c4countdown

It really has been Ladies Week on Countdown, with Nick and Aliraza being the only males in front of the camera this week so far. And appropriately, a woman has won all 4 games, and without meaning to rhyme, her name is Victoria James. She's had some good wins and some not so good wins, but if she wins one more game, she may be the number 2 seed even with a loss on Monday. Can she do it? It's the last day of Kate Humble in dictionary corner, and I'm glad to see her go just because of my internal struggle of whether I liked her or not. Hopefully with the next guest it's clear cut whether I like this person or not, but we shall see.

Join Thomas for the recap later.

Jordan F wrote:without meaning to rhyme, her name is Victoria James.

That doesn't rhyme.

Ryan Taylor wrote:

Jordan F wrote:without meaning to rhyme, her name is Victoria James.

That doesn't rhyme.

But if Nick's been pronouncing her name wrong the entire time? What if the s is silent and "Jame" sounds exactly like "rhyme?" I've thought of these things Ryan, I've thought them out deeply.

Jordan F wrote:It really has been Ladies Week on Countdown, with Nick and Aliraza being the only males in front of the camera this week so far. And appropriately, a woman has won all 4 games, and without meaning to rhyme, her name is Victoria James. She's had some good wins and some not so good wins, but if she wins one more game, she may be the number 2 seed even with a loss on Monday. Can she do it? It's the last day of Kate Humble in dictionary corner, and I'm glad to see her go just because of my internal struggle of whether I liked her or not. Hopefully with the next guest it's clear cut whether I like this person or not, but we shall see.

Join Thomas for the recap later.

According to the audience ticket schedule it should be Gloria Hunniford next week and Anne Widdecome after that, whereupon I think you'll conclude that Kate Humble is not so bad after all. Hunniford in particular falls into that smug self-satisfied group that was so prominent in RW's days (think Sheridan Morley, Brandreth, Martin Jarvis et al). IMHO of course....

Edit: although just to be fair I should add that she has had to deal very publicly with the death of her daughter (Caron Keating), every parent's nightmare, and has subsequently raised a lot of money for people who have to care for cancer patients.

"Many a mickle makes a muckle". If you look this up, you'll discover that it's nonsense. I expected far better of Susie than this.

I thought Dave was a bit unlucky there today. Victoria definitely looked a bit off today and was definitely grateful for that horrible last numbers round.

CANOPIES in round 2 (although I'm still mortified by finally knowing what CAPONISE means now ), TROPHIES in round 3, INJURE in round 6, TRIGONAL in round 7 and TUTORED in round 11.

2nd Numbers Alt.: ((75 + 5) x 2) + (9 x 3) = 187

Anyone else noticed that the first 2 numbers games had the exact same numbers come up in the exact same order

Keith Bennett wrote:According to the audience ticket schedule it should be Gloria Hunniford next week and Anne Widdecome after that

Uh oh.

3rd numbers: ((25x5x3)+7)x2+5 = 769

Victoria might not have been quite as good today as she was recently (maybe because of tiredness) but the lightning quick conundrum solve was pretty impressive.

Jack Worsley wrote: Victoria might not have been quite as good today as she was recently (maybe because of tiredness).

Yeah might have been a late one. She's definitely faded last 2 episodes, she was really quite sharp on Wednesday

James Robinson wrote:Anyone else noticed that the first 2 numbers games had the exact same numbers come up in the exact same order

No! That's crazy. I'm usually all over stuff like this.
Maybe with the human factor they were just shuffled really badly.

Edit: Jack, if she goes far there's a good chance you could play her in finals. Looking forward to 30 points in the numbers?

Andy Platt wrote:

Jack Worsley wrote:
Edit: Jack, if she goes far there's a good chance you could play her in finals. Looking forward to 30 points in the numbers?

If she gets to the finals, she might want to work on her numbers a bit. I'd definitely fancy my chances of getting a few points off her on the numbers, might even ditch four large as I'd probably have a decent enough chance of winning on the numbers rounds anyway. If her numbers were anywhere near as good as her letters rounds and conundrums, she could easily have got a couple of centuries by now.

