- By considering the sum of a geometric series, or otherwise, show that
\[\sum_{r=1}^{\infty} rx^{r-1} = \frac{1}{(1-x)^2} \quad \text{for } |x| < 1.\]
- Ali plays a game with a fair \(2k\)-sided die. He rolls the die until the first \(2k\) appears. Ali wins if all the numbers he rolls are even.
- Find the probability that Ali wins the game.
If Ali wins the game, he earns £1 for each roll, including the final one. If he loses, he earns nothing.
- Find Ali's expected earnings from playing the game.
- Find a simplified expression for
\[1 + 2\binom{n}{1}x + 3\binom{n}{2}x^2 + \ldots + (n+1)x^n,\]
where \(n\) is a positive integer.
- Zen plays a different game with a fair \(2k\)-sided die. She rolls the die until the first \(2k\) appears, and wins if the numbers rolled are strictly increasing in size. For example, if \(k = 3\), she wins if she rolls 2, 6 or 1, 4, 5, 6, but not if she rolls 1, 4, 2, 6 or 1, 3, 3, 6.
If Zen wins the game, she earns £1 for each roll, including the final one. If she loses, she earns nothing.
Find Zen's expected earnings from playing the game.
- Using the approximation
\[\left(1 + \frac{1}{n}\right)^n \approx e \quad \text{for large } n,\]
show that, when \(k\) is large, Zen's expected earnings are a little over 35\% more than Ali's expected earnings.
Show Solution
- Note that,
\begin{align*}
&& \sum_{r = 0}^\infty x^r &= \frac{1}{1-x} && |x| < 1\\
\underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \sum_{r = 0}^\infty rx^{r-1} &= \frac{1}{(1-x)^2} && |x| < 1\\
&& \sum_{r = 1}^\infty rx^{r-1} &= \frac{1}{(1-x)^2} && |x| < 1\\
\end{align*}
- \begin{align*}
&& \mathbb{P}(\text{Ali wins in }s\text{ rounds}) &= \left ( \frac{k-1}{2k} \right)^{s-1} \frac{1}{2k} \\
\Rightarrow && \mathbb{P}(\text{Ali wins}) &= \sum_{s=1}^\infty \mathbb{P}(\text{Ali wins in }s\text{ rounds}) \\
&&&=\sum_{s=1}^\infty \left ( \frac{k-1}{2k} \right)^{s-1} \frac{1}{2k} \\
&&&= \frac{1}{2k} \sum_{s=0}^\infty \left ( \frac{k-1}{2k} \right)^{s} \\
&&&= \frac{1}{2k} \frac{1}{1 - \frac{k-1}{2k}} \\
&&&= \frac{1}{2k - (k-1)} \\
&&&= \frac{1}{k+1}
\end{align*}
- \begin{align*}
\mathbb{E}(\text{Ali score}) &= \sum_{s=1}^{\infty} s \mathbb{P}(\text{Ali wins in }s\text{ rounds}) \\
&= \sum_{s=1}^{\infty} s \left ( \frac{k-1}{2k} \right)^{s-1} \frac{1}{2k} \\
&= \frac{1}{2k} \frac{1}{\left (1 - \frac{k-1}{2k} \right)^2} \\
&= \frac{2k}{(k+1)^2}
\end{align*}
- \begin{align*}
&& (1+x)^{n} &= \sum_{k=0}^n \binom{n}{k} x^k \\
\Rightarrow && x(1+x)^n &= \sum_{k=0}^n \binom{n}{k} x^{k+1} \\
\Rightarrow && (1+x)^n + nx(1+x)^{n-1} &= \sum_{k=0}^n (k+1)\binom{n}{k} x^k \\
\Rightarrow && (1+x)^{n-1}(1+(n+1)x) &= 1 + 2\binom{n}{1}x + 3\binom{n}{2}x^2 + \ldots + (n+1)x^n
\end{align*}
- \begin{align*}
\mathbb{E}(\text{Zen score}) &= \sum_{s=1}^{2k} s \mathbb{P} \left ( \text{Zen gets }s\text{ numbers in increasing order ending with }2k \right) \\
&= \sum_{s=1}^{2k} s \binom{2k-1}{s-1} \frac{1}{(2k)^s} \\
&= \frac{1}{2k}\sum_{s=0}^{2k-1} (s+1) \binom{2k-1}{s} \frac{1}{(2k)^s} \\
&= \frac{1}{2k} \left ( 1 + \frac{1}{2k} \right)^{2k-2} \left ( 1 + (2k-1+1) \frac{1}{2k} \right) \\
&= \frac{1}{k}\left ( 1 + \frac{1}{2k} \right)^{2k-2}
\end{align*}
- Therefore as \(k \to \infty\)
\begin{align*}
\frac{\mathbb{E}(\text{Zen score})}{\mathbb{E}(\text{Ali score}) } &= \frac{1}{k}\left ( 1 + \frac{1}{2k} \right)^{2k-2} \big / \frac{2k}{(k+1)^2} \\
&= \frac{(k+1)^2}{2k^2} \cdot \left ( 1 + \frac{1}{2k} \right)^{2k} \cdot \left ( 1 + \frac{1}{2k} \right)^{-2} \\
&\to \frac12 e \approx 2.7/2 = 1.35
\end{align*} ie Zen's expected earnings are \(\approx 35\%\) more.
