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If I have 1 in 20 probability of a theft...

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    The sharing and pooling of risk is still, however, crucial. in the real world, the pattern of losses (cars stolen or houses burning down) is unsteady. That said, the following is an illustration of the application of the law of large numbers.

    Suppose that statistics in a particular country show that, on average, 1 car in 20 is stolen each year. If the thefts are independent of one another, an insurer who had only insured 20 cars would discover that there was a 1 in 4 chance that 2 or more would be stolen, which would double their expected outlay on claims. This does not qualify as good business practice.

    However, if 100,000 cars are insured, the probability that more than 10,200 (or less than 9,800) will be stolen is only about 1%.

    Sourced from Business Knowledge for IT in Insurance
    ISBN 978 0955 412 431
    Page 120.

    Would you kindly explain the logic behind the above statement?

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    I have two coins, probability of 2 heads is 1 in 4. I have 100,000 coins, probabiltiy of 100,000 heads? For practical sake is 0.

    EDIT: oh sorry, were you referring to the bolded bit?
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    Well, I think this relates to statistics not just a guess, however I was asking for confirmation because statistics is not my speciality.
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    they probably assumed that the events that each car is stolen are all independent (i.e. if one car is stolen, it does not make it more or less likely that another car is too)
    They probably assumed all cars are equally likely to be stolen (i.e. that supercar with no security system is just as likely to be stolen as that bust up corsa)

    Then, if you wish to find the probability that say 1 car in 3 is stolen. and the cars are ABC. There are 3 ways this could happen - only A is stolen, only B stolen, or only C stolen. Each way has the same probability p(1-p)(1-p), where p = 1/20. (this is the part which requires the assumptions)
    If you wanted to find the probability that 2 in 4 were stolen, you find there are 6 ways this can happen, all with probability (1-p)p^2 (by the two assumptions at the top)

    We call this setup, the binomial distribution. Look it up. There is a nice way of calculating these results when the numbers get large, and it's hard to count the permutations in your head.
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    If you are strictly talking about the part in bold, this isn't "really" a mathematical question, more one of business/economics. If you have a 1 in 4 chance of having to pay out two claims, you will have to have a higher capital reserve (or alternatively, you could lose profits by laying off some of the insurance risk to a reinsurer).
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    An example of the law of large numbers (mentioned above) is flipping a coin. After a very large number of flips, we expect the proportion of heads : tails to be around 50 : 50.

    In this specific case, imagine you insure 20 cars. The average rate of theft is 1 in 20, so you would reasonably guess that only 1 car will be stolen.

    However, as the article says, there is a 1 in 4 chance that 2 or more cars will be stolen. So in the relatively likely case that 2 cars are stolen, you end up paying out twice as much as you budgeted - which is a huge variance on how much you expected to pay out.

    Now imagine you insure 100,000 cars, as the article says. You expect for 5,000 to be stolen. The likelihood is that the amount of actually stolen cars will be very close to this - the probability that the amount of stolen cars will double to 10,000, like in the previous case, is extremely small (almost 0). The likelihood of the amount of cars stolen being between 4,800 and 5,200 is ~99%. So, in proportion to how much you budget for 5,000 cars to be stolen, it is likely that the actual amount you will pay out in claims will be very close to the amount you budget.

    Basically, the small variations before - just one extra car stolen - had a huge effect on the final payout, in proportion to how much was budgeted. When the numbers get bigger, one extra car stolen is almost irrelevant, compared to the amount that was budgeted.

    The laws of large numbers say that, for a given probability distribution, the mean of the observations from this distribution converges to the expectation of that distribution. In the case of flipping a coin, we know that the probability distribution is 0.5 heads, 0.5 tails. So we should see a 50 : 50 split between heads and tails, over a long period of time. However, in the short term, it is easy to see that we could feasibly have a run of 5 heads in a row - this is what the article talks about. As the number of cars insured increases, the number of cars stolen "settles down" to the average - it gets much more unlikely that twice as many cars as budgeted for get stolen.

    Mathematically, we can approximate the car insurance case by treating each car as a 20 sided die. Rolling a "1" is equivalent to the car being stolen, and rolling from 2 to 20 is equivalent to the car not being stolen. This type of situation is best represented using the binomial distribution, with the number of trials being the number of cars insured, and the probability of success (in this case, a car being stolen, which is equivalent to rolling a "1") being 0.05. If you're interested in some statistics, then take a look at the binomial distribution - this will allow you to calculate the probabilities that are used in the article.


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