please empty your brain below
dg writes: Blogged here.
dg writes: Added, thanks.
I’ve often wondered who created it and how.
Dg was it you??
dg writes: No, it dates back to the 1980s.
dg writes: Added, thanks.
As to the 50/50 thing, there is also how frequently different letters in the alphabet are used, although a map of railways means that these results may well be different from normal, because of all the station names that contain 'Central', 'East', 'North', 'South', 'Town', 'West' etc.
Hounslow East, Hounslow Central, Hounslow West, Hatton Cross, Heathrow Terminal 4, and lastly Heathrow Terminals 1,2,3.
Ken is this the run of stations you mention?
dg writes: Famously, there are 10.
Not necessarily: after leaving Moorfields on the way back to Bidston you go round the loop, passing through Liverpool Lime Street and Liverpool Central, before reaching James Street. Thus there are nine (not seven) in reverse alphabetical order
Is this due to enforcement of the Bow Omnibus Halt Sign M (alternate Wednesdays) Rule?
Yes, and yes. Thanks
However, DG said "This is similar to flipping a coin until you get a different result, so a long chain isn't very likely" The only doubtful word here is "so". A long chain isn't likely, and his argument might help to convince us why. The arithmetic/geometric business makes it even less likely, so we have an a fortiori situation.
If you take the direct line to Weybridge, you'll not call at any station with a Q in it.
But if you take the other route, you don't need to go any further than Brentford to find all the letters except Z.
But the first station after Vauxhall is Queenstown Road (Battersea), which has a Q in it, and Gregg is correct.
Letter frequency makes no difference, as it merely the stations' positions in the sequence that matters.
I cannot add anything meaningful to this post, so will probably incur a type character colour change.
But this is what living on the edge is all about.
That's not true, because stations aren't divided equally between the two halves of the alphabet.
For example, 36 more tube stations start with a letter from the first half of the alphabet than the second half. In this particular case the odds Timbo mentions are near enough 2/1, not 3/1.
We can now view this as a permutation problem. Given a shuffle of the numbers 1 to n (with all permutations equally likely), what is the probability the shuffle is in order. If n=1, then the probability is 1. Otherwise there are 2 ordered shuffles (1 2 3 ... and n n-1 .. 2 1) with n! possible shuffles. So for n>=2, the probability is 2/n!.
For a wordy approach, consider a race with each runner equally likely to finish first. If one runner isn't recorded on the timing screens due to a technical issue, even if we know the relative position of all the other runners on the timing screen that one runner is still as likely to have won as before. So in a race of n runners, there is a 1/n chance the runner not on the screen finished first and (n-1)/n chance the leader on the timing screens finished first. The other runners on the timing screen have 0 probability of being first - we know someone who ran it faster.
To then consider a chain of stations, given a chain of n ordered stations, the chance the next one beats (either later if chain is A-Z or earlier if Z-A) is 1/(n+1) - this is both necessary and sufficient for it to be a chain of n+1 stations. So with 2 ordered stations, it's 1/3 chance we then have a chain of 3, 1/4 given a chain of 3 we have a chain of 4, i.e. 2/n! as above.
Note 1: With duplicate station names, it's easiest to give them a unique position in the numerical ordering and then add the probability that only the duplicate station numbers were out of place. This would double the probability if one station was duplicated in the chain, but having two stations with the same name adjacent and treating them as different entities for this seems ridiculous.
Note 2: The less mathematical approach is more often considered as how likely you are to get a personal best if your ability remains constant. (Assumes your results could be strictly ordered best to worst) the probability of your nth try being your personal best is 1/n. At any point, you'd expect at least one more personal best if you tried enough further attempts since 1 + 1/2 + 1/3 + ... never converges.
2 stations are always in alphabetical order.
3 stations are in alphabetical order one third of the time (because there are 6 possible combinations, two of which are in order)
ABC✓ ACB✗ BAC✗ BCA✗ CAB✗ CBA✓
4 stations are in alphabetical order one twelfth of the time (because there are 24 possible combinations, two of which are in order).
The general case is that only two of the permutations can be in order, (one forwards, one backwards), hence the probability is 2/n!.
2 stations: 2/2 = 1
3 stations: 2/6 = 1/3
4 stations: 2/24 = 1/12
5 stations: 2/120 = 1/60
6 stations: 2/720 = 1/360
7 stations: 2/2520 = 1/1260
7 stations should only appear in order 0.08% of the time... but there are a heck of a lot of chains of 7 stations on the London rail map, which is why it manages to occur once.
I'll stick to drawing straight lines on a map. I can do that!
Stratford-on-Avon (Parkway) to Jewellery Quarter covers all 26 letters of the alphabet in 20 stops
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