\(A\) and \(B\) both toss the same biased coin. The probability that the coin shows heads is \(p\), where \(0 < p < 1\), and the probability that it shows tails is \(q = 1 - p\).
Let \(X\) be the number of times \(A\) tosses the coin until it shows heads. Let \(Y\) be the number of times \(B\) tosses the coin until it shows heads.
- The random variable \(S\) is defined by \(S = X + Y\) and the random variable \(T\) is the maximum of \(X\) and \(Y\). Find an expression for \(\mathrm{P}(S = s)\) and show that
\[ \mathrm{P}(T = t) = pq^{t-1}(2 - q^{t-1} - q^t). \]
- The random variable \(U\) is defined by \(U = |X - Y|\), and the random variable \(W\) is the minimum of \(X\) and \(Y\). Find expressions for \(\mathrm{P}(U = u)\) and \(\mathrm{P}(W = w)\).
- Show that \(\mathrm{P}(S = 2 \text{ and } T = 3) \neq \mathrm{P}(S = 2) \times \mathrm{P}(T = 3)\).
- Show that \(U\) and \(W\) are independent, and show that no other pair of the four variables \(S\), \(T\), \(U\) and \(W\) are independent.
- Two people adopt the following procedure for deciding where to go for a cup of tea: either to a hotel or to a tea shop. Each person has a coin which has a probability \(p\) of showing heads and \(q\) of showing tails (where \(p+q = 1\)). In each round of the procedure, both people toss their coins once. If both coins show heads, then both people go to the hotel; if both coins show tails, then both people go to the tea shop; otherwise, they continue to the next round. This process is repeated until a decision is made.
Show that the probability that they make a decision on the \(n\)th round is
$$(q^2 + p^2)(2qp)^{n-1}.$$
Show also that the probability that they make a decision on or before the \(n\)th round is at least
$$1 - \frac{1}{2^n}$$
whatever the value of \(p\).
- Three people adopt the following procedure for deciding where to go for a cup of tea: either to a hotel or to a tea shop. Each person has a coin which has a probability \(p\) of showing heads and \(q\) of showing tails (where \(p + q = 1\)). In the first round of the procedure, all three people toss their coins once. If all three coins show heads, then all three people go to the hotel; if all three coins show tails, then all three people go to the tea shop; otherwise, they continue to the next round.
In the next round the two people whose coins showed the same face toss again, but the third person just turns over his or her coin. If all three coins show heads, then all three people go to the hotel; if all three coins show tails, then all three people go to the tea shop; otherwise, they go to the third round.
Show that the probability that they make a decision on or before the second round is at least \(\frac{7}{16}\), whatever the value of \(p\).
Show Solution
- The probability they don't make a decision in a round is \(qp + pq = 2qp\) (TH and HT). The probability they make a decision in a round is \(q^2+p^2\) (TT and HH). Therefore the probability they make a decision in the \(n\)th round is:
\[ (q^2+p^2)(2qp)^{n-1} \]
by having \(n-1\) failures and one success.
The probability they make a decision on or before the \(n\)th round is the \(1-\) the probability they don't, ie \(1 - (2qp)^n\). Notice that \(\sqrt{qp} \leq \frac{p+1}{2} = \frac12 \Rightarrow qp \leq \frac14\) so \(1-(2pq)^n \leq 1 - \frac1{2^n}\)
- The probability it's decided in the first round is \(p^3 + q^3\) (HHH, TTT). The probability it's decided in the second round is \(3p^2q \cdot p^2 + 3qq^2 \cdot q^2 = 3pq(p^3+q^3)\) (HHT -> HHH) and (TTH -> TTT) with reorderings).
Therefore the probability of making a decision in the first or second round is \((p^3+q^3)(1 + 3pq)\) which is minimised when \(p = q\) by Muirhead (or whatever your favourite inequality is). So \(\frac{2}{8} \cdot \left ( 1 + \frac{3}{4} \right) = \frac{7}{16}\)
In a game, I toss a coin repeatedly. The probability, \(p\), that the coin shows Heads on any given toss is given by
\[ p= \frac N{N+1} \,, \]
where \(N\) is a positive integer. The outcomes of any two tosses are independent.
The game has two versions. In each version, I can choose to stop playing after any number of tosses, in which case I win £\(H\), where \(H\) is the number of Heads I have tossed. However, the game may end before that, in which case I win nothing.
- In version 1, the game ends when the coin first shows Tails (if I haven't stopped playing before that).
I decide from the start to toss the coin until a total of \(h\) Heads have been shown, unless the game ends before then. Find, in terms of \(h\) and \(p\), an expression for my expected winnings and show that I can maximise my expected winnings by choosing \(h=N\).
- In version 2, the game ends when the coin shows Tails on two consecutive tosses (if I haven't stopped playing before that).
I decide from the start to toss the coin until a total of \(h\) Heads have been shown, unless the game ends before then. Show that my expected winnings are
\[ \frac{ hN^h (N+2)^h}{(N+1)^{2h}} \,.\]
In the case \(N=2\,\), use the approximation \(\log_3 2 \approx 0.63\) to show that the maximum value of my expected winnings is approximately £3.
