>“With approximately 70 auto burglaries a day”
That doesn’t seem like a crisis in a city the size of San Francisco.
That doesn’t seem like a crisis in a city the size of San Francisco.
This doesn't obviously address the issue that I doubt there's even 400k cars there which more than doubles the likelyhood. That's not nothing, it's over 25k stolen cars in a year and an enormous burden on the local police force.
Of course, with both cars and bikes, police have much more important to do. That's the invariant: whatever the crime, there is surely something more worthy of their attention out there, so nothing is ever done.
I'm only half-joking.
Naturally, it's the one window that even my Haynes Book says "don't try replacing it yourself; hire a professional". Thanks, guy-who-desperately-wanted-my-jumper-cables-and-jump-box-out-of-the-back-of-my-car. Real considerate of you.
Smash and grab makes sense if you don't have those tools and know no one, not even pedestrians, react to broken glass or car alarms in that part of town.
edit: slightly disappointed I was downvoted for this; I literally just googled how many cars there were in SF and tried to work out the percentage. Am I factually incorrect? If not, can someone please explain what I've done wrong so I don't do it again, please.
For calculating probabilities, it's much easier to work in fractions than percentages, so I'll convert 0.01% to 0.0001 and equivalent for other percentages for the remainder of this comment.
If every day of the year, 0.01% of cars are broken into, the probably of being broken into is 0.0001 and the probably of being safe is 0.9999. To calculate the odds for a year, you need to take the being safe probability to the 365th power (0.9999^365). That gives 0.9642. This is the probability of being safe from break-ins for a year. You can subtract from 1 for the annual odds of being broken into: 0.0358.
0.01% per day is 3.58% per year. Also can continue to get longer term odds. Break-in chance is 30% over 10 years. At 19 years, it's 50/50 whether you'll be broken into. At this term, you get a big deviation from the incorrect calculation, which would give 0.01 * 19 * 365 = 69.35% chance of break-in over 19 years.
Of course, this ignores the uneven distribution of break-ins, etc., etc.
(70*365) / 500,000 = 0.0511
You have about a 5% chance of having your car stolen in SF annually.
Or to replicate your process:
70 / 500,000 = 0.00014
1 - 0.00014 = 0.99986
0.99986^365 = 0.950180248299
Again, about a 5% annual rate.