Giles had a few of these in his octorun.

Because he did better in the CoC than Edward, and Giles probably shouldn't have won his series final, statistically.

Because he was trophyless (I don't feel that's the #1 factor) and just look at Conor's amount of maxes. I think it was 86/90, which overshadows the rest. Amazing.

and therefore have a higher % of max*

Example: two people play an octorun which has exactly the same maximum, in this case, it's 1000. Player 1 scores 964 and gets 118/120 maxes. Player 2 scores 965 and gets 85/120 maxes. As you can see, the only way in which Player 1 could have dropped 36 points from two rounds is if he/she had failed ...

Well, this brings me to another point of mine - eras. If Dylan, or Jen, had been on a few years ago at the same standard now, then they'd have won their series, for sure. So they're unlucky that they were both in a solid series. Also, I've found the perfect reason as to why collection of maxes is a ...

My top 3 would be: 1) Conor Travers 2) Dylan Taylor 3) Jack Hurst I feel that winning a series doesn't mean much, as Dylan has better statistics than any other winner of a regular series, but did he win it? No. Just look, he has the highest amount of maxes from a series by some way - 140/65, and the...

... Or he was for his octorun. I'll have to do the further calculations now.

You are right, I am a big fan of Dylan. Still, that has nothing to do with the conundrum rounds being eliminated altogether, as the thing with the timing of them makes it extremely difficult to measure. And, I did do a leaderboard (I think) of what the "impossible to get lower than" scores...

If we were to do the letters and numbers thing, then the top 3 would look like this: 1) Dylan Taylor: 96/112 (85.7142857143%) 2) Jen Steadman: 93/112 (83.0357142857) 3) Edward McCullagh: 89/112 (79.4642857143%) ... This means that it's a guarantee that the 3 above would have score in at least the am...

Yeah, this is an issue. I wonder if anyone's thought about going through someone's entire octorun with the ODO, making changes to valid/invalid words. Dylan's run would change, as he'd have OWLERIES allowed, giving him 982 and the highest % of maximum achieved. I'm sure plenty others would be affect...

Yeah, as someone mentioned at the end of Jen's run, your amount of maximums is definitely a better way of measuring than % of overall maximum achieved. This is because, say, in a certain round, you could lose 18 points with that way of measuring. The collection of maximums is better as whenever you ...

If we are to do that, then the points will remain the same, but the percentages will probably increase, as there's a lower maximum available. The equation for this is: ("impossible to get lower than" score) / (octorun maximum - 80 points) x 100 = percentage. Here's the top 5: 1) Dylan Tayl...

Yeah, the conundrum aspect is definitely debatable. If we were to include the points got from those, it would look like this: 1) Dylan Taylor - 873 2) Jennifer Steadman - 846 3) Giles Hutchings - 818 4) Edward McCullagh - 804 5) Jack Hurst - 797 But, this goes back to my point of "impossible to...

which you got in those rounds*, "impossible to get lower than"*

I've just done some research on something which I'm sure many will have briefly thought about doing (apologies if it already has and I'm an idiot). It's the calculation for what your score would've been if your opponent maxed every single round throughout. So, the way to do this is by taking all the...

No, I didn't mention "numbers" once in it

It's impossible to get 100% max percentage if you lose, though. And no, I'm not, Jon.

How, Gavin?

How about a table which shows the top 10 players who have the highest % of max from their time on the show in a series? They can have played however many games. The equation for this should be: = (Total Points scored / Total max available) x 100. An example, let's look at Jack Hurst's run. He had 12...