Show Solution
- Since we either win \(h\) or \(0\), to calculate the expected winnings we just need to calculate the probability that we get \(h\) consecutive heads, therefore:
\begin{align*}
&& \mathbb{E}(\text{winnings}) &= E_h \\
&&&= h \cdot \left ( \frac{N}{N+1} \right)^h \\
&& \frac{E_{h+1}}{E_h} &= \frac{h+1}{h }\left ( \frac{N}{N+1} \right)
\end{align*}
Therefore \(E_h\) is increasing if \(h \leq N\), so we can maximise our winnings by taking \(h = N\). (In fact, we could take \(h = N\) or \(h = N+1\), but arguably \(h = N\) is better as we have the same expected value but lower variance).
- We can have up to \(h\) tails appearing (if we imagine slots for tails of the form \(\underbrace{\_H\_H\_H\_\cdots\_H}_{h\text{ spaces and }h\, H}\) so, we have
\begin{align*}
&& \mathbb{P}(\text{wins}) &= \sum_{t = 0}^h \mathbb{P}(\text{wins and } t\text{ tails}) \\
&&&= \sum_{t = 0}^h\binom{h}{t} \left ( \frac{N}{N+1} \right)^h\left ( \frac{1}{N+1} \right)^t \\
&&&= \left ( \frac{N}{N+1} \right)^h \sum_{t = 0}^h\binom{h}{t}\left ( \frac{1}{N+1} \right)^t \cdot 1^{h-t} \\
&&&= \left ( \frac{N}{N+1} \right)^h \left ( 1 + \left ( \frac{1}{N+1} \right) \right)^h \\
&&&= \left ( \frac{N}{N+1} \right)^h \left ( \frac{N+2}{N+1}\right)^h \\
&&&= \frac{N^h(N+2)^h}{(N+1)^{2h}} \\
\Rightarrow && \E(\text{winnings}) &= h \cdot \frac{N^h(N+2)^h}{(N+1)^{2h}}
\end{align*}
If \(N = 2\), we have
\begin{align*}
&& \E(\text{winnings}) &= E_h \\
&&&= h \cdot \frac{2^h\cdot2^{2h}}{3^{2h}}\\
&&&= h \cdot \frac{2^{3h}}{3^{2h}} \\
\Rightarrow && \frac{E_{h+1}}{E_h} &= \frac{h+1}{h} \frac{8}{9} \\
\end{align*}
Therefore to maximise the winnings we should take \(h = 8\), and the expected winnings will be:
\begin{align*}
&& E_8 &= 8 \cdot \frac{2^{24}}{3^{16}} \\
\Rightarrow && \log_3 E_8 &= 27 \log_3 2 - 16 \\
&&&\approx 24 \cdot 0.63 - 16 \\
&&&\approx 17 - 16 \\
&&&\approx 1 \\
\Rightarrow && E_8 &\approx 3
\end{align*}
A biased coin has probability \(p\) of showing a head
and probability \(q\) of showing a tail, where \(p\ne0\), \(q\ne0\)
and \(p\ne q\). When the coin is tossed repeatedly, runs occur.
A straight run of length \(n\) is a sequence of \(n\) consecutive
heads or \(n\) consecutive tails. An alternating run of length
\(n\) is a sequence of length \(n\) alternating between heads and tails. An
alternating run can start with either a head or a tail.
Let \(S\) be the length of the longest straight run beginning with the
first toss and let \(A\) be the length of the longest
alternating run beginning with the
first toss.
- Explain why \(\P(A=1)=p^2+q^2\) and find \(\P(S=1)\). Show that
\(\P(S=1)<\P(A=1)\).
- Show that
\(\P(S=2)= \P(A=2)\)
and determine the relationship between
\(\P(S=3)\) and \( \P(A=3)\).
- Show that, for \(n>1\), \(\P(S=2n)>\P(A=2n)\) and determine
the corresponding relationship between \(\P(S=2n+1)\) and \(\P(A=2n+1)\).
[You are advised not to use \(p+q=1\) in this part.]
Show Solution
- The only way \(A = 1\) is if we get \(HH\) or \(TT\) which has probability \(p^2+q^2\). The only way we get \(S=1\) is if we have \(HT\) to \(TH\), ie \(2pq\).
Since \((p-q)^2 = p^2 + q^2 - 2pq >0\) we must have \(\mathbb{P}(A=1) > \mathbb{P}(S=1)\).
