Assignment 3

The sample space for this question is the set $S$ of all $\binom{107}{5}$ $5$-element subsets of $\{1,\ldots,107\}$. An element in $S$ represents the numbers $\{z_1,\ldots,z_5\}$ chosen by the machine. This is a uniform sample space, so $\Pr(\omega)=1/|S|=1/\binom{107}{5}$ for each $\omega\in S$. Let $A$ be the event "you win the Big Jackpot". Then $\Pr(A)=\Pr(\{x_1,\ldots,x_5\})=1/\binom{107}{5}$.

Let $B$ be the event "you win a Little Jackpot" and, for each $i\in\{1,\ldots,5\}$, let $B_i$ be the event "$\{x_1,\ldots,x_5\}\cap\{z_1,\ldots,z_5\}=\{x_1,\ldots,x_5\}\setminus\{x_i\}$. In words, $B_i$ is the event "all your numbers matched except for $x_i$". Clearly the events $B_1,\ldots,B_5$ are pairwise disjoint. Therefore $\Pr(B)=\Pr(\bigcup_{i=1}^5 B_i)=\sum_{i=1}^5 \Pr(B_i)$, by the Sum Rule.

Now we just need to figure out $\Pr(B_i)=|B_i|/|S|=|B_i|/\binom{107}{5}$. Without loss of generality, we focus on $\Pr(B_5)$. In words, the event $B_5$ happens when $\{z_1,\ldots,z_5\}$ is a $5$-element subset of $\{1,\ldots,107\}\setminus\{x_5\}$ that contains $\{x_1,\ldots,x_4\}$. In other words $\{z_1,\ldots,z_5\}=\{x_1,\ldots,x_4,z_5\}$, where $z_5$ is any integer in the $102$-element set $\{1,\ldots,107\}\setminus\{x_1,\ldots,x_5\}$. Therefore $|B_5|=\binom{102}{1}=102$. There is nothing special about $x_5$ in this argument, so \[ \Pr(B_1)=\Pr(B_2)=\cdots=\Pr(B_5)=|B_5|/\binom{107}{5}=102/\binom{107}{5} \enspace . \] Now we can finish off what we started with \[ \Pr(B)=\Pr\left(\bigcup_{i=1}^5 B_i\right)=\sum_{i=1}^5 \Pr(B_i)=5\cdot\Pr(B_5)=510/\binom{107}{5} \enspace . \]

Let $C$ be the event "you win a Bonus Jackpot" and, for each $i\in\{1,\ldots,5\}$, let $C_i$ be the event "$\{z_1,\ldots,z_5\}=\{x_1,\ldots,x_5,y\}\setminus\{x_i\}$. In words, $C_i$ is the event "you win the Bonus Jackpot because all your numbers matched except for $x_i$ and the machine picked your bonus number $y$". Just like the last question, $C_1,\ldots,C_5$ are pairwise disjoint, so $\Pr(C)=\Pr(\bigcup_{i=1}^5 C_i)=\sum_{i=1}^5 \Pr(C_i)$, by the Sum Rule. Also, like the last question, we can focus on $\Pr(C_5)$ since there is nothing special about $x_5$.

But now, $C_5$ is easy to describe. It happens precisely when $\{z_1,\ldots,z_5\}=\{x_1,\ldots,x_4,y\}$. Therefore, $|C_5|=1$ and $\Pr(C_5)=1/\binom{107}{5}$. Therefore, \[ \Pr(C)=\Pr\left(\bigcup_{i=1}^5 C_i\right)=\sum_{i=1}^5 \Pr(C_i)=5\cdot\Pr(C_5)=5/\binom{107}{5} \enspace . \]

The sample space $S$ here represents the $5$ cards in your hand, and it's a uniformly random $5$-element subset of the $95$-element set $\{1,\ldots,100\}\setminus \{56, 55, 46, 1, 33\}$. Therefore $|S|=\binom{95}{5}$ and $\Pr(\omega)=1/|S|=1/\binom{95}{5}$ for each $\omega\in S$.