James Robinson wrote:Anyone else noticed that the first 2 numbers games had the exact same numbers come up in the exact same order

Does anyone remember this game? Perilously close to the hat-trick.

This was one of Rachel's first games and IIRC she was tending to pick the same pattern from the board each time. That certainly wasn't the reason this time.

Probability of two fair, random, one-large games having the same selection in the same order: (1/4) * (2/20) * (2/19) * (2/18) * (2/17) * (2/16) = 32/7441920 = 1/232560. That assumes all the numbers picked are different, though. For selections where you get two of the same number I think the probability is slightly lower.

Jack Worsley wrote:3rd numbers: ((25x5x3)+7)x2+5 = 769

Victoria might not have been quite as good today as she was recently (maybe because of tiredness) but the lightning quick conundrum solve was pretty impressive.

It was impressive, but after the conundrum was revealed Nick did faff about for a bit before starting the clock.

Confirm it's Gloria H. next week, recorded January 18, I was there for the final 3 of the week in the p.m.. Smug? Maybe, and I have mixed feelings about her, but she usually has some good banter with Dudley in front of the audience beforehand.

Graeme Cole wrote:Probability of two fair, random, one-large games having the same selection in the same order: (1/4) * (2/20) * (2/19) * (2/18) * (2/17) * (2/16) = 32/7441920 = 1/232560. That assumes all the numbers picked are different, though. For selections where you get two of the same number I think the probability is slightly lower.

I often get these wrong, but if you assume all the numbers are different, isn't it (1/4) * (1/10) * (1/9) * (1/8) * (1/7) * (1/6) = 1/120960?
And the chances of all the numbers being different is 1 - ((18/19) * (16/18) * (14/17) * (12/16)) = 48%.

Jack Worsley wrote:

Andy Platt wrote:

Jack Worsley wrote:
Edit: Jack, if she goes far there's a good chance you could play her in finals. Looking forward to 30 points in the numbers?

If she gets to the finals, she might want to work on her numbers a bit. I'd definitely fancy my chances of getting a few points off her on the numbers, might even ditch four large as I'd probably have a decent enough chance of winning on the numbers rounds anyway. If her numbers were anywhere near as good as her letters rounds and conundrums, she could easily have got a couple of centuries by now.

If you get to the finals as well. I was in a similar position and I wasn't comfortable until my name went bold on the wiki standings.

David Williams wrote:

Graeme Cole wrote:Probability of two fair, random, one-large games having the same selection in the same order: (1/4) * (2/20) * (2/19) * (2/18) * (2/17) * (2/16) = 32/7441920 = 1/232560. That assumes all the numbers picked are different, though. For selections where you get two of the same number I think the probability is slightly lower.

I often get these wrong, but if you assume all the numbers are different, isn't it (1/4) * (1/10) * (1/9) * (1/8) * (1/7) * (1/6) = 1/120960?
And the chances of all the numbers being different is 1 - ((18/19) * (16/18) * (14/17) * (12/16)) = 48%.

If the selection in one round is, say, 75 7 2 5 3 9, then in the next numbers round, the probability of picking a 75 is 1/4, then the probability of picking a 7 next is 2/20 (because there are two sevens out of 20 smalls), then the probability of picking a two next is 2/19 (there are two twos out of 19 smalls left), and so on.

Of course, if the selection was 75 7 7 5 5 9, then the probability of getting the same selection in the same order next time would be something like 1/4 * 2/20 * 1/19 * 2/18 * 1/17 * 2/16, so it doesn't work if there are repetitions.

Graeme Cole wrote:If the selection in one round is, say, 75 7 2 5 3 9, then in the next numbers round, the probability of picking a 75 is 1/4, then the probability of picking a 7 next is 2/20 (because there are two sevens out of 20 smalls), then the probability of picking a two next is 2/19 (there are two twos out of 19 smalls left), and so on.

Yes, always a mistake to try to think clearly after midnight. I could see how you'd arrived at your answer, but decided that on the third pick there were only 18 smalls available, because one of them was the other 7, invalidating your assumption. But that's nonsense!