\documentclass{article}
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\begin{document}
\title{Point values in the game ``Target''}
\author{Greg Stoll}
{\Large Point values in the game ``Target''}
\tableofcontents[section,subsection]
\section{Introduction}
\href{http://boardgamegeek.com/game/2049}{Target} is a game where you draw
cards to try to satisfy the target cards. Each of the cards has a separate
point value, and I wondered how "true" those points were. So, my goal was to
calculate the number of 6 card hands (in the game your hand limit is 5, but
you draw a card and then can turn sets in) that satisfy each of the conditions
and see if there was a way to derive how points were allocated. It was also a
good excuse to calculate (using combinatorics) the exact number of hands that
satisfy each condition, and to check those numbers using a Python script.
\section{Setup}
Target cards are 0-9 in four different suits (purple, green, red, blue) plus
1 wild suit in each number. 0 and 9 have one of each suit and all other numbers
have two of each suit for a total of $2(4+1)+8(2\cdot4+1)=82$ cards.
The total number of possible hands is \({82\choose 6}=350161812\). \\
\section{Graph}
\label{graph}
The x-axis represents the probability that a random 6-card hand will satisfy
the target card. The y-axis is just the number of cards you have to turn in
to fulfill the target card, and the color of the dot (and the corresponding
text) is how many points it's worth. So, dots in the upper-left hand corner
are the hardest to get and take the most cards, and you would expect these to
be worth the most points (which they generally are).\\
\includegraphics[0mm,0mm][640px,480px]{targetpoints.png}\\
There are some interesting results here - all the 5 point targets are at the
left and towards the top, as expected, but there are some interesting outliers.
At 2 points, 5 odd cards seems undervalued; it's the only target that takes more
than 3 cards that's only worth 2 points (and it has a pretty low probability
to boot). 4 card straight including a 3 and skip straight both seem undervalued
since mixed straight is more likely and is worth more points. 3 card straight
including a 2 looks similarly undervalued. (or maybe 3 cards totalling <= 4 is
overvalued?)
One possibility is that the designers of the game looked at the probabilities
with 5 card hands instead of 6 cards as I've done. Another is that they chose
to fiddle with the point values to account for the fact that certain target
cards are easier to turn in together - if you have 5 odd cards you have a
pretty good chance of having a skip straight, etc.
\section{Results}
I was able to calculate almost all the results symbolically. Here are the
results. Click on a hand type to see the derivation of that number. These
results were derived by hand (except the ones in bold) and checked with a
\hyperref[calculateprobs]{Python script}.\\
\begin{flushleft}
\begin{tabular}{lllrr}
& Hand type & Points & Possible hands & Probability\\
Flush \\
& \label{flush3cardspeftop}\hyperref[flush3cardspef]{3 card, specific suit} & 2 & 116241489 & 33.197\%\\
& \label{flush4cardanytop}\hyperref[flush4cardany]{4 card, any suit} & 3 & 122384088 & 34.951\% \\
& \label{flush5cardanytop}\hyperref[flush5cardany]{5 card, any suit} & 5 & 22272138 & 6.361\% \\
Straight \\
& \label{straight3cardwith2top}\hyperref[straight3cardwith2]{3 card including a 2} & 2 & 72355734 & 20.664\%\\
& \label{straight3cardwith4top}\hyperref[straight3cardwith4]{3 card including a 4} & 2 & 81222993 & 23.196\%\\
& \label{straight3cardwith5top}\hyperref[straight3cardwith5]{3 card including a 5} & 2 & 81222993 & 23.196\%\\
& \label{straight3cardwith7top}\hyperref[straight3cardwith7]{3 card including a 7} & 2 & 72355734 & 20.664\%\\
& \label{straight4cardwith3top}\hyperref[straight4cardwith3]{4 card including a 3} & 3 & 36204327 & 10.339\%\\
& \label{straight4cardwith6top}\hyperref[straight4cardwith6]{4 card including a 6} & 3 & 36204327 & 10.339\%\\
& \label{straight5cardtop}\hyperref[straight5card]{5 card} & 5 & 15149349 & 4.326\%\\
Straight Flush \\
& \label{straightflush3cardspeftop}\hyperref[straightflush3cardspef]{3 card, specific suit} & 4 & 12588978 & 3.595\%\\
& \label{straightflush3cardanytop}\hyperref[straightflush3cardany]{\bf 3 card, any suit} & 3 & 45888079 & 13.105\%\\
