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Probability problem - A and B are college teams in overtime

  • Thread starter bslnerd00
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Homework Statement



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.

Homework Equations


none

The Attempt at a Solution


a) P(A wins) = 0.345, P(B wins) = 0.436, P(tie) = 0.219

Hi, I found out the probabilities each time wins. I can't figure out the answers to B or C however. Any ideas?

Thanks
 

Answers and Replies

  • #2
D H
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A wins if A wins on the first round OR if the first round ends in a tie and A wins on the second round OR if the first two rounds both end in a tie and A wins on the third round OR ...

See if you can write that mathematically. Hint: What is ##\sum_{n=0}^{\infty} x^n## ? (This has a nice simple expression).


The expected number of rounds is 1*P(game ends on round 1) + 2*P(ends on round 2) + 3*P(ends on round 3) + ... .

See if you can write that mathematically. Hint: What is ##\sum_{n=0}^{\infty} n x^n## ? (This too has a nice simple expression).
 
  • #3
Ray Vickson
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Homework Statement



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.

Homework Equations


none

The Attempt at a Solution


a) P(A wins) = 0.345, P(B wins) = 0.436, P(tie) = 0.219

Hi, I found out the probabilities each time wins. I can't figure out the answers to B or C however. Any ideas?

Thanks
Besides the method suggested by DH, you can look at it recursively. Let a, b and t be the one-round probabilities that A wins, B wins, or there is a tie.

Just before starting, let x = probability that A wins eventually. If A or B win in round 1 we are done; otherwise (if there is a tie) we start again. So we get x = a + t*x; can you see why?

Similarly, let N = number of rounds until a win. If someone wins on round 1 we have N = 1; otherwise we start again and have N = 1 + N', where N' has the same probability distribution as N; that is, it is an independent, probabilistic copy of N. So: N = 1 if no tie; N = 1 + N' if tie. Take expectations to get an equation ford EN (noting that EN' = EN).
 

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