My Math Forum Probability question-- need help

 October 5th, 2013, 11:10 AM #1 Newbie   Joined: Oct 2013 Posts: 1 Thanks: 0 Probability question-- need help A and B are college football teams that have gone into overtime. In the first round A will go first with the following possible outcomes: no score; 3 points; 6 points; 7 points; 8 points; a turnover where B wins (note in this case the game ends immediately). The probabilities of these happening are: .2, .3, .1, .3, .09, .01. B then follows with the following conditional outcomes: if A scored 0–B ties with probability .1; B wins with probability .88; A wins with probability .02. if A scored 3–B ties with probability .3; B wins with probability .6; A wins with probability .1. if A scored 6–B ties with probability .01; B wins with probability .4; A wins with probability .59. if A scored 7–B ties with probability .3; B wins with probability .1; A wins with probability .6. if A scored 8–B ties with probability .2; A wins with probability .8 If the teams are tied after the first round, they go to a second round and continue until a team wins. a) Find the probability that: A wins in the first round; B wins in the first round; they’re tied after the first round. b) Find the probability that A wins. c) Find the expected number of rounds. I figured out part a) 0.345, 0.436, and 0.219. Not sure how to do parts b and c. Help! thanks
 October 7th, 2013, 04:08 PM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications Re: Probability question-- need help Hi MathNoob123, and welcome to the forums. I agree with your calculations for part a). Let's use those probabilities to find b) and c): $\text{b)}$ The probability of A winning is the probability of A winning in the first round (0.345) plus that of the second round, and so on. The probability of a tie (0.219) factors in as follows: $P_{\text{A wins}}=0.345+(0.219)(0.345)+(0.219)^2(0.345)+(0.21 9)^3(0.345)+\ldots$ $P_{\text{A wins}}=0.345(1+0.219+0.219^2+0.219^3+\ldots)$ The infinite sum inside the parenthesis is: $\frac{1}{1-0.219} \$ so: $P_{\text{A wins}}=0.345 \cdot \frac{1}{1-0.219} \approx 0.442$ $\text{c)}$ The expected value is the sum of the round number times that round's probability of ending the game: $E_{\text{rounds}}=1(1-0.219)+2(0.219)(1-0.219)+3(0.219)^2(1-0.219)+ 4(0.219)^3(1-0.219) + \ldots$ $E_{\text{rounds}}=(1-0.219)(1+2(0.219)+3(0.219)^2+ 4(0.219)^3 + \ldots )$ The infinite sum inside the parenthesis is: $\frac{1}{(1-0.219)^2} \$ so: $E_{\text{rounds}}=(1-0.219) \cdot \frac{1}{(1-0.219)^2}=\frac{1}{1-0.219} \approx 1.280$ One comment, although I am not an expert: The statement "if A scored 0–B ties with probability .1; B wins with probability .88; A wins with probability .02." seems incomplete. The only way that A can win in that round is if B turns the ball over, and that is not specifically stated. I assumed that this is the case.

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