Year: 2021
Paper: 2
Question Number: 12
Course: LFM Stats And Pure
Section: Probability Definitions
No solution available for this problem.
Candidates were generally well prepared for many of the questions on this paper, with the questions requiring more standard operations seeing the greatest levels of success. Candidates need to ensure that solutions to the questions are supported by sufficient evidence of the mathematical steps, for example when proving a given result or deducing the properties of graphs that are to be sketched. In a significant number of steps there were marks lost through simple errors such as mistakes in arithmetic or confusion of sine and cosine functions, so it is important for candidates to maintain accuracy in their solutions to these questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item A game for two players, $A$ and $B$, can be won by player $A$, with probability $p_A$, won by player $B$, with probability $p_B$, where $0 < p_A + p_B < 1$, or drawn. A match consists of a series of games and is won by the first player to win a game. Show that the probability that $A$ wins the match is
\[
\frac{p_A}{p_A + p_B}.
\]
\item A second game for two players, $A$ and $B$, can be won by player $A$, with probability~$p$, or won by player $B$, with probability $q = 1 - p$. A match consists of a series of games and is won by the first player to have won two more games than the other. Show that the match is won after an even number of games, and that the probability that $A$ wins the match is
\[
\frac{p^2}{p^2 + q^2}.
\]
\item A third game, for only one player, consists of a series of rounds. The player starts the game with one token, wins the game if they have four tokens at the end of a round and loses the game if they have no tokens at the end of a round. There are two versions of the game. In the cautious version, in each round where the player has any tokens, the player wins one token with probability $p$ and loses one token with probability $q = 1 - p$. In the bold version, in each round where the player has any tokens, the player's tokens are doubled in number with probability $p$ and all lost with probability $q = 1 - p$.
In each of the two versions of the game, find the probability that the player wins.
Hence show that the player is more likely to win in the cautious version if $1 > p > \tfrac{1}{2}$ and more likely to win in the bold version if $0 < p < \tfrac{1}{2}$.
\end{questionparts}
Many candidates were able to reach the required probability in the first part of this question, although many ignored drawn matches instead making an argument that the probability can be found by dividing the probability that A wins this game by the probability that someone wins on this game. While this argument is possible, generally far more justification was needed than candidates provided. Those who identified the necessary sequences were able to successfully reach the result in a well-justified way. In part (ii) a small number of candidates assumed that the number of games in a match would always be even rather than showing why this must be true. Of the other candidates, many were able to explain why this is the case. Relatively few candidates failed to spot that the games in parts (ii) and (iii) could be reduced to the same game as in (i). In (ii), many candidates attempted a combinatorial argument, but a significant number failed to observe that there are two ways to order each pair where each of the players wins one of the games. In part (iii) most candidates were able to derive the probability of winning the bold game. Most of those who reached the end of this part used logical implications in the wrong direction, for example showing that if the player is more likely to win the cautious version, then the given inequality holds.