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Obviously the various permutations are: AA BB ABB BAB ABA BAA. Nevertheless, I'm still confused because I can reason to myself both ways so any help would be much appreciated.

3 Answers 3

3

Assuming each set is independent from those before, we can see, that it doesn't matter, who wins first. The decision to play a third set will be made, by the end of the second. If the player of the first (whoever that is) has won (which is 50%), the game is over. If he lost the second set (again 50%) we will see another one.

So the odds for winning this bet are 50-50!

Edit: The number of permutations does not really matter here. You could (if you want to go by that way) extend the two-set-wins by a third set (assuming they played a third set and determined afterwards, that the game was won before that). By that, you will get $AAA$, $AAB$, $BBA$, $BBB$ as two set wins and $ABA$, $ABB$, $BAA$, $BAB$ as games over the whole distance.

2

The disjoint-event probabilities are:

  • $P(AA)=\frac12\cdot\frac12=\frac14$
  • $P(BB)=\frac12\cdot\frac12=\frac14$
  • $P(ABA)=\frac12\cdot\frac12\cdot\frac12=\frac18$
  • $P(ABB)=\frac12\cdot\frac12\cdot\frac12=\frac18$
  • $P(BAA)=\frac12\cdot\frac12\cdot\frac12=\frac18$
  • $P(BAB)=\frac12\cdot\frac12\cdot\frac12=\frac18$

Hence:

  • The probability of a $2$-set match is $\frac14+\frac14=\frac12$
  • The probability of a $3$-set match is $\frac18+\frac18+\frac18+\frac18=\frac12$

Hence a $2$-set match and a $3$-set match are equally likely to happen.

0

Probability of each 2-set game is $\left(\frac{1}{2}\right)^2=\frac{1}{4}$, and probability of each 3-set game is $\left(\frac{1}{2}\right)^3=\frac{1}{8}$.

There are $2$ 2-set games, so probability of 2-set game is $2\cdot\frac{1}{4}=\frac{1}{2}$, and $4$ 3-set games, so probability of 3-set game is $4\cdot\frac{1}{8}=\frac{1}{2}$.

Therefore chance to win this bet is $50\%$, no matter of your choice.