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\centerline{Homework 10  MORALLY Due April 28} 


\begin{enumerate}
\item
(0 points)
What is your name.

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\item
(10 points) 

\begin{enumerate}
\item 
(10 points) 
Show that 

$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$

with a combinatorial proof. 

(Hint: Show that the RHS solves the question of how many ways to choose $k$ objects
out of $n$ objects.) 

\item
(0 points but you'll need this for a later problem on this HW set
and maybe later in life as well.) 

By convention $(\forall n\ge 0)[\binom{n}{0}=1]$ and  $(\forall k\ge 1)[\binom{0}{k}=0]$.

From Part 1 that $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$.

Use these two equations to write a program that will do the following:

Given $N,K$ outputs $\binom{k}{n}$ for all $0\le k\le K$ and $0\le n\le N$. 

Run this program for $N=52$ and $K=6$. 

\end{enumerate}

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\item 
(30 points) 
Oliver-Poker is poker with a normal deck, but 
each player gets SIX cards.

For all questions here answer it BOTH in terms of binomial coefficient
(e.g., 

$$\frac{\binom{12}{8}}{{\binom{52}{6}}}$$

)

and as an actual number to four places (e.g., 0.2192). 

(For that you will use the output of the program you wrote
in Problem 2.) 

\begin{enumerate}
\item
(10 points) 
What is the probability of getting three 2-of-a-kinds?

\newcommand{\he}{\heartsuit}
\newcommand{\spa}{\spadesuit}
\newcommand{\cl}{\clubsuit}
\newcommand{\di}{\diamondsuit}

Example: $(2\he, 2\spa, 4\he, 4\di, 8\spa, 8\cl)$ is three 2-of-a-kind. 

Counterexample: $(2\he, 2\spa, 2\cl, 2\di, 8\spa, 8\cl)$ DOES NOT count as three 2-of-a-kind. 

\item 
(10 points) 
What is the probability of getting a 2-of-a-kind and a 4-of-a-kind? 

Example: $(2\he, 2\sp, 2\he, 2\di, 8\sp, 8\cl)$ is a 4-of-a-kind and a 2-of-a-kind. 

\item 
(10 points) 
What is the probability of getting a two 3-of-a-kind?

Example: $(2\he, 2\spa, 2\he, 8\di, 8\spa, 8\cl)$ is two 3-of-a-kind. 

\item
(0 points) Once its approved Oliver (also called {\it Poptart}) will be teaching a
STIC on Poker (real poker, not this problem). Consider taking it!!

\end{enumerate}

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\item 
(30 points)
Show that there is no way to load two 8-sided dice to get fair sums. 


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\item 
(30 points)

\begin{enumerate}
\item
(0 points but you have to do it)
Bill throws 2 fair 8-sided dice that are labeled in the normal way
(each dice has faces with 1,2,3,4,5,6,7,8). 
For each $0\le 2\le 16$, give the probability of the sum of the dice being $i$. 


\item 
(30 points) Give two 8-sided dice that are NOT labelled $(1,2,3,4,5,6,7,8)$.
which, when rolled, give the same probabilities from Part $a$. 
(The labels have to all be natural numbers that are $\ge 1$.) 



\end{enumerate}
\end{enumerate}

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