Let $A$ be the event "your highest card is lower than $56$". Then $A$ occurs if and only if your cards are a $5$-element subset of the $51$-card set $\{1,\ldots,54\}\setminus\{1,46,33\}$. The number of such subsets is $|A|=\binom{51}{5}$, so \[ \Pr(A)=\frac{|A|}{|S|}=\frac{\binom{51}{5}}{\binom{95}{5}} = \frac{55930}{1107249} \approx 0.040542612329723865 \enspace . \]

This is similar to the last question. Let $B$ be the event "your lowest card is higher than $56$". Then $B$ contains every $5$-element subset of $44$-element set $\{57,\ldots,100\}$, so $|B|=\binom{44}{5}$ and \[ \Pr(B)=\frac{\binom{44}{5}}{\binom{95}{5}} = \frac{155144}{8277217} \approx 0.018743497965560164 \enspace . \]

Let $C$ be the event "your second highest card is lower than $55$". We break $C$ up into two disjoint events, so that we can use the Sum Rule:

1. $C_0$ is the event "all your cards are lower than $55$". This is actually the same as the event $A$ in the first part of the problem, so $\Pr(C_0)=\Pr(A)=\binom{51}{5}/\binom{100}{5}$.
2. $C_1$ is the event "one of your cards is higher than $55$ and the other $4$ are lower than $55$". We can compute $|C_1|$ using the Product Rule. We can choose one number $z_1$ from the set $44$-element set $\{57,\ldots,100\}$ and then choose $z_2,\ldots,z_5$ to be a $4$-element subset of the $51$-element set $\{1,2,\ldots,54\}\setminus\{1,33,46\}$. Therefore, $|C_1|=44\cdot\binom{51}{4}$.

Therefore \begin{align*} \Pr(C) & = \Pr(C_0\cup C_1) \\ & = \Pr(C_0)+\Pr(C_1) & \text{(Sum Rule)} \\ & = \frac{\binom{51}{5}}{\binom{100}{5}} + \frac{|C_1|}{|S|} \\ & = \frac{\binom{51}{5}}{\binom{95}{5}} + \frac{44\cdot\binom{51}{4}}{\binom{95}{5}} \\ & = \frac{1906380}{8277217} & \text{(Using Python)} \\ & \approx 0.2303165423837505 & \text{(Using a calculator)} \end{align*}

For this question, the sample space $S:=\{1,\ldots,8\}^8$, $|S|=8^8$ and this is a uniform probability space, so $\Pr(\omega)=1/|S|=1/8^8$ for each $\omega\in S$. Let $E$ be the event "I win the game". The easiest thing to do here is to compute $\Pr(\overline{E})$ and use the Complement Rule. If we never roll an $8$, then each of the $8$ roles is in $\{1,\ldots,7\}$, so $\overline{E}=\{1,\ldots,7\}^8$ and $\Pr(\overline{E})=|\overline{E}|/|S|=(7/8)^8$. Therefore \[ \Pr(E)=1-\Pr(\overline{E}) = 1-\frac{|\overline{E}|}{|S|}= 1-\left(\frac{7}{8}\right)^8 = \frac{11012415}{16777216}\approx 0.6563910841941833 \]

For this question, the sample space $S:=\{1,\ldots,8\}^{16}$, $|S|=8^{16}$ and this is a uniform probability space, so $\Pr(\omega)=1/|S|=1/8^{16}$ for each $\omega\in S$. Let $E$ be the event "I win the game". The easiest thing to do here is to compute $\Pr(\overline{E})$ and use the Complement Rule. To compute $\Pr(\overline{E})$, we will partition $\overline{E}$ into two disjoint events and use the Sum Rule:

1. Let $\overline{E}_0$ be the event "I roll 8 zero times". Then $|\overline{E}_0|=7^{16}$.
2. Let $\overline{E}_1$ be the event "I roll 8 exactly once". We can compute $|\overline{E}_1|$ using the Product Rule: First choose which of the $16$ rolls will be an $8$ and then a value in $\{1,\ldots,7\}$ for each of the $15$ other rolls. Therefore, $|\overline{E}_1|=16\cdot 7^{15}$.

