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AppendixII.tex
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AppendixII.tex
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\section{Appedix II: Getting the odds}
\subsection{Yatzee}
\subsubsection{rolling 0 dice}
This now is nice and easy there is ony one way for the dice to do anything, and thats to stay the same. So there is a 1 in column 5.
Total rolls: $6^0=1$
\subsubsection{rolling 1 die}
Five of the six sides do not match the other 4 dice, so for column 4 there's 5 possibilities, and there's one way to match the other dice, so for column 5 theres 1 possibility.
Total rolls: $6^1=6$
\subsubsection{rolling 2 dice}
There's $5^2$ ways to not match, ${2 \choose 1}\times5$ ways to match one of the two, and 1 way to match both.
Total rolls: $6^2=36$
\subsubsection{rolling 3 dice}
Three dice is the one state that's a little tricky, because if the three dice match and aren't the number I was saving I should switch numbers.
There's 1 way to roll a Yahtzee, ${3 \choose 2}\times5$ ways to move to rolling 1 die, ${3 \choose 1}\times5^2$ ways to rolling 2 dice plus an additional 5 ways where the dice match but don't match the two that were saved, to a total of ${3 \choose 1}\times5^2+5$. Finally theres $5^3-5$ ways to stay at rolling 3 dice.
Total rolls: $6^3=216$
\subsubsection{rolling 5 dice}
Five dice is quite nice, because all the cases are special in the same way, simply that there isn't a number that it going for.
There are 6 ways to straight roll a Yahtzee.
There are 6 ways to pick a number times ${5 choose 4}$ ways to pick what dice will be that number times 5 things the other die could be, so $6\times{5 \choose 4} \times5 = 150$.
There are two ways to get a three of a kind, a full house which occurs $6\times{5 \choose 3}\times5\times{2 \choose 2}=300$ ways,and three of a kind and two non matching numbers which occurs $6\times{5 \choose 3} \times5\times4=1200$ ways, so in all it can occur 1500 ways.
There are also two ways to get two of a kind, two pairs and another number $6\times{5 \choose 2}\times5\times{4 \choose 2}=1800$ and then there's the two pair and no other matching number case so $6\times{5 \choose 2}\times5\times4\times3=3600$ so there's $1800+3600=5400$ ways
There are $6\times5\times4\times3\times2=720$ ways to have no matching numbers and then to stay in the current state
Total rolls: $6^5=7776$
\subsubsection{Resulting matricies}
Plugging all of these numbers into their places results in the following matrix.
\[
\left(
\begin{array}{c c c c c}
720 & 5400 & 1500 & 150 & 6 \\
0 & 120 & 80 & 15 & 1 \\
0 & 0 & 25 & 10 & 1 \\
0 & 0 & 0 & 5 & 1 \\
0 & 0 & 0 & 0 & 1
\end{array}
\right)
\]
This is then multiplied by the inverse of the diagonal scaling matrix:
\[
\left(
\begin{array}{c c c c c c}
7776& 0 &0 &0 &0\\
0 & 216 &0 &0 &0\\
0 &0 &36 &0 &0\\
0 &0 &0 &6 &0\\
0 &0 &0 &0 &1
\end{array}
\right)
\]
\subsection{Rolling for a specific number}
This one can be fully described in a simple formula, moving from rolling $n$ dice to $r$ dice, given that $n \ge r$ is ${n \choose r} \times 5^r$. This makes getting the matrix much easier. Also the number to scale by is $6^n$
\subsubsection{Resulting matricies}
\[
\left(
\begin{array}{c c c c c c}
3125 & 3125 & 1250 & 250 & 25 & 1\\
0 & 625 & 500 & 150 & 20 & 1\\
0 & 0 & 125 & 75 & 15 & 1\\
0 & 0 & 0 & 25 & 10 & 1\\
0 & 0 & 0 & 0 & 5 & 1\\
0 & 0 & 0 & 0 & 0 & 1
\end{array}
\right)
\]
\[
\left(
\begin{array}{c c c c c c}
7776 &0&0&0&0&0\\
0 & 1296&0&0&0&0\\
0 &0&216&0&0&0\\
0 &0&0&36&0&0\\
0 &0&0&0&6&0\\
0&0&0&0&0&1
\end{array}
\right)
\]
\subsection{Full House}
\subsubsection{Holding no dice}
The only way to stay rolling no dice is if all are different so $6\times5\times4\times3\times2=720$
The way to go to a single is only if the first two match and the restare unique, so ${5 \choose 2}\times6\times5\times4\times3=3600$.
To get 3 of a number and 0 of any other is a Yahtzee, and there's 6 ways to do that.
The 3 of one and 1 of another has ${5 \choose 3}\times6\times5\times4 = 1200 $ ways the first ones can match and the last two don't, and another ${5 \choose 4}\times6\times5=150$ ways the four dice will match with one odd ball, for a total of 1350 possibilities.
Then there's ${5 \choose 2}\times6\times{3 \choose 2}\times5\times2 = 1800$ ways to get exactly 2 pair.
Finally there's ${5 \choose 3}\times6\times5 =300$ ways to staright roll a Yahtzee.
\subsubsection{Holding a Pair and a single}
There are 2 ways to move to a Full House from here, getting a pair that matches the single, 1 possiblity, or having each die match one of the numbers, 2 possibilities, for a total of 3 possibilities.
There are two wayst o move to two pair, one is matching the single and not the double, ${2 \choose 1} \times 1 \times 4 = 8$ possibilities, and the other is having the pair match a number other than the single so $ 4 $ possibilities, bringing it to a total of 12.
Then there's ${2 \choose 1} \times1 \times 5= 10$ ways to move to the 3 and 1 case.
Finally there's $ 11$ways to stay in the current state.
\subsubsection{Holding 3 of a kind}
To move to a full house one of 5 pairs is needed, so there's 5 ways to move to full house, and then to stay an pair that matches the three is needed, so 1 way to stay. In addition there are $5^2=25$ ways to moving to holding 3 of a number and 1 of another.
\subsubsection{Holding 3 of one number and 1 of another}
To change state a specific number is needed, so 1 chance to move up, 5 to stay
\subsubsection{Holding 2 pair}
If you are holding two pairs then the die can be any of two of six, so 4 chances to stay and 2 to move.
\subsubsection{Full House!}
Easy case, if you have a full house stick with it!!
\subsubsection{Resulting matricies}
\[
\left(
\begin{array}{c c c c c c}
720 & 3600 & 6 & 1350 & 1800 & 300\\
0 & 11 & 0 & 10 & 12 & 3\\
0 & 0 & 1 & 30 & 0 & 5\\
0 & 0 & 0 & 5 & 0 & 1\\
0 & 0 & 0 & 0 & 4 & 2\\
0 & 0 & 0 & 0 & 0 & 1\\
\end{array}
\right)
\]
\[
\left(
\begin{array}{c c c c c c}
7776&0 & 0 & 0 &0 &0\\
0 &36 & 0 & 0 &0 &0\\
0 &0 & 36 & 0 &0 &0\\
0 &0 & 0 & 6 &0 &0\\
0 &0 & 0 & 0 &6 &0\\
0 &0 & 0 & 0 &0 &1
\end{array}
\right)
\]