- \(\,\) \begin{align*}
\mathbb{P}(S=2) &= p^2q + q^2p \\
\mathbb{P}(A=2) &= pq^2 + qp^2 = \mathbb{P}(S=2) \\
\\
\mathbb{P}(S=3) &= p^3q + q^3p = pq(p^2+q^2) \\
\mathbb{P}(A=3) &= pqp^2 + qpq^2 = pq(p^2+q^2) = \mathbb{P}(S=3)
\end{align*}
- For \(n > 1\) we must have
\begin{align*}
&& \mathbb{P}(S = 2n) &= p^{2n}q + q^{2n}p \\
&& \mathbb{P}(A=2n) &= (pq)^{n}q + (qp)^{n}p \\
&&&= p^nq^{n+1} + q^np^{n+1} \\
&& \mathbb{P}(S = 2n) &> \mathbb{P}(A = 2n) \\
\Leftrightarrow && p^{2n}q + q^{2n}p & > p^nq^{n+1} + q^np^{n+1}\\
\Leftrightarrow && 0 & < p^{2n}q+q^{2n}p - p^nq^{n+1} -q^np^{n+1}\\
&&&= (p^n-q^n)(qp^n - pq^n)
\end{align*}
which is clearly true.
\begin{align*}
&& \mathbb{P}(S=2n+1) &= p^{2n+1}q + q^{2n+1}p \\
&& \mathbb{P}(A=2n+1) &= (pq)^{n}p^2 + (qp)^{n}q^2 \\
&&&= p^{n+2}q^n + q^{n+2}p^n
\end{align*}
The same factoring logic shows that \(\mathbb{P}(S = 2n+1) > \mathbb{P}(A=2n+1)\)
The infinite series \(S\) is given by
\[
S = 1 + (1 + d)r + (1 + 2d)r^2 + \cdots + (1+nd)r^n +\cdots\;
,\]
for \(\vert r \vert <1\,\).
By considering \(S - rS\), or otherwise, prove that
\[
S = \frac 1{1-r} + \frac {rd}{(1-r)^2}
\,.\]
Arthur and Boadicea shoot arrows at a target. The probability that an
arrow shot by Arthur hits the target is \(a\); the probability that an arrow
shot by Boadicea hits the target is \(b\). Each shot is independent of all
others. Prove that the expected number of shots it takes Arthur to hit
the target is \(1/a\).
Arthur and Boadicea now have a contest. They take alternate shots,
with Arthur going first. The winner is the one who hits the target first.
The probability
that Arthur wins the contest is \(\alpha\) and
the probability that Boadicea wins is
\(\beta\). Show that
\[
\alpha = \frac a {1-a'b'}\,,
\]
where \(a' = 1-a\) and \(b'=1-b\), and find \(\beta\).
Show that the expected number of shots in the contest is
\(\displaystyle \frac \alpha a + \frac \beta b\,.\)
Show Solution
Notice that
\begin{align*}
&& S - rS &= 1 + dr + dr^2 + \cdots \\
&&&= 1 + dr(1 + r+r^2+ \cdots) \\
&&&= 1 + \frac{rd}{1-r} \\
\Rightarrow && S &= \frac{1}{1-r} + \frac{rd}{(1-r)^2}
\end{align*}
The number of shots Arthur takes is \(\textrm{Geo}(a)\), so it's expectation is \(1/a\).
The probability Arthur wins is:
\begin{align*}
\alpha &= a + a'b'a + (a'b')^2a + \cdots \\
&= a(1+a'b' + \cdots) \\
&= \frac{a}{1-a'b'} \\
\\
\beta &= a'b + a'b'a'b + \cdots \\
&= a'b(1+b'a' + (b'a')^2 + \cdots ) \\
&= \frac{a'b}{1-a'b'}
\end{align*}
The expected number of shots in the contest is:
\begin{align*}
E &= a + 2a'b + 3a'b'a + 4a'b'a'b + \cdots \\
&= a(1 + 3a'b' + 5(a'b')^2 + \cdots) + 2a'b(1 + 2(a'b') + 3(a'b')^2 + \cdots) \\
&= a \left ( \frac{1}{1-a'b'} + \frac{2a'b'}{(1-a'b')^2} \right) + 2a'b \left ( \frac{1}{1-a'b'} + \frac{a'b'}{(1-a'b')^2}\right) \\
&= \frac{a}{1-a'b'} \left (1 + \frac{2a'b'}{(1-a'b')} \right) + 2\frac{a'b}{1-a'b'} \left ( 1 + \frac{a'b'}{(1-a'b')}\right) \\
&= \alpha \frac{1+a'b'}{1-a'b'} + \beta \frac{2}{1-a'b'} \\
&= \alpha \frac{1+1-a-b+ab}{1-a'b'} + \beta \frac{2}{1-a'b'} \\
\end{align*}
I have two identical dice. When I throw either one of them, the
probability of it showing a 6 is \(p\) and the probability of it not showing a 6 is \(q\), where \(p+q=1\). As an experiment to determine \(p\), I throw the dice simultaneously until at least one die shows a 6. If both dice show a six on this throw, I stop. If just one die shows a six, I throw the other die until it shows a 6 and then stop.
- Show that the probability that I stop after \(r\) throws
is \(pq^{r-1}(2-q^{r-1}-q^r)\), and find an expression for the
expected number of throws.
[{\bf Note:} You may use the result $\ds \sum_{r=0}^\infty rx^r =
x(1-x)^{-2}\(.]
- In a large number of such experiments, the mean number of throws was \)m\(. Find an estimate for \)p\( in terms of \)m$.
