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You go gambling with a pair of loaded dice. Because of this, your odds of winning are 53% on every throw. Assuming the game pays 2:1 and you keep betting the same amount, how many games do you need to play to ensure an 80% likelihood of winning money?

I am lost with where to start on this problem. I would appreciate some help so I can figure it out. Thanks.

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    $\begingroup$ Hi rdittmer, welcome to the site! As this is a self-study question, could you please add the self-study tag? Thank you. $\endgroup$
    – COOLSerdash
    Commented Oct 31, 2013 at 7:00
  • $\begingroup$ Given OP's comment under the answer confirming the source of the question, I've added the self-study tag myself. $\endgroup$
    – Glen_b
    Commented Oct 31, 2013 at 7:06
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    $\begingroup$ Start simply. What's the probability of coming out ahead on one game? $\endgroup$
    – Glen_b
    Commented Oct 31, 2013 at 7:08
  • $\begingroup$ The probability of winning on any given throw is 53%. What I am having trouble with is where does the payout come in and the likelihood of winning money? $\endgroup$
    – rdittmer
    Commented Oct 31, 2013 at 7:14

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I assume this is a "self-study" problem. Here are the hints for you:

  1. Suppose you play N throws. What is the probability to win in exactly M throws?
  2. How many throws you need to win money?
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  • $\begingroup$ It's actually a homework question for a stats class. The professor doesn't really teach anything and just advertises for his research lab for an hour twice a week. $\endgroup$
    – rdittmer
    Commented Oct 31, 2013 at 6:35

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