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For a sports league, the objective is to pick games that end in a tie. The bettor picks 8 games from a list of 45 or more. For each game, the bettor gets 3 points if there is a tie, 2 points if the visiting team wins, and 1.5 points if the home team wins. The set of chosen 8 games with the highest point total wins. In each game, the probability of the home team's winning is 0.5, the visiting team's winning is 0.4, and 0.1 for a tie. How does the bettor compute the probability her point total will be equal to or greater than 22 ?

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  • $\begingroup$ What is the bettor's strategy? Does she wish to maximize her expected return (which leads to one answer); or to maximize the chance of getting a return of 22 or more (which leads to a different answer); or to maximize her chance of getting more points than all other bettors; or--as you suggest at the outset--to maximize the number of tied games where she bet on a tie? And if the "objective is to pick games that end in a tie," there doesn't seem to be anything one can do if all games have these random outcomes with the same probability. Is that really the objective? $\endgroup$
    – whuber
    Commented Jul 20, 2022 at 14:52
  • $\begingroup$ Thanks for your inquiry - our bettor wants to maximize the chance of getting a return of 22 points or more. Obviously, for example, were she lucky enough to choose 8 ties, her point total would be 24. But that would be only one result that satisfies the criterion. Other combinations can give 22 or more points (a lot of 3's would be needed!), but how do we compute the number of these combinations and weave in the win-loss-tie probabilities? $\endgroup$ Commented Jul 20, 2022 at 16:09
  • $\begingroup$ Please see my comment to the first answer: it seeks clarification of how the betting and payoffs work, because there appear to be at least two different interpretations. $\endgroup$
    – whuber
    Commented Jul 20, 2022 at 22:31

1 Answer 1

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With the given information, computing the probability makes sense only if the bettor blindly chooses the 8 games.

Let $\{H,V,T\}$ be the numbers of home team wins, visiting team wins, and ties in the bettor's 8 selected games. There are $3^8=6561$ possible sequences of 3 outcomes for 8 games.

Only 4 sets satisfy $1.5H+2V+3T\geq22$: $\{0,0,8\}$, $\{0,1,7\}$, $\{1,0,7\}$, and $\{0,2,6\}$. The numbers of sequences corresponding to each set are, respectively, $\binom{8}{0}=1$, $\binom{8}{1}=8$, $\binom{8}{1}=8$, and $\binom{8}{2}=28$.

The probability of observing any given sequence is $0.5^H0.4^V0.1^T$.

If the bettor chooses the 8 games blindly, the probability of the point total being greater than or equal to 22 is the sum of the probability of each of the four sets identified multiplied by the number of sequences that would result in the set:

$1\cdot0.5^00.4^00.1^8+8\cdot0.5^00.4^10.1^7+8\cdot0.5^10.4^00.1^7+28\cdot0.5^00.4^20.1^6=5.21\cdot10^{-6}$.

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  • $\begingroup$ I don't understand how you can answer this question without specifying how the bettor bets! $\endgroup$
    – whuber
    Commented Jul 20, 2022 at 17:50
  • $\begingroup$ @whuber As I stated, the only way I could make sense of the question was if we assume the absence of a strategy. This isn't necessarily absurd, so it seemed a plausible assumption. Imagine a competition with a 10-dollar entry fee that pays out 10k dollars for anyone who scores 22 or better. A reasonable question may be something like "what is my expected payoff if my ability to pick high point value games is indistinguishable from blind selection?" $\endgroup$
    – jblood94
    Commented Jul 20, 2022 at 18:13
  • $\begingroup$ I see now there are at least two ways to interpret the question: (1) in my interpretation, the bettor specifies whether the outcome will be win, lose, or tie, and wins zero unless the specified outcome happens. (2) In your interpretation, the bettor merely commits money and wins some amount regardless of the outcome. Your answer makes perfect sense in this interpretation (+1). $\endgroup$
    – whuber
    Commented Jul 20, 2022 at 22:30
  • $\begingroup$ Yes. I see what you mean now. I guess sometimes the best way to get clarification on a question is to make an assumption and attempt an answer. $\endgroup$
    – jblood94
    Commented Jul 20, 2022 at 23:51
  • $\begingroup$ Awesome, jblood94! I would never have been able to figure out this analysis approach. Many thanks! $\endgroup$ Commented Jul 21, 2022 at 23:19

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