Statistics: Expectations Question

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The discussion revolves around calculating expected profits and fair game costs in probability scenarios. For the insurance problem, the expected profit for the company is calculated as $300 (premium) minus $40 (expected loss from a 0.002 chance of a $20,000 payout), resulting in an expected profit of $260. In the dice game scenario, the user initially miscalculated the probability of rolling doubles, which should be 1/6 instead of 1/12, affecting the fair price to play the game. After clarification, the correct amount to charge for the game to be fair is determined to be $0.83. The thread concludes with users helping each other understand the calculations involved in these expectation problems.
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Hi guys, I'm having a hard time figuring out these two expectation problems below. Can someone please help me?

An insurance company insures a person's antique coin collection worth $20,000 for an annual premium of $300. If the company figures that the probability of the collection being stolen is 0.002, what will be the company's expected profit?

For this problem, here's what I got:
x...300...?
p(x)...0.998...0.002

I can't seem to figure out what to put in the ? (loss). Maybe i just don't understand the wording of the problem... I mean, it's saying that if the antique does get lost, the company would pay the insurer $20,000? If not, then there's really no way of figuring out what the company's loss is since it doesn't say it...

The answer is: $260.

If a person rolls doubles when he tosses two dice, he wins $5. For the game to be fair, how much should the person pay to play the game?

For this problem, I have:

x....5 ...?
p(x)...1/12...11/12

The reason I got 1/12 for the winning side is because there's 3 chances of getting doubles (rolling a 4 & 6, 5 & 6, and 6 & 6), and the total possible outcome is 36 (6 x 6)

And for the 11/12, I just subtracted 12/12 from 1/12. And this is as far as I got...

The answer is: $0.83Any help would be appreciated. Thanks in advance!
 
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Clearly, is there is a 0.002 chance of it being stolen - then that is the probability they have to pay out $20,000. So the expected loss is (20,000)*(0.002) = $40. Take that away from the $300 they would make otherwise... and the expected profit is $260, as desired.

As for the second problem - what is "doubles"? I was assuming that "doubles" means double 1s, double 2s, etc. In which case there are six possibilities of getting this, which means he stands 1/6 of a chance of doing so. Try this hint instead.
 
Thanks, got the right answer!
 
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