1994 Paper 2 Q12

Year: 1994
Paper: 2
Question Number: 12

Course: LFM Stats And Pure
Section: Conditional Probability

Difficulty: 1600.0 Banger: 1499.3

probability conditional probability dice geometric series craps repeated trials case analysis

Problem

Calamity Jane sits down to play the game of craps with Buffalo Bill. In this game she rolls two fair dice. If, on the first throw, the sum of the dice is \(2,3\) or \(12\) she loses, while if it is \(7\) or \(11\) she wins. Otherwise Calamity continues to roll the dice until either the first sum is repeated, in which case she wins, or the sum is \(7\), in which case she loses. Find the probability that she wins on the first throw. Given that she throws more than once, show that the probability that she wins on the \(n\)th throw is \[ \frac{1}{48}\left(\frac{3}{4}\right)^{n-2}+\frac{1}{27}\left(\frac{13}{18}\right)^{n-2}+\frac{25}{432}\left(\frac{25}{36}\right)^{n-2}. \] Given that she throws more than \(m\) times, where \(m>1,\) what is the probability that she wins on the \(n\)th throw?

No solution available for this problem.

Rating Information

Difficulty Rating: 1600.0

Difficulty Comparisons: 0

Banger Rating: 1499.3

Banger Comparisons: 2

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Problem source

Calamity Jane sits down to play the game of craps with Buffalo Bill.
In this game she rolls two fair dice. If, on the first throw, the
sum of the dice is $2,3$ or $12$ she loses, while if it is $7$
or $11$ she wins. Otherwise Calamity continues to roll the dice until
either the first sum is repeated, in which case she wins, or the sum
is $7$, in which case she loses. Find the probability that she wins
on the first throw. 

Given that she throws more than once, show that the probability that
she wins on the $n$th throw is 
\[
\frac{1}{48}\left(\frac{3}{4}\right)^{n-2}+\frac{1}{27}\left(\frac{13}{18}\right)^{n-2}+\frac{25}{432}\left(\frac{25}{36}\right)^{n-2}.
\]
Given that she throws more than $m$ times, where $m>1,$ what is
the probability that she wins on the $n$th throw?

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