We finish with \begin{align*} \Pr(E) & =1-\Pr(\overline{E}) & \text{(Complement Rule)} \\ & = 1-(\Pr(\overline{E}_0) + \Pr(\overline{E}_1)) & \text{(Sum Rule)} \\ & = 1-\left(\frac{|\overline{E}_0| + |\overline{E}_1|}{|S|}\right) & \text{(Uniform probability space)} \\ & = 1-\left(\frac{7^{16} + 16\cdot 7^{15}}{8^{16}}\right) \\ & = 1-\left(\frac{172281061981967}{281474976710656}\right) & \text{(Python)} \\ & \approx 0.6120652855016111 \end{align*}

There are different ways to setup the sample space here, but one of the easiest is to define the bottle set $B=\{I_1,\ldots,I_{10},L_1,\ldots,L_{10}\}$. Then playing the game corresponds to choosing an ordered $4$-element subset of $B$ which represents the order in which four bottles of beer are drank (drunk?). The size of the resulting sample space $S$ is then $20\cdot 19\cdot 18\cdot 17=20!/16!$. (We showed this when proving that $\binom{n}{k}=n!/(k!(n-k)!)$.) This is a uniform sample space, which will make things easier.

Let $P$ be the event "Pat wins the game of First IPA Wins". Partition $P$ into the two disjoint events:

1. $P_1$ is the event "Pat wins in the first round". Then $|P_1|=10\cdot 19\times 18\times 17=\tfrac{1}{2}|S|$ since $P_1$ occurs precisely when Pat chooses one of the $10$ IPA bottles in the first round, after which anything goes.
2. $P_2$ is the event "Pat wins in the second round". Then $|P_2|=10\cdot 9\cdot 10\cdot 17$ since $P_2$ happens when Pat drinks one of the 10 lagers, then Michiel drinks one of the 9 remaining lagers, then Pat drinks one of the 10 IPA, after which anything goes.

Since $P_1$ and $P_2$ are disjoint and $P=P_1\cup P_2$, \begin{align*} \Pr(P) & =\Pr(P_1\cup P_2) \\ & = \frac{|P_1\cup P_2|}{|S|} \\ & = \frac{|P_1|+|P_2|}{|S|} \\ & = 1/2 + \frac{10\cdot 9\cdot 10\cdot 17}{20\cdot 19\cdot 18\cdot 17} \\ & = 12/19 \approx 0.6315789 \end{align*}

Let $M$ be the event "Michiel wins the game of First IPA Wins". We partition $M$ into two events so that we can use the Sum Rule.

1. $M_1$ is the event "Michiel wins in the first round". Then $|M_1|=10\cdot 10\cdot 18\cdot 17$ since $M_1$ occurs when Pat drinks one of the 10 lagers, then Michiel drinks one of the ten IPA, after which anything goes.
2. $M_2$ is the event "Michiel wins in the second round". Then $|M_2|=10\cdot 9\cdot 8\cdot 10$ since $M_2$ occurs when Pat drinks one of the 10 lagers then Michiel drinks one of the $9$ remaining lagers, then Pat drinks one of the $8$ remaining lagers, then Michiel drinks one of the $10$ IPA.

Since $M_1$ and $M_2$ are disjoint and $M=M_1\cup M_2$, we proceed as above to get \[ \Pr(M)=\frac{|M_1|+|M_2|}{|S|} = \frac{10\cdot 10\cdot 18\cdot 17+10\cdot 9\cdot 8\cdot 10}{20\cdot 19\cdot 18\cdot 17} = 105/323 \approx 0.32507 \]

Let $P$ be the event "Pat wins the game of Second IPA Wins". We will use the Sum Rule again, with the following two events:

1. $P_1$ is the event "Pat drinks the first IPA and the second IPA". Then $|P_1|=10\cdot 10\cdot 9\cdot 17$ since $A_1$ occurs when Pat drinks one of the 10 IPA, then Michiel drinks one of the ten lagers, then Pat drinks one of the 9 remaining IPAs, then anything goes.
2. $P_2$ is the event "Michiel drinks the first IPA and Pat drinks the second IPA". Then $|P_2|=10\cdot 10\cdot 9\cdot 17$ since $A_2$ occurs when Pat drinks one of the 10 lagers, then Michiel drinks one of the 19 IPAs then Pat drinks one of the 9 remaining IPAs then anything goes.