& \label{straightflush4cardanytop}\hyperref[straightflush4cardany]{\bf 4 card, any suit} & 5 & 5054955 & 1.444\%\\
Poker \\
& \label{twopairtop}\hyperref[twopair]{2 pair} & 4 & 92445176 & 26.401\%\\
& \label{kind3top}\hyperref[kind3]{3 of a kind} & 4 & 45800298 & 13.080\%\\
& \label{fullhousetop}\hyperref[fullhouse]{Full house} & 5 & 12646280 & 3.612\%\\
Misc \\
& \label{fiveoddtop}\hyperref[fiveodd]{5 odd cards} & 2 & 35221706 & 10.059\%\\
& \label{fiveeventop}\hyperref[fiveeven]{5 even cards} & 2 & 35221706 & 10.059\%\\
& \label{total4top}\hyperref[total4]{3 cards totalling $\leq$ 4} & 3 & 73617980 & 21.024\%\\
& \label{total23top}\hyperref[total23]{3 cards totalling $\geq$ 23} & 3 & 73617980 & 21.024\%\\
& \label{skipstraighttop}\hyperref[skipstraight]{Skip straight} & 3 & 35894502 & 10.251\%\\
& \label{mixedstraighttop}\hyperref[mixedstraight]{\bf Mixed straight} & 4 & 62519949 & 17.855\%\\
\end{tabular}
\end{flushleft}
\section{Derivation}
\subsection{Flushes}
\subsubsection{3 card flush, specific suit}
\label{flush3cardspef}
Without loss of generality, let the suit be G(reen). There are 28 cards that
can be G - 18 G and 10 W(ild), and so there are $82-28=54$ cards that cannot be
G. To avoid double-counting, we split up the possibilities:
\begin{itemize}
\item 6 card flush: \({28\choose{6}}=376740\)
\item 5 card flush: \({28\choose 5}{54\choose 1}=5307120\)
\item 4 card flush: \({28\choose 4}{54\choose 2}=29299725\)
\item 3 card flush: \({28\choose 3}{54\choose 3}=81257904\)
\end{itemize}
So the total is $376740+5307120+29299725+81257904=\mathbf{116241489}$.\\
\hyperref[flush3cardspeftop]{Back to table}
\subsubsection{4 card flush, any suit}
\label{flush4cardany}
We break it down by the number of wilds in the hand. Note that there are
10 wilds and 18 cards of each suit, which means there are 54 cards not in a
given suit.
\begin{itemize}
\item 0 wilds: There are 4 choices for the suit the flush is in, and in a given suit there are \({18\choose6}+{18\choose5}{54\choose1}+{18\choose4}{54\choose2}\)
possibilities, which gives \(4({18\choose6}+{18\choose5}{54\choose1}+{18\choose4}{54\choose2})=19440384\).
\item 1 wild: There are 10 choices for which wild we have. We only need 3 cards
in any one suit to have a flush, and since there are only 5 cards left we don't
need to worry about overcounting. So there are 4 choices for a suit and
\({18\choose5}+{18\choose4}{54\choose1}+{18\choose3}{54\choose2}\) choices for
the cards in that suit, which gives
\(10\cdot4({18\choose5}+{18\choose4}{54\choose1}+{18\choose3}{54\choose2})=53660160\).
\item 2 wilds: There are \(10\choose2\) choices for which wilds we have. There
are 4 cards left and we only need 2 of the same suit to make a flush, so only
1 card in each suit will fail to give a flush. There are $18^4$ failures,
so there are \({72\choose4}-18^4\) possibilities, which gives
\({10\choose2}({72\choose4}-18^4)=41571630\).
\item 3 wilds: There are \(10\choose3\) choices for which wilds we have, and any
remaining set of cards will give a 4 card flush, so there are \(72\choose3\)
possibilities, which gives \({10\choose3}{72\choose3}=7156800\).
\item 4 wilds: There are \(10\choose4\) choices for which wilds we have, and
any remaining set of cards will give a 4 card flush, so there are a total of
\({10\choose4}{72\choose2}=536760\) possibilities.
\item 5 wilds: There are \(10\choose5\) choices for which wilds we have, and
any remaining set of cards will give a 4 card flush, so there are a total of
\({10\choose5}{72\choose1}=18144\) possibilities.
\item 6 wilds: There are \(10\choose6\) possibilities.
\end{itemize}
So, the total is $19440384+53660160+41571630+7156800+536760+18144+210=\mathbf{122384088}$.\\
\hyperref[flush4cardanytop]{Back to table}
\subsubsection{5 card flush, any suit}
\label{flush5cardany}
Similarly, we break it down by the number of wilds in the hand.
\begin{itemize}
\item 0 wilds: There are 4 choices for the suit the flush is in, and in a given
suit there are \({18\choose6}+{18\choose5}{54\choose1}\) possibilities, which
gives \(4({18\choose6}+{18\choose5}{54\choose1})=1924944\).
\item 1 wild: There are 10 choices for which wild we have, and 4 choices for the
suit the flush is in. We need 4 out of the remaining 5 cards to be in that suit,
so there are \({18\choose5}+{18\choose4}{54\choose1}