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Assume games are independent

For a single game:

P(A wins) = 0.4
P(B wins) = 0.6

Now suppose that to win, you have to win at least two out of three games. What's the probability of A still winning?

I wanted to split this into outcomes:

The probability of A winning all of the games is $0.4^3=0.064$.

The probability of A winning two of three games is $0.4^20.6=0.096$.

The probability of A winning one game is $0.6^20.4=0.144$.

The probability of A winning no games is $0.6^3=0.216$.

I thought this covered all of the cases, but the sum of these four probabilities is equal to $0.52$, not $1$. Why?

Then I considered the fact that the second and third outcomes can occur in multiple ways. Specifically, A can win two games in three different ways, and they can win one game in three different ways. Then $0.064+0.096\cdot3+0.144\cdot3+0.216=1$.

But then I realized another problem - I'm assuming order doesn't matter. It does, though. If A wins the first two games, they automatically win - the third game doesn't matter. And there's actually only two ways that A can win one game - they can't only win the third game because if B went up 2-0, the game would've ended. Now that I realized this complexity in the problem, I'm a bit stumped.

Machetes0602
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    Since each game has two outcomes, over three games there are $2^3 = 8$ possible permutations of outcomes. Out of those 8, there are 4 possible ways A can win best of three (i) Losing the first game then winning the last two, (ii) Losing #2 but winning 1 and 3, (iii) Losing #3 but winning 1 and 2, (iv) Winning all three. Compute the probabilities of those four events and you'll have the answer. – ecnmetrician Nov 05 '21 at 21:00
  • If you want to consider the possibility that the game stops after someone already score two, then just add up the probabilities of (i) and (ii). – ecnmetrician Nov 05 '21 at 21:03

1 Answers1

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I think you already got the correct answer, and you're second guessing yourself. The very first statement you make is

Assume games are independent

But then at the end you say

I'm assuming order doesn't matter. It does, though.

The first quote says that the games are independent. The second quote says that the games are dependent. They can't be both. If this is a homework question and the question specifically says "assume games are independent," then you already know how to find the answer.

If the question does not explicitly say that games are independent, you may want to clarify that, because you're right, in a real-world scenario they would not be independent in most tournament styles. And if they are not assumed independent, you will need to calculate conditional probabilities, which will indeed be a little more difficult.

AJV
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  • It is a homework Q and it does say to assume independence, but it just didn't really make sense for me to use $0.4^3$ to calculate A winning in a 2-0 sweep when that 3rd game would never happen. Maybe I'm overthinking it though – Machetes0602 Nov 05 '21 at 19:19
  • I guess it never says that the third game would never happen and I'm just thinking in a practical real world context like sports – Machetes0602 Nov 05 '21 at 19:20
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    I think you're trying to make it more realistic than it really is. The question says "suppose that to win, you have to win at least two out of three games." It doesn't say anything about it being a knock-out tournament. In fact, read it more carefully. "At least two" seems to imply the possibility of winning three, which is impossible in a knock-out tournament. I think you're on the right track already. – AJV Nov 05 '21 at 20:24