Stuck with part (ii) of this question, could someone explain how they got the answer:

(b) Bill and Ben play a game in which a fair die is rolled. Bill wins the game if the die shows 1, 2, 3 or 4. Ben wins the game if the die shows 5 or 6.

(i) If they play the game 6 times, find in simplest fraction form the probability that Bill wins two games more than Ben. (answer: 80/243)

(ii) If they continue playing until one wins two games more than the other, find in simplest fraction form the probability that Bill is the eventual winner. (answer: 4/5)

Note that the first part was a binomial probability but the second part is NOT. This is because in the first part, we don't care about the ordering of the wins; we just care that Bill wins 4 out of the 6. The second part imposes a restriction that actually impacts ordering.

\[ \text{Note obviously that there must be more than one game}\\ \text{for one person to win twice more than the other.} \]

Denote \(L\) for Bill and \(N\) for Ben.

\[ \text{If only two games are played and Bill wins,}\\ \text{the only possibility is }LL\\ \text{which occurs with probability }\left( \frac23 \right)^2.\]

\[ \text{Now observe that there cannot have been only three games played either. }\]

See that NONE of the following configurations are possible:

- \(LLL\) - can't happen because Bill would've won on the second game and hence the third game doesn't happen

- \(LLN\) - as above

- \(LNL\) - score becomes 2 to 1, so the games must continue

- \(NLL\) - as above

- \(LNN\) - score becomes 1 to 2, so the games must continue

- \(NLN\) - as above

- \(NNL\) - can't happen because Ben would've won on the second game and hence the third game doesn't happen

- \(NNN\) - as above

\[ \text{So the next possibility is when Bill wins on game }\textbf{four}.\\ \text{Whilst not really necessary, we illustrate all 16 cases to make it extremely obvious.} \]

- \(LLLL\) - can't happen because Bill would've already won on the second game

- \(LLLN\) - as above

- \(LLNL\) - as above

- \(LLNN\) - as above

- \(LNLL\) -

**a favourable outcome!**- \(LNLN\) - score becomes 2 to 2, so the games must continue

- \(LNNL\) - as above

- \(LNNN\) - Ben wins here because the score is 1 to 3

- \(NLLL\) -

**a favourable outcome!**- \(NLLN\) - score becomes 2 to 2 as well

- \(NLNL\) - as above

- \(NLNN\) - Ben wins here as well on game 4

- \(NNLL\) - Ben will already have won on game 2, so this cannot happen

- \(NNLN\) - as above

- \(NNNL\) - as above

- \(NNNN\) - as above

\[ \text{So the possibilities are }LNLL\text{ and }NLLL\\ \text{with probability }2\times \left(\frac13\right)\left(\frac23\right) \times \left( \frac23\right)^2. \]

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Now this will become annoying to deal with for 5 games, because then I'd have to write out 32 cases. So let's see if we can spot the pattern.

The idea is that the issues stemming from winning on the 3rd game will occur

**again** on winning on the 5th game. Basically, either the game will already have been won, or the score will be 3-to-2 or 2-to-3 instead. Similarly, for the 7th game, either it will already have won or the score becomes 4-to-3 or 3-to-4. So in general, we

**never** have the possibility of winning on an odd number of games, so we may discard it.

As for the evens, let's try figuring out the go with Bill winning on the 6th game with

**out** mapping out all the cases for this one. (Which would suck because there are 64 of them.) Basically we recycle what's above.

\[ \text{For Bill to win on game }\textbf{six},\\ \text{firstly we require the game has progressed that far.}\\ \text{This would mean that after the fourth game,}\\ \text{the match's outcome has }\textbf{still}\text{ not been decided.} \]

\[ \text{If we look above, we see that there are 4 cases where the game continues on game four:}\\ LNLN,\, LNNL,\, NLNL,\, NLLN. \]

\[ \text{But also, if the game continues into game four, we required it to not have been finished by game two.}\\ \text{Hence at the end of game 2, we must've had }LN\text{ or }NL.\]

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\[ \text{Indeed there is a pattern.}\\ \text{The idea is that we always require the game to be unfinished on the }(2n-2)\text{th turn}\\ \text{for it to finish on the }2n\text{th turn.} \]

To formally prove this we'd require induction, so let's not.

\[ \text{More importantly however is this.}\\ \text{Note that at each }2n\text{th stage, the previous sequences must've involved}\\ \text{a sequence of }n\,\,\boxed{NL}\text{'s or }\boxed{LN}\text{'s!} \]

So for example, if instead the game were finished on the

**8th** game, it must've been unfinished after the 6th. But that would require

**three** \(NL\)'s or \(LN\)'s appearing. In fact, the possibilities are

- \(LNLNLN\)

- \(LNLNNL\)

- \(LNNLLN\)

- \(LNNLNL\)

- \(NLLNLN\)

- \(NLLNNL\)

- \(NLNLLN\)

- \(NLNLNL\)

\[ \text{The idea is that if the game ends on the }2n\text{th turn,}\\ \text{it must be that we've had a sequence of }n-1\, \boxed{LN}\text{'s}\text{ or }\boxed{NL}\text{'s.}\\ \text{Note also that they're independent of one another - one being }LN\text{ does NOT mean the other must be }LN\text{ or }NL. \]

\[ \text{Therefore we realise that to win on the }2n\text{th game, we have }\boxed{2^{n-1}}\text{ choices}\text{ for the ordering of the previous }LN\text{ and }NL\text{ pairs.}\\ \text{So the probability that the game is unfinished on the }2(n-1)\text{th turn is}\\ 2^{n-1} \left(\frac13 \right)^{n-1} \left( \frac23\right)^{n-1} \]

\[ \text{And of course, the icing on the cake is that the last two games must always go to Bill.}\\ \text{(Should be obvious, but make sure you understand why!)}\\ \text{This, of course, occurs with probability }\left( \frac23 \right)^2 .\]

\[ \text{So the probability of winning on exactly the }2n\text{th turn is}\\ 2^{n-1} \left(\frac13 \right)^{n-1} \left( \frac23\right)^{n-1} \left( \frac23\right)^2 \]

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\[ \text{Hence as we require the probability of winning on the 2nd, 4th, 6th, 8th or so on-th turn,}\\ \text{the probability we wish to compute is}\\ \begin{align*} &\quad\sum_{n=1}^\infty 2^{n-1} \left(\frac13 \right)^{n-1} \left( \frac23\right)^{n-1} \left( \frac23\right)^2\\&= \frac49 \sum_{n=1}^\infty \left( \frac49 \right)^{n-1} \tag{simplifying}\\ &= \frac49 \left(1+\frac49 + \left( \frac49 \right)^2 + \dots \right) \\ &= \frac49 \times \frac{1}{1-\frac49} \tag{recognising geometric series}\\ &= \frac45\end{align*} \]