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June 23rd, 2008, 11:40 AM   #1
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probability of winning a bet with 3 rolls 0-999

A friend and I both play an online game and included in the game is a /random command which generates a number from 0 to 999. He has offered to pay me some ingame money if he is unable to roll over 550 on 2 out of 3 rolls, and if he can I have to pay him.

Now my stats is pretty rusty, but I think it goes something like this

the probability of him winning 2/3 rolls is (0.449*0.449*0.551)*3 + (0.449*0.449*0.449)
which comes to ~42.3%

So if Im right I have the upper hand, can anyone help me out because I'm not at all certain.
Thanks!
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June 23rd, 2008, 11:43 AM   #2
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Re: probability of winning a bet with 3 rolls 0-999

hmm I couldnt find an edit button, but I should specify and say that he has to roll 550 or above on at least 2 out of the 3 rolls, so can roll 2 or 3 550+ numbers and win
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June 25th, 2008, 09:43 AM   #3
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Re: probability of winning a bet with 3 rolls 0-999

From a Bayesian perspective, what is the subjective probability that your friend is better at statistics than you and offering you a sucker bet?
(just kidding)

your calculation is right
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June 27th, 2008, 06:41 PM   #4
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Re: probability of winning a bet with 3 rolls 0-999

Good reasoning, correct setup, correct calculation -- though I would have rounded it to 42.4%.

If your friend wagered 1000 units that he could do it, you should accept the bet if he'll take a bet of 1350 or less from you. (Equal bets would be favorable for you.) This assumes that you're risk neutral and your friend doesn't cheat.
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