Tukukkk

This problem is from my teacher and I think their answer is wrong.

The problem is in the context of table tennis.

The players in the tournament final are Ani and Bertha. The score in the game is drawn at 20-20. The final game will continue until one player has scored two more points than the other. It is known from previous games between Ani and Bertha that the probability of Ani winning each point is 0.6. Find the probability that Ani will win the game after exactly 8 more points.

I think that this means that, over the next 6 games on the 2nd, 4th and 6th game Ani and Bertha need to have a draw. For each draw there are two possible paths. Ani wins then Bertha or Bertha then Ani. After the 6th game Ani just needs to win twice to win after exactly 8 points.

However my teacher says that:

If Bertha wins the first game, it is not possible for Ani to win after exactly 8 points.

and also asserts that there is only one path to the desired outcome. Using this they find the probability of Ani winning after exactly 8 points to be 0.005. (

Here is the linked image:

I found an alternate answer using multiple paths.

$P(text{Ani winning a game})=0.6$

$P(text{Bertha winning a game})=0.4$

$P(text{Ani win after 8 points})=2dot(0.4 cdot 0.6)cdot2dot(0.4 cdot 0.6)cdot2dot(0.4 cdot 0.6)cdot(0.6cdot0.6)=0.040 (2sf)$

After I found this answer I asked my teacher if the proposed answer was correct. My teacher replied saying that there was nothing wrong with it.

Am I missing something painfully obvious and if so what? or is the teacher's answer incorrect?

I think your teacher is wrong. He shows BAA as a win for Annie, but it's not. He has confused "ahead by two" with "win two in a row".

Ani needs to be have two more points than Bertha exactly on the eighth game (not earlier, and never two less). She does not just need to win two games in a row, but to be tied then win the last two games.

So since we are staring at a tie and need to avoid an being ahead or behind by two after the second game, the first two games may be a win followed by a loss, or a loss followed by a win.

Either way leaves Ani and Bertha tied again at the end.   And so likewise for the next two games, and the next pair.   Then as stated, the final two games need both be wins.

There are eight ways to do this, each giving Ani three losses and five wins. (Also, four among these paths do have Bertha ahead on the first game, so your teacher supposed erraneously.)

$$mathsf P(text{W on 8th})= 2^3cdot0.4^3cdot0.6^5$$

As you found.