Show Solution
- \(\,\) \begin{align*}
\mathbb{P}(\text{stop after r}) &= \mathbb{P}(\text{both stop at r}) + 2\mathbb{P}(\text{first stops before r second stops at r})\\
&= (q^2)^{r-1} p^2 + 2\cdot q^{r-1} p\cdot(1-q^{r-1}) \\
&= q^{r-1}p\left (2-2q^{r-1}+pq^{r-1} \right) \\
&= q^{r-1}p\left (2-q^{r-1}(1+p+q-p) \right) \\
&= q^{r-1}p\left (2-q^{r-1}-q^r\right) \\
\end{align*}
\begin{align*}
\E[\text{throws}] &= \sum_{r=1}^{\infty} r \mathbb{P}(\text{stop after r}) \\
&= \sum_{r=1}^{\infty} r q^{r-1}p\left (2-q^{r-1}-q^r\right) \\
&= \sum_{r=1}^{\infty} 2r q^{r-1}p-\sum_{r=1}^{\infty}r pq^{2r-2}-\sum_{r=1}^{\infty}r q^{2r-1}p \\
&=2p \sum_{r=1}^{\infty} r q^{r-1}-pq^{-2}\sum_{r=1}^{\infty}r q^{2r}-pq^{-1}\sum_{r=1}^{\infty}r q^{2r} \\
&= 2p(1-q)^{-2} - pq^{-2}q^2(1-q^2)^{-2}-pq^{-1}q^2(1-q^2)^{-2} \\
&= 2pp^{-2} -p(1+q)(1-q^2)^{-2} \\
&= 2p^{-1}-p(1+q)(1+q)^{-2}p^{-2} \\
&= 2p^{-1}-p^{-1}(1+q)^{-1} \\
&= \frac{2(1+q)-1}{p(1+q)} \\
&= \frac{1+2q}{p(1+q)} \\
&= \frac{3-2p}{p(2-p)}
\end{align*}
- \(\,\) \begin{align*}
&& m &= \frac{3-2p}{p(2-p)} \\
\Rightarrow && 2mp-mp^2 &= 3-2p \\
\Rightarrow && 0 &= mp^2-(2m+2)p + 3 \\
\Rightarrow && p &= \frac{2m+2 \pm \sqrt{(2m+2)^2-12m}}{2m} \\
&&&= \frac{m+1- \sqrt{m^2-m + 1}}{m} \\
\end{align*}
If we are looking for an approximation, we could say \(p^2 \approx 0\) and \(p \approx \frac{3}{2(m+1)}\)
A men's endurance competition has an unlimited number of rounds. In each round, a competitor has, independently, a probability \(p\) of making it through the round; otherwise, he fails the round. Once a competitor fails a round, he drops out of the competition; before he drops out, he takes part in every round. The grand prize is awarded to any competitor who makes it through a round which all the other remaining competitors fail; if all the remaining competitors fail at the same round the grand prize is not awarded. If the competition begins with three competitors, find the probability that:
- all three drop out in the same round;
- two of them drop out in round \(r\) (with \(r \ge 2\)) and the third in an earlier round;
- the grand prize is awarded.
Show Solution
- This is the same as the sum of the probability that they all drop out in the \(k\)th round for all values of \(k\), ie
\begin{align*}
\mathbb{P}(\text{all drop in same round}) &= \sum_{k=0}^\infty \mathbb{P}(\text{all drop out in the }k+1\text{th round}) \\
&= \sum_{k=0}^{\infty}(p^k(1-p))^3 \\
&= (1-p)^3 \sum_{k=0}^{\infty}p^{3k} \\
&= \frac{(1-p)^3}{1-p^3} \\
&= \frac{1+3p(1-p)(p-(1-p))-p^3}{1-p^3} \\
&= \frac{1-p^3-3p(1-p)(1-2p)}{1-p^3}
\end{align*}
- There are \(3\) ways to choose the person who drops out earlier, and then they can drop out in round \(0, 1, \cdots r-1\)
\begin{align*}
\mathbb{P}(\text{exactly two drop out in round }r\text{ and one before}) &= 3\sum_{k=0}^{r-2} (p^{r-1}(1-p))^2p^k(1-p) \\
&= 3p^{2r-2}(1-p)^3 \sum_{k=0}^{r-2}p^k \\
&= 3p^{2r-2}(1-p)^3 \frac{1-p^{r-1}}{1-p} \\
&= 3p^{2r-2}(1-p)^2(1-p^{r-1})
\end{align*}
- The probability exactly \(2\) finish after the third
\begin{align*}
\mathbb{P}(\text{exactly two drop out after third}) &= \sum_{r=2}^{\infty}\mathbb{P}(\text{exactly two drop out in round }r\text{ and one before}) \\
&= \sum_{r=2}^{\infty}3p^{2r-2}(1-p)^2(1-p^{r-1}) \\
&= 3(1-p)^2p^{-2}\sum_{r=2}^{\infty}(p^{2r}-p^{3r-1}) \\
&= 3(1-p)^2p^{-2} \left( \frac{p^4}{1-p^2} - \frac{p^5}{1-p^3} \right) \\
&= \frac{3(1-p)^2(p^2(1-p^3)-p^3(1-p^2))}{(1-p^2)(1-p^3)}\\
&= \frac{3(1-p)^3p^2}{(1-p^2)(1-p^3)}\\
\end{align*}
Therefore the probability the grand prize is not awarded is
\begin{align*}
P &= 1 - \frac{(1-p)^3}{1-p^3} - \frac{3(1-p)^3p^2}{(1-p^2)(1-p^3)} \\
&= \frac{(1-p^3)(1-p^2) - (1-p)^3(1-p^2)-3(1-p)^3p^2}{(1-p^2)(1-p^3)} \\
&= \frac{(1-p^3)(1-p^2) - (1-p)^3(1+2p^2)}{(1-p^2)(1-p^3)} \\
\end{align*}
If a football match ends in a draw, there may be a "penalty shoot-out". Initially the teams each take 5 shots at goal. If one team scores more times than the other, then that team wins. If the scores are level, the teams take shots alternately until one team scores and the other team does not score, both teams having taken the same number of shots. The team that scores wins.
