(Blah Blah Blah, maybe utter balls)
As of January 2010 and according to http://ojp.nationalrail.co.uk/en/s/seasonticket/tickets the cost of a season ticket from Edinburgh to Dalmeny (return) is:
| 7 Days | £18.80 |
| 1 Months | £72.20 |
| 3 Months | £216.60 |
| 6 Months | £433.20 |
| 12 Months | £752.00 |
Since I'm currently paying per week, my current cost is around 52 * 18.80 = 977.6. So, 977.6 - 752 = 225.6 is the saving if I travelled all year.
However, I don't. I have around 27days of work holidays a year and this
expands into roughly 27 * 7/5 actual days of holiday (because of the weekend). Since with a weekly ticket I can decide to skip payment on these weeks, this means I actually only use the train for this many weeks a year:
52 - ((27 * 7/5) / 7) = 52 - 5.2 = 46.6
This means the saving is:
46.6 * 18.80 - 752 = 124.08
So, still positive, but not as much as you might think. I will also lose as much as £752 worth of value if I lose this ticket during the year.
Given that I have an equal chance each day of losing the ticket (I carry it in my wallet), how big would this chance have to be to make it not worthwhile? That is, when does my expected loss become greater than my expected saving?
I can frame this is a bet, where, if I win, I never lose the card and I get £752. However, I can also lose on day 1, or day 2, or day 365. On each of these days, I lose a different amount, depending on how far through the year I got. So, I either:
lose on day one
OR (don't lose on day 1 AND lose on day 2)
OR (don't lose on day 1 AND don't lose on day 2 AND lose on day 3)
...
OR (dont' lose on day 1..364 AND lose on day 365)
OR (never lose on any day)
= 1.0, because it covers all possibilities
The chance of losing on a single day is equal to any other day, so call it L. The chance of keeping my ticket for one day is 1.0 - L, call this K. This makes the above become:
L
+ K * L
+ K^2 * L
...
+ K^364 * L
+ K^365
The expected winnings are the sum of value I get from travelling for one day, call this T:
L * 0T
+ K * L * T
+ K^2 * L * 2T
...
+ K^364 * L * 364T
+ K^365
Let's just consider the head of this sequence, where I always lose at least once:
(L * 0T) + (K * L * T) + (K^2 * L * 2T) + ... + (K^364 * L * 364T)
= L((K * T) + (K^2 * 2T) + ... + (K^364 * 364T))