Joe S. answered • 01/31/15
Tutor
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Taught Business Stats 18 years; teaching awards & exc. student evals
The stated answer, (p^2) / (1 - 2p + p^2), is an odds ratio rather than a probability. It cannot be a probability since it evaluates to more than 1 if p > .5.
Here is the logic for the solution. The sequence ends when one party has 2 wins more than the other one. Therefore the sequence can end only after an even number of “games” (or plays or rounds). For example, with only one play, the score will be 1 to 0 and neither party wins. If the sum of two numbers is odd, then one must by odd and the other must be even. Therefore the difference must be odd, and so cannot be 2.
After 2 plays this can happen:
A wins with conditional probability P(A | exactly 2 games) = p^2, winning 2 to 0
B wins with probability P(B | exactly 2 games) = q^2, winning 0 to 2
The game is tied, with probability 2pq.
These probabilities add to 1, as a check.
After 3 games, either A or B is ahead by one, either 2 to 1 or 1 to 2, but the sequence continues for at least another game.
On the 4th game the results are similar to what happened after 2 plays:
A wins 3 to 1 with conditional probability P(A | 4 games) = p^2 (i.e., 2 consecutive wins after the tie.)
B wins 1 to 3 “ P(B | exactly 4 games) = q^2 (2 consecutive B's after the tie)
tied 2 to 2 “ P(tie | exactly 4 games) = 2pq (one each of A and B after the tie)
The insight is that if the game is tied at any point it has the same effect as starting over with the game tied 0 to 0.
Let P(A) be the unconditional probability that A wins after any number of games.
P(A) = P(A wins in exactly 2 games) + P(A wins | tied after 2 games)
= p^2 + 2pq*P(A) (since a tie is the same as starting over)
P(A)*(1+2pq) = p^2
P(A) = (p^2) / (1 + 2pq)
Similarly, P(B) = (q^2) / (1 + 2pq)
The odds ratio is P(A) / P(B) = (p/q)^2
The denominator of the given solution is (1-2p+p^2) = (1-p)^2 = q^2, so the given solution is (p^2)/(q^2) = (p/q)^2.
Here is the logic for the solution. The sequence ends when one party has 2 wins more than the other one. Therefore the sequence can end only after an even number of “games” (or plays or rounds). For example, with only one play, the score will be 1 to 0 and neither party wins. If the sum of two numbers is odd, then one must by odd and the other must be even. Therefore the difference must be odd, and so cannot be 2.
After 2 plays this can happen:
A wins with conditional probability P(A | exactly 2 games) = p^2, winning 2 to 0
B wins with probability P(B | exactly 2 games) = q^2, winning 0 to 2
The game is tied, with probability 2pq.
These probabilities add to 1, as a check.
After 3 games, either A or B is ahead by one, either 2 to 1 or 1 to 2, but the sequence continues for at least another game.
On the 4th game the results are similar to what happened after 2 plays:
A wins 3 to 1 with conditional probability P(A | 4 games) = p^2 (i.e., 2 consecutive wins after the tie.)
B wins 1 to 3 “ P(B | exactly 4 games) = q^2 (2 consecutive B's after the tie)
tied 2 to 2 “ P(tie | exactly 4 games) = 2pq (one each of A and B after the tie)
The insight is that if the game is tied at any point it has the same effect as starting over with the game tied 0 to 0.
Let P(A) be the unconditional probability that A wins after any number of games.
P(A) = P(A wins in exactly 2 games) + P(A wins | tied after 2 games)
= p^2 + 2pq*P(A) (since a tie is the same as starting over)
P(A)*(1+2pq) = p^2
P(A) = (p^2) / (1 + 2pq)
Similarly, P(B) = (q^2) / (1 + 2pq)
The odds ratio is P(A) / P(B) = (p/q)^2
The denominator of the given solution is (1-2p+p^2) = (1-p)^2 = q^2, so the given solution is (p^2)/(q^2) = (p/q)^2.
It should say "a series of independent games" instead of "a serious of independent games".
Simeon N.
01/29/16