cockbeard wrote:

Right I think I might get you

Spoiler:

1+(1/100)+(1/200)+(1/300), up to (1/4900), that's a hella lotta steps, does it get any faster when he crosses half way, as most of the stretchiness is behind him?

I don't think your end point is right - he has way further to go than that!  I can't really get my head round what changes at half way, but I'm sure something must!  It's all very counter-intuitive!

The Daddy wrote:

This is a bit like the Achilles and the tortoise puzzle isn't it? I use that when I introduce mechanics at A level. The caterpillar guy moves with the stretch, so at the beginning of day 2 he's still 1% across the bridge even though it's lengthened. And so on.

Zeno's paradox?  A little bit like that.  And yes the 1% thing is right.  But what's the answer to the question?

If the bridge only extends ahead of him, then after a full day he is 1cm from the start with 199cm to go, then 2cm from the start with 298cm to go, and he never gets to the end.

If it extends both ways, then after that first day he's 51cm from the start with 149cm to go, and then 102cm from the start with 198cm to go... and my brain tells me he still never gets to the end. I think.

And I've realised I kinda misread the question.

I don't think he gets 51 in the first leg, because he's travelled 1cm then the bridge doubles, so he should at that point be 1.5 cm in, walks 1 cm, so 2.5cm at which point the bridge extends by 50%, so he's 3.75cm, before moving 1cm, at which point the bridge extends by 1/3, etc

Gah. I've got as far as

x = 100(y+1) - ((something)y + 1)

where x is the distance left to travel, and y is the number of days, but I'm realising how much my brain has lost its grasp on this through lack of practice. The (something) is a percentage gain with each stretch, but I can't get there.

He makes it if the bridge is 2cm long so I expect yes.

Edit: Bollocks.

It's not just that though. Because each growth moves him a little bit further.

I'm still not sure he ever gets there, though, because the growth is always more than the distance travelled, and I don't think he ever gets to within 1cm (by which time he's off before the bridge extends).

Wrong!

As above, the proportion of the bridge he's covered is 

1/100 x (1+1/2+1/3+1/4+...)

= 1/100 x (1 + 1/2 + (1/3+1/4) + (1/5+1/6+1/7+1/8 ) + (1/9+...1/16) + ...)

> 1/100 x (1 + 1/2 + 1/2 + 1/2 + 1/2 + ...)

So he will definitely make it.  It's just the nth half-percent of progress takes him order 2^n moves, and yeah 2^200 is like fucking enormous...

Very counter-intuitive though, I still can't quite get my head round it.

I have another question about maths by the way.  And one about dice...

I only wandered in here by mistake btw.

acemuzzy wrote:

Wrong! As above, the proportion of the bridge he's covered is  1/100 x (1+1/2+1/3+1/4+...) = 1/100 x (1 + 1/2 + (1/3+1/4) + (1/5+1/6+1/7+1/8 ) + (1/9+...1/16) + ...) > 1/100 x (1 + 1/2 + 1/2 + 1/2 + 1/2 + ...) So he will definitely make it.  It's just the nth half-percent of progress takes him order 2^n moves, and yeah 2^200 is like fucking enormous... Very counter-intuitive though, I still can't quite get my head round it. I have another question about maths by the way.  And one about dice...

I read it as the distance increased in front of him, but it's stretching isn't it, so it increases in front and behind. Doh.

No but someone mentioned tortoises, and like rabbits it's always sums if series.

hahaha

Nowhere near as complex as some of the things I've seen in this thread, and apologies if it's already been posted: the half and double method.

It's used in casinos as a way for croupiers to calculate roulette payouts a bit quicker.

So, in roulette, a bet on one number (straight up as it's known) pays 35 to 1. A bet on the line between two numbers (a split) pays 17 to 1. So, if you have 3 chips on number 1 and the ball drops there, you'll win 105 chips, plus the 3 you had bet. If you had 3 on the line separating 1 and 2 and the ball dropped into number 1, you'd win 51.

If there is an equal and even amount of chips on the straight up and the split, the dealer can use the half and double method. For example: 8 chips on number 5, and 8 on the line separating 5 and 6. Rather than doing the two calculations separately then adding them together (8x35 then 8x17), you half the 8, then double it, then put those numbers together, giving you 416. (Half of 8=4, double it=16).

Have i explained that correctly? I was a dealer for a year before it clicked. So, for ten on the number and ten on the split, it'd be 520. And so on.

Next, I'll try and explain how we calculated payouts on punto banco (a form of Baccarat) so quickly.