Two teams, Team A and Team B, take part in a penalty shoot-out.
Their probabilities of scoring when they take a single shot are \(p_A\) and \(p_B\) respectively.
Explain why the probability \(\alpha\) of neither side having won at the end of the initial \(10\)-shot period is given by
$$\alpha =\sum_{i=0}^5\binom{5}{i}^2(1-p_A)^i(1-p_B)^i\,p_A^{5-i}p_B^{5-i}.$$
Show that the expected number of shots taken is \(\displaystyle 10+ \frac{2\alpha}\beta\;,\) where \(\beta=p_A+p_B-2p_Ap_B\,.\)
Show Solution
Note that in the first \(10\)-short period the number of goals scored by each team is \(B(5, \p_i)\). For them to be equal they must both have scored the same number of goals, ie
\begin{align*}
&& \alpha &= \sum_{i=0}^5 \mathbb{P}(\text{both teams score }5-i) \\
&&&= \sum_{i=0}^5 \binom{5}{i} (1-p_A)^ip_A^{5-i} \binom{5}{i} (1-p_B)^i p_B^{5-i} \\
&&&= \sum_{i=0}^5 \binom{5}{i} ^2(1-p_A)^i (1-p_B)^i p_A^{5-i} p_B^{5-i} \\
\end{align*}
Suppose we make it to the end of the shoot out with scores tied. The probability that we finish each round is \(p_A(1-p_B) + p_B(1-p_A)\) (the probability \(A\) wins or \(B\) wins). This is \(p_A + p_B - 2p_Ap_B = \beta\)). Therefore the number of additional rounds is geometric with parameter \(\beta\) and the expected number of rounds is \(\frac{1}{\beta}\). Each round has two shots, and there is a probability \(\alpha\) of this occuring, ie \(\frac{2\alpha}{\beta}\). Added to the \(10\) guaranteed shots we get the desired result
The life of a certain species of elementary particles
can be described as follows. Each particle has a life time
of \(T\) seconds, after which it disintegrates into \(X\) particles
of the same species, where \(X\) is a random variable with
binomial distribution \(\mathrm{B}(2,p)\,\).
A population of these particles starts with the creation
of a single such particle at \(t=0\,\). Let \(X_n\) be
the number of particles in existence in the time interval
\(nT < t < (n+1)T\,\), where \(n=1\,\), \(2\,\), \(\ldots\).
Show that \(\P(X_1=2 \mbox { and } X_2=2) = 6p^4q^2\;\), where \(q=1-p\,\).
Find the possible values of \(p\) if it is known that
\(\P(X_1=2 \vert X_2=2) =9/25\,\).
Explain briefly why \(\E(X_n) =2p\E(X_{n-1})\) and hence determine \(\E(X_n)\)
in terms of \(p\).
Show that for one of the values of \(p\) found above
\(\lim_{n \to \infty}\E(X_n) = 0\)
and that for the other
\(\lim_{n \to \infty}\E(X_n) = + \infty\,\).
Show Solution
Notice that we can see the total number generated as \(X_n \sim B(2X_{n-1},p)\), since a Binomial is a sum of independent Bernoullis, and there are two Bernoullis per particle.
\begin{align*}
&& \mathbb{P}(X_1=2 \mbox { and } X_2=2) &= \underbrace{p^2}_{\text{two generated in first iteration}} \cdot \underbrace{\binom{4}{2}p^2q^2}_{\text{two generated from the first two}} \\
&&&= 6p^4q^2
\end{align*}
\begin{align*}
&& \mathbb{P})(X_1 = 2 |X_2 = 2) &= \frac{ \mathbb{P}(X_1=2 \mbox { and } X_2=2) }{ \mathbb{P}( X_2=2) } \\
&&&= \frac{6p^4q^2}{6p^4q^2+2pq \cdot p^2} \\
&&&= \frac{3pq}{3pq+1} \\
\Rightarrow && \frac{9}{25} &= \frac{3pq}{3pq+1} \\
\Rightarrow && 27pq + 9 &= 75pq \\
\Rightarrow && 9 &= 48pq \\
\Rightarrow && pq &= \frac{3}{16} \\
\Rightarrow && 0 &= p^2 - p + \frac3{16} \\
\Rightarrow && p &= \frac14, \frac34
\end{align*}
By the same reasoning about the Bernoullis, we must have \(\E[X_n] = \E[\E[X_n | X_{n-1}]] = \E[2pX_{n-1}] = 2p \E[X_{n-1}]\) therefore \(\E[X_n] = (2p)^n\).
