Probability of winning streak within series of trials

Question

This is q2 from here, but the provided solution was unenlightening/incorrect. Also, questions have been asked regarding a variant of q2, but they differ from mine.

To encourage Elmer's promising tennis career, his father offers a prize if he wins at least 2 tennis sets in a row in a three-set series to be played with his father and the club champion alternately. He is given the choice between playing:

• Father; Champion; Father.
• Champion; Father; Champion.

Given $P[``\text{E beats F''}] = 0.8$ & $P[``\text{E beats C''}] = 0.4$, which series should Elmer choose to play?

---

This is how I calculated the probabilities:

• $P[``\text{E's winning streak}\geq\text{2''}] = P[``\text{E beats C''}]\left(P[``\text{E beats F''}]\right)^2\\+ 2P[``\text{E beats F''}]P[``\text{E beats C''}]P[``\text{F beats E''}]\\= (0.4)(0.8)^2 + (0.8)(0.4)(0.2) = 0.384$

• $P[``\text{E's winning streak}\geq\text{2''}] = P[``\text{E beats F''}]\left(P[``\text{E beats C''}]\right)^2\\+ 2P[``\text{E beats C''}]P[``\text{E beats F''}]P[``\text{C beats E''}]\\= (0.8)(0.4)^2 + (0.4)(0.8)(0.6) = 0.512$

So Elmer should choose to play the second series.

---

My 4 questions are:

1. Was my thinking correct?
2. Are my calculated probabilities, as well as my conclusion, correct?
3. Poisson, binomial, geometric, & hypergeometric are the only distributions that I am familiar w/, & I don't think any of them can be applied to this problem (at least not directly). Am I mistaken?
4. If you replied no to q3, is there some other probability distribution formula that I could use to solve this (semi-)directly?

Answer 1

Yes your reasoning is correct and none of the distributions you have listed appear to be applicable. Alternatively, you can use the inclusion-exclusion principal. Let $X$ denote the longest winning streak in the $3$ game set. If the order is Father-Champion-Father then

$$\mathsf P(X\geq 2)=0.8\cdot0.4+0.4\cdot0.8-0.8\cdot0.4\cdot0.8=0.384$$

If the order is Champion-Father-Champion then

$$\mathsf P(X\geq 2)=0.4\cdot0.8+0.8\cdot0.4-0.4\cdot0.8\cdot0.4=0.512$$

Hence Champion-Father-Champion would be the more advantageous ordering. This can be confirmed with simulation using R statistical software.

father <- rbinom(10^6,1,.8)
champ <- rbinom(10^6,1,.4)
father2 <- rbinom(10^6,1,.8)
mean((father==1 & champ==1)|(champ==1 & father2==1))

> 0.384032

champ <- rbinom(10^6,1,.4)
father <- rbinom(10^6,1,.8)
champ2 <- rbinom(10^6,1,.4)
mean((champ==1 & father==1)|(father==1 & champ2==1))

> 0.511856