2007 iTest Problems/Problem 12

This problem is phrased oddly, leading to the non-standard result of $\frac{3}{2}$ instead of the standard $\frac{9}{8}$ .

Label the teams A,B. If A wins the series, it necessarily wins the $\underline{last}$ game. A three game series with A winning occurs 1 way (AAA), a four game series occurs ${3\choose 1}=3$ ways (AABA,ABAA,BAAA), and a five game tournament occurs ${4\choose 2}=6$ ways (AABBA,ABABA,BAABA,ABBAA,BABAA,BBAAA). These series occur in the proportions 1:3:6. The problem is phrased to imply that the actual probabilities occur in these proportions as well - leading to an expected value of $\frac{(0*1+1*3+2*6)}{(1+3+6)}=\frac{3}{2}$ . This assumption leads to the events (AAA) or (BBB) occurring with $p=\tfrac{1}{10}$ Which requires each game to be $\underline{negatively}$ correlated with the previous one (i.e. A is less likely to win a game if it won the previous game).

A more standard approach would be to assume the games are uncorrelated and that each team is equally likely to win. Then the probabilities would occur in the proportions $1*(\tfrac{1}{2})^3:3*(\tfrac{1}{2})^4:6*(\tfrac{1}{2})^5$ or $2:3:3$ . These probabilities would imply that A wins the series with probability $1*(\tfrac{1}{2})^3+3*(\tfrac{1}{2})^4+6*(\tfrac{1}{2})^5=\frac{1}{2}$ , and an expected value for the losing team of $\frac{(0*2+1*3+2*3)}{(2+3+3)}=\frac{9}{8}$ .

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2007 iTest Problems/Problem 12