Random 15 Challenge - Round 1 - 11th July 2009

[Soup Dragon]

Round 1 of the 2009/10 Random 15 championship takes place this Saturday.
The start time will be at 13:00 at Queensway Station.
This round will include also include any DLR stations in Z1-2.

Specific rulings for DLR stations:
Bank - Will be classed as one station and can be visited by Central, Northern or DLR
Bow Church / Bow Road will count as two seperate station.
Canary Wharf - will count as two seperate stations Canary Wharf (Jubilee) and Canary Wharf (DLR), this being a slightly mute point as the Jubilee station is closed this weekend and hence excluded from the draw.
Tower Hill / Tower Gateway will count as two seperate stations.

Joker Rulings
Each competitor has one joker to play throughout the entire championship, you must advise the BERC that you wish to play your joker when you hand your time in at the pub, don't be put off if other competitors are already at the pub, they might have finished very close to the pub or maybe they aborted, therefore no discussions should take place amongst competitors about their times until everyone has reported to the BERC. The BERC must decide on playing his joker before gathering anyone elses time.

Point Scoring
1st = 10pts
2nd = 8pts
3rd = 6pts
4th = 5pts
5th = 4pts
6th = 3pts
7th = 2pts
8th = 1pt

[jamesthegill]

The BERC must decide on playing his joker before gathering anyone else's time.

And presumably notify somebody else that he's playing his joker, rather than announcing that he's come first "oh and by the way, I played my joker on that round. I decided it before all of you handed in your times, honest."

[Soup Dragon]

The BERC must decide on playing his joker before gathering anyone else's time.

And presumably notify somebody else that he's playing his joker, rather than announcing that he's come first "oh and by the way, I played my joker on that round. I decided it before all of you handed in your times, honest."

Perhaps if I give someone a sealed envelope stating that i have/haven't played my joker, this can be used to verify.

[Soup Dragon]

The R15 has just been generated by Ping Pon and the starting station for Saturday is.... Queensway start time 13:00, please try to be at the station 15 minutes before the start.

[greatkingrat]

I have just worked out there is a 7.2% chance that you will not have to visit any DLR stations at all on this R15

[Going Underground]

See you all at Queensway for 12:45hrs then.....

[Edgemaster]

I hope to be there.

[mps247]

I'm not sure if you want confirmation of participation, but just in case - I'll be there!

[Going Underground]

I'm not sure if you want confirmation of participation, but just in case - I'll be there!

I guess it's always nice to know who is coming so we know if we are waiting for anyone or not

[mps247]

I have just worked out there is a 7.2% chance that you will not have to visit any DLR stations at all on this R15

I got a different answer of 6.1%. The difference isn't very substantial though. Here are the probabilities I calculated copied and pasted from a quick and dirty computer program, where k is the number of DLR stations drawn:

P( k = 0 ) = 6.07587%
P( k = 1 ) = 20.0146%
P( k = 2 ) = 29.0504%
P( k = 3 ) = 24.5811%
P( k = 4 ) = 13.5196%
P( k = 5 ) = 5.10173%
P( k = 6 ) = 1.35887%
P( k = 7 ) = 0.258833%
P( k = 8 ) = 0.0353224%
P( k = 9 ) = 0.0034252%
P( k = 10 ) = 0.000231433%
P( k = 11 ) = 1.05197e-05%
P( k = 12 ) = 3.02557e-07%
P( k = 13 ) = 4.8997e-09%
P( k = 14 ) = 3.34762e-11%

So, there is a 87% chance that we will end up with 1-4 DLR stations.

Anybody interested in the maths, please read on for a brief description (you might be able to point out any mistakes I've made)... at first I was thinking of using the binomial distribution to calculate this. However, as drawn stations are not replaced, a more accurate result could be derived by modelling a more complicated hypergeometric distribution:

P(k) = f(k;N;m;n)

where N = 139 (the total number of stations in the sample, ie. 115 tube stations and 24 DLR stations), m = 24 (the total number of DLR stations in the sample), and n = 14 (the number of draws remaining). The total number of tube stations excludes Queensway, as this has already been drawn, and any stations that are unavailable due to engineering works.

The function f(k;N;m;n) can be expressed in terms of binomial coefficients, which can be written as:

P(k) = (m! / (k! * (m-k)!) * (N-m)! / ((n-k)! * (N-m-n+k)!)) / N! / (n! * (N-n)!)

Ermmm... yes, that isn't so easy to understand... sorry. There is no such thing as a bad student - only a bad teacher.

If anybody spots a mistake in any of that, please let me know!

Edit: 115 tube stations, not 116 as I previously posted.

[greatkingrat]

I used the same formula with slightly different numbers. I have N=141 and m=23. I counted Bank as a tube station rather than a DLR station and maybe I missed off one of the closed stations.

I had the following stations as closed
Bermondsey
Bow Road
Brixton
Bromley-by-Bow
Canada Water
Canary Wharf (Jubilee)
Highbury & Islington
North Greenwich
Pimlico
Southwark
Stepney Green
Vauxhall
Queensway (starting station)

[tubeguru]

This is fascinating.

[mps247]

I used the same formula with slightly different numbers. I have N=141 and m=23. I counted Bank as a tube station rather than a DLR station and maybe I missed off one of the closed stations.

Yes, N should be 141, not 139. However, m is definitely 24 (I counted Bank as a tube station, so unless there is a DLR closure I don't know about it, there are 24 DLR stations left in zones 1 and 2. Tower Gateway and Pudding Mill Lane are probably the easiest to miss when counting them).

So, with new values of N = 141 and m = 24, the probability table looks like:

P( k = 0 ) = 6.36271%
P( k = 1 ) = 20.5565%
P( k = 2 ) = 29.2685%
P( k = 3 ) = 24.2984%
P( k = 4 ) = 13.1143%
P( k = 5 ) = 4.85715%
P( k = 6 ) = 1.26999%
P( k = 7 ) = 0.237504%
P( k = 8 ) = 0.0318277%
P( k = 9 ) = 0.00303121%
P( k = 10 ) = 0.000201187%
P( k = 11 ) = 8.98441e-06%
P( k = 12 ) = 2.53907e-07%
P( k = 13 ) = 4.04096e-09%
P( k = 14 ) = 2.71371e-11%

(I did use an input value of m = 23 just to check I got the same answer as you for k = 0. I got 7.2%, so I'm confident the program is working correctly.)

In testing, there has been only one instance where I saw no DLR stations. Most commonly there is 2, and a few times I have seen 4

This is expected from what I have above. The chances of 4 DLR stations being drawn is just over twice that of none of them making an appearance.

Most importantly, there is a 0.00000000003% chance of the remaining 14 stations being on the DLR! A very quick challenge this Saturday!

This is fascinating.

If you wish, I can give a lecture series.

[greatkingrat]

The station missing was Limehouse. Actually this is a problem with Matt's R15 generator as it does not seem to be in the system at all.

I don't know if this means the stations need to be redrawn?

[Mitchell&BrownLook]

This is fascinating.

If you wish, I can give a lecture series.

Please do Hassan. I look forward to more unadulterated sarcasm from the King of the subject.

I'll also add that 0.0000000000003% is equivalent to a 1 in one quadrillion chance. Winning Euro Millions is 1 in 76,275,360.