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This problem is from khan academy:

John just bought a brand new cell phone and is considering buying a warranty. The warranty costs 200 euros and is worth 1000 euros if his phone breaks. John estimates that there is a 10%, percent chance of his phone breaking.

Find the expected value of buying the warranty.

Step by step guide to the solution

Now, the reason why I created this question is because I am unsure of what the conclusion is. I assume that since the expected value is -100 euros, it means that we can expect to lose 100 euros by buying this warrenty; thus, we should not buy the warrenty.

  1. Is my assumption (conclusion) correct?

expected value function, where x = the probability of braking phone

Intersection with the x axis

  1. Would it be correct to assume that:

if the probability of the phone braking is over 20%, we should buy a warrenty, and simiarily avoid buying one if the probability is less than 20%.

Ryan Cameron
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  • @MorganRodgers If we look at it as a casino game. So you have 10% chance of winning 800, and 90% chance of losing 200, would it be correct to say, that we over time would lose money; thus we should avoid playing the game all together. back to our warranty case. We should avoid buying the warrenty ... Right? As we over time would lose money on buying these warrenties, right? – Ryan Cameron Oct 11 '18 at 22:34

1 Answers1

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The expected overall value of the warranty, as calculated, is -100 Euros.

On the other hand, assuming that without the warranty John will still purchase a new phone if his current one breaks, then there is a 90% chance John will pay nothing and a 10% chance he will have to pay 1000 Euros, making the expected value of not buying the warranty also -100 Euros.

In this case, as it turns out, there is no change in the value between buying and not buying the warranty so based on these calculations it doesn't matter if he does or not. However, if John valued certain outcomes differently (e.g. he doesn't want to have to pay large chunks of money, or he hates filling out forms), then you would need to used a different measure of the expected utility of the two options.

ConMan
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  • The second paragraph computes the expected cost to John if he doesn't buy the warranty. An accurate comparison would be with the expected cost if he does buy the warranty, namely 200 Euros. The quoted "expected overall value of the warranty" of $-100$ euros is the difference between these. It's the expectation of how much worse off John is when he buys the warranty. – Andreas Blass Oct 11 '18 at 22:31
  • No, the calculation in the solution is the expected cost of buying the warranty - if his phone does break, he's down 200 but up 1000 with probability .1, and if it doesn't he's down 200 with probability .9, giving an overall expected value of -100. – ConMan Oct 11 '18 at 22:34
  • If you're going to count "up 1000" if he has the warranty and the phone breaks, because he has a nice new phone worth 1000, then in the second paragraph of your answer you should also have included the nice new phone, balancing out the 1000 that he had to pay. – Andreas Blass Oct 11 '18 at 22:47
  • You might be right. It's probably more accurate to say that "If he has a warranty and the phone breaks they pay him 1000 to buy a new phone, which offsets the 1000 lost value of the broken phone", whereas in my scenario "If he has no warranty and the phone breaks then he has lost 1000 worth of phone". So I probably need to rearrange my calculations to be on the same basis as the KA ones. – ConMan Oct 11 '18 at 23:24
  • The simple way to see the foolishness is to follow your assumption that he will buy a replacement if the phone breaks. At the end of the warranty he has the same phone whether it has broken or not, regardless of whether he bought the warranty. If he buys it he $+800$ if it breaks and $-200$ if not. A $10%$ chance makes it a clear losing bet. – Ross Millikan Oct 12 '18 at 02:26