# Thread: What's the maximum amount you should pay to play a game?

1. ## What's the maximum amount you should pay to play a game?

You're playing a game where you have to throw a ball into a basket five times. You only have five tries. The probability of making the ball into the basket for any of the five tries is 60% or 6/10. If you make all five of the balls in, you win $1000. If you make four of the balls in, you win 500$. If you make three of the balls in, you win 100$. If you make one or two balls in, you get nothing. What's the maximum amount of money one should be willing to pay so that, given these certain probabilities, he or she will not risk losing any money? This is just a hypothetical question I made up... I think the answer must be over .6^5 x 1000. I'm not sure how to factor in the 2nd and 3rd prizes though, if they make much of a difference at all. Note:I'm only a senior in high school and I think these types of question are at college level; if you could clearly explain how these types of questions are solved, that'd be great. I'm not very familiar with all the symbol lingo. 2. This fits a binomial probability distribution with p = 0.60 and n = 5 p(0) = 0.40^5 p(1) = 5*(0.40^4)*0.60 p(2) = ?? p(3) = ?? p(4) = 5*0.40*(0.60^4) p(5) = 0.60^5 I left you a couple to fill in. After specifying the distribution, calculating the expected value is simple enough: 0*p(0) + 0*p(1) + 0*p(2) + 100*p(3) + 500*p(4) + 1000*p(5) Note: "much of a difference" is not a good concept. There is a unique answer. If you don't care about part of the distribution, it shouldn't be in the universe of probabilities. In this case, four balls contributes 129.50 and three balls contributes 34.60. together, quite a bit more than the measly 78.00 contributed by the top prize. By the way: Excellent question. Keep exploring!! 3. Originally Posted by megatonfist You're playing a game where you have to throw a ball into a basket five times. You only have five tries. The probability of making the ball into the basket for any of the five tries is 60% or 6/10. If you make all five of the balls in, you win$1000. If you make four of the balls in, you win 500$. If you make three of the balls in, you win 100$. If you make one or two balls in, you get nothing. What's the maximum amount of money one should be willing to pay so that, given these certain probabilities, he or she will not risk losing any money?
Nothing, since there is always the possibility of scoring zero there is always the risk of losing money if you pay any positive amount to play.

Now if you want to know how much you should pay so the expected return is 0 (which is break even on average) that is a different question and I'm sure someone will answer that, but it is not what you asked.

CB