Then, \[ \Pr(P) = \Pr(P_1\cup P_2) =\frac{|P_1\cup P_2|}{|S|} =\frac{10\cdot 10\cdot 9\cdot 17+10\cdot 10\cdot 9\cdot 17}{20\cdot 19\cdot 18\cdot 17} = \frac{5}{19} \approx 0.2631578947368421 \enspace . \]

Let $M$ be the event "Michiel wins the game of Second IPA Wins". We will use the Sum Rule again, with the following two events:

1. $M_0$ is the event "Michiel wins in the first round". Then $|M_0|=10\cdot 9\cdot 18\cdot 17$ since $M_0$ occurs when Pat drinks one of the $10$ IPA, then Michiel drinks one of the $9$ remaining IPA, after which anything goes.
2. $M_1$ is the event "Michiel drinks the first IPA and the second IPA". Then $|M_1|=10\cdot 10\cdot 9\cdot 9$.
3. $M_2$ is the event "Pat drinks the first IPA and Michiel drinks the second IPA". We can further partition $M_2$ into two events $M_{2,1}$ and $M_{2,2}$ depending on whether Pat drinks the IPA in the first or second round.
   1. $|M_{2,1}|=10\cdot 10\cdot 9\cdot 9$ since $M_{2,1}$ occurs when Pat drinks one of the 10 IPAs, then Michiel drinks one of the 10 lagers, then Pat drinks one of the 9 remaining lagers, then Michiel drinks one of the 9 remaining IPAs.
   2. $|M_{2,2}|=10\cdot 9\cdot 10\cdot 9$ since $M_{2,2}$ occurs when Pat drinks one of the 10 lagers, then Michiel drinks one of the $9$ remaining lagers, then Pat drinks one of the 10 IPAs, then Michiel drinks one of the 9 remaining IPAs.

(Notice that $|M_1|=|M_{2,1}|=|M_{2,2}|$.) Then \begin{align*} \Pr(M) & = \Pr(M_0\cup M_1\cup M_{2,1}\cup M_{2,2}) \\ & = \frac{|M_0\cup M_1\cup M_{2,1}\cup M_{2,2}|}{|S|} \\ & = \frac{|M_0|+|M_1|+|M_{2,1}|+|M_{2,2}|}{|S|} & \text{(Sum Rule)} \\ & = \frac{10\cdot 9\cdot 18\cdot 17 + 3\cdot 10\cdot 10\cdot 9\cdot 9}{20\cdot 19\cdot 18\cdot 17} \\ & = 144/323 \approx 0.4458204334365325 \\ \end{align*}

The event $A$ is equivalent to "the three capsules I bought contain $2$ or $3$ red horses".

The complementary event $\overline{A}$ is "the three capsules I bought contain $2$ or $3$ brown horses". Under hypothesis 0, there is complete symmetry between red and brown, so $\Pr_0(A)=\Pr_0(\overline{A})$. Since $\Pr_0(A)+\Pr_0(\overline{A})=1$, this means $\Pr_0(A)=\Pr_0(\overline{A})=1/2$.

The analysis under hypothesis $H_1$ takes a bit more work. Split $A$ into the two events $A_2$ and $A_3$ depending on whether I get $2$ or $3$ red horses, respectively. Then \[ \Pr_1(A_2)=\frac{3\times 900\times 899\times 100}{1000\times 999\times 998} \] and \[ \Pr_1(A_3) = \frac{900\times 899\times 898}{1000\times 999\times 998} \] so \[ \Pr_1(A)=\Pr_1(A_2) + \Pr_1(A_3) = \frac{3\times 900\times 899\times 100+900\times 899\times 898}{1000\times 999\times 998} = \frac{538501}{553890} \approx 0.9722165050822366 \]

There's really only one reasonable test. Buy three capsules. If they all contain red horses, guess $H_1$, otherwise guess $H_0$. In the language of the previous answer, if event $A_3$ occurs then guess $H_1$, otherwise guess $H_0$.

If $H_1$ is correct, then we're running the experiment with probability function $\Pr_1$, and we already know: \[ \Pr_1(A_3) = \frac{900\times 899\times 898}{1000\times 999\times 998} = \frac{403651}{553890} \approx 0.7287566123237466 > 1/2 \enspace . \]

If $H_0$ is correct, then we're running the experiment with probability function $\Pr_0$ and we get \[ \Pr_0(A_3) = \frac{500\times 499\times 498}{1000\times 999\times 998} = \frac{83}{666} \approx 0.12462462462462462 < 1/2 \enspace . \] so $\Pr_0(\overline{A}_3)=1-\Pr_0(A_3)>1/2$.