If \(p = \frac14\) then \(\E[X_n] = \frac1{2^n} \to 0\)
If \(p = \frac34\) then \(\E[X_n] = \left(\frac32 \right)^n \to \infty\)
I have \(k\) different keys on my key ring. When I
come home at night I try one key after another until I
find the key that fits my front door. What is the probability
that I find the correct key in exactly \(n\) attempts in
each of the following three cases?
- At each attempt,
I choose a key that I have not tried
before but otherwise each choice is equally likely.
- At each attempt,
I choose a key from all my
keys and each of the \(k\) choices is equally likely.
- At the first attempt,
I choose from all my
keys and each of the \(k\) choices is equally likely. Thereafter,
I choose from the keys that I did not try the previous time
but otherwise each choice is equally likely.
The game of Cambridge Whispers starts with the first participant Albert
flipping an
un-biased coin and whispering to his neighbour Bertha
whether it fell `heads' or `tails'. Bertha then whispers this information to
her neighbour, and so on. The game ends when the final player Zebedee whispers
to Albert and the game is won, by all players, if what Albert hears is
correct. The acoustics are such that the listeners have,
independently at each stage, only
a probability of 2/3 of hearing correctly what is said. Find the probability
that the game is won when there are just three players.
By considering the binomial expansion of \((a+b)^n+(a-b)^n\), or otherwise,
find a concise expression for the probability \(P\) that the game is won
when is it played by \(n\) players each having a probability \(p\) of hearing
correctly.
% Show in particular that, if \(n\) is even,
%\(P(n,1/10) = P(n,9/10)\).% How do you explain this apparent anomaly?
To avoid the trauma of a lost game, the rules are now modified to require
Albert to whisper to Bertha what he
hears from Zebedee, and so keep the game going, if what he hears
from Zebedee is not correct. Find the expected total number of times that
Albert whispers to Bertha before the modified game ends.
\noindent
[You may use without proof the fact that \(\sum_1^\infty kx^{k-1}=(1-x)^{-2}\)
for \(\vert x\vert<1\).]
The makers of Cruncho (`The Cereal Which Cares') are giving away a
series of cards depicting \(n\) great mathematicians. Each packet of
Cruncho contains one picture chosen at random. Show that when I have
collected \(r\) different cards the expected number of packets I must
open to find a new card is \(n/(n-r)\) where \(0\leqslant r\leqslant n-1.\)
Show by means of a diagram, or otherwise, that
\[
\frac{1}{r+1}\leqslant\int_{r}^{r+1}\frac{1}{x}\,\mathrm{d}x\leqslant\frac{1}{r}
\]
and deduce that
\[
\sum_{r=2}^{n}\frac{1}{r}\leqslant\ln n\leqslant\sum_{r=1}^{n-1}\frac{1}{r}
\]
for all \(n\geqslant2.\)
My children will give me no peace until we have the complete set of
cards, but I am the only person in our household prepared to eat Cruncho
and my spouse will only buy the stuff if I eat it. If \(n\) is large,
roughly how many packets must I expect to consume before we have the
set?
An unbiased twelve-sided die has its faces marked \(A,A,A,B,B,B,B,B,B,B,B,B.\)
In a series of throws of the die the first \(M\) throws show \(A,\) the next \(N\) throws show \(B\) and the \((M+N+1)\)th throw shows \(A\).
Write down the probability that \(M=m\) and \(N=n\), where \(m\geqslant0\) and \(n\geqslant1.\) Find
- the marginal distributions of \(M\) and \(N\),
- the mean values of \(M\) and \(N\).
Investigate whether \(M\) and \(N\) are independent.
Find the probability that \(N\) is greater than a given integer \(k\), where \(k\geqslant1,\) and find \(\mathrm{P}(N > M).\) Find also \(\mathrm{P}(N=M)\)
and show that \(\mathrm{P}(N < M)=\frac{1}{52}.\)
Show Solution
\begin{align*}
\mathbb{P}(M = m, N = n) &= \left ( \frac{3}{12} \right)^m \left ( \frac{9}{12} \right)^n \frac{3}{12} \\
&= \frac{3^n}{4^{m+n+1}}
\end{align*}
- \begin{align*}
\mathbb{P}(M = m) &= \sum_{n = 1}^{\infty} \mathbb{P}(M=m,N=n) \\
&= \sum_{n = 1}^{\infty} \frac{3^n}{4^{m+n+1}} \\
&= \frac{1}{4^{m+1}} \sum_{n = 1}^{\infty} \left ( \frac34\right)^n \\
&= \frac{1}{4^{m+1}} \frac{3/4}{1/4} \\
&= \frac{3}{4^{m+1}} \\
\\
\mathbb{P}(N = n) &= \sum_{m = 0}^{\infty} \mathbb{P}(M=m,N=n) \\
&= \sum_{m = 0}^{\infty} \frac{3^n}{4^{m+n+1}} \\
&= \frac{3^n}{4^{n+1}} \sum_{m = 0}^{\infty} \left ( \frac14\right)^n \\
&= \frac{3^n}{4^{n+1}} \frac{1}{3/4} \\
&= \frac{3^{n-1}}{4^{n}} \\
\end{align*}
- \(M+1 \sim Geo(\frac34) \Rightarrow \mathbb{E}(M) = \frac43 -1 = \frac13\)
\(N \sim Geo(\frac14) \Rightarrow \mathbb{E}(N) = 4\)
\(M,N\) are independent since \(\mathbb{P}(M = m, N =n ) = \mathbb{P}(M=m)\mathbb{P}(N=n)\)
\begin{align*}
\mathbb{P}(N > k) &= \sum_{n=k+1}^{\infty} \mathbb{P}(N = n) \\
&= \sum_{n=k+1}^{\infty} \frac{3^{n-1}}{4^{n}} \\
&= \frac{3^k}{4^{k+1}} \sum_{n = 0}^{\infty} \left ( \frac34\right)^n \\
&= \frac{3^k}{4^{k+1}} \frac{1}{1/4} \\
&= \frac{3^k}{4^k}
\end{align*}
\begin{align*}
\mathbb{P}(N > M) &= \sum_{m=0}^{\infty} \mathbb{P}(N > m) \mathbb{P}(M = m) \\
&= \sum_{m=0}^{\infty} \left (\frac34 \right)^m \frac{3}{4^{m+1}}\\
&=\sum_{m=0}^{\infty} \frac{3^{m+1}}{4^{2m+1}}\\
&= \frac{3}{4} \frac{1}{13/16} \\
&= \frac{12}{13} \\
\\
\mathbb{P}(N=M) &= \sum_{m=1}^{\infty} \mathbb{P}(N=m, M=m) \\
&= \sum_{m=1}^{\infty} \frac{3^m}{4^{2m+1}} \\
&= \frac{3}{64} \sum_{m=0}^{\infty} \left ( \frac{3}{16} \right)^m \\
&= \frac{3}{64} \frac{1}{13/16} \\
&= \frac{3}{52}\\
\\
\mathbb{P}(N < M) &= 1 - \frac34 - \frac3{52} \\
&= 1 - \frac{48}{52} - \frac{3}{52} \\
&= 1 - \frac{51}{52} \\
&= \frac{1}{52}
\end{align*}
\(A,B\) and \(C\) play a table tennis tournament. The winner of the tournament will be the first person to win two games in a row. In any game, whoever is not playing acts as a referee, and each playerhas equal chance of winning the game. The first game of the tournament is played between \(A\) and \(B\), with \(C\) as referee. Thereafter, if the tournament is still undecided at the end of any game, the winner and referee of that game play the next game. The tournament is recorded by listing in order the winners of each game, so that, for example, \(ACC\) records a three-game tournament won by \(C\), the first game having been won by \(A\). Determine which of the following sequences of letters could be the record of a complete tournament, giving brief reasons for your answers:
- \(ACB\),
- \(ABB\),
- \(ACBB\).
Find the probability that the tournament is still undecided after 5 games have been played. Find also the probabilities that each of \(A,B\) and \(C\) wins in 5 or fewer games.
Show that the probability that \(A\) wins eventually is \(\frac{5}{14}\), and find the corresponding probabilities for \(B\) and \(C\).
Show Solution
- \(ACB\) is not a complete tournament since no-one has won two matches.
- \(ABB\) is not a possible complete tournament since it implies \(B\) won game 2, which is between \(A\) (winner of game 1) and \(C\) (referee of game 1).
- \(ACBB\) is a valid tournament, \(A\) beat \(B\), then \(C\) beat \(A\), then \(B\) beat \(C\) and finally \(B\) beat \(A\) to win.
After the first game there is always someone playing for the tournament, so for there to be no result after 5 games, 4 games must have gone against the leader, so the probability is \(\frac{1}{2^4} = \frac{1}{16}\).
If \(A\) wins their first game, they can either win in two games (WW) or in five games (WLRWW). The probability of this is \(\frac14 + \frac1{16} = \frac{5}{16}\). Similarly \(B\) has exactly the same chance as \(A\) since everything about them is symmetric, ie a probability of \(\frac5{16}\) of winning. Since there is a \(\frac{15}{16}\) chance the tournament is decided after 5 games, the remaining \(\frac{5}{16}\) must be \(C\)'s chance of winning.
After the first game is played, there's \(3\) states for each player. King (about to win if they win, becomes Ref if they lose), Challenger (needs to win to become king) and Ref (who becomes Challenger if Challenger wins).
\begin{align*}
\P(\text{King wins}) &= \frac{1}{2} + \frac{1}{2}\P(\text{Ref wins})\\
\P(\text{Challenger wins}) &= \frac{1}{2} \P(\text{King wins}) \\
\P(\text{Ref wins}) &= \frac{1}{2} \P(\text{Challenger wins}) \\
\end{align*}
\(p_K = \frac12 + \frac12 (\frac12 \frac12 p_K) \Rightarrow \frac78 p_K = \frac12 \Rightarrow p_K = \frac47, p_C = \frac27, p_R = \frac17\).
\(A\) has \(\frac12\) of being king, \(\frac12\) of being ref after the first match, so \(\frac12 \frac47 + \frac12 \frac17 = \frac{5}{14}\). Similarly \(B\) has \(\frac5{14}\) chance of winning, but unfortunately \(C\) must be the challenger after the first match and only has \(\frac27 = \frac4{14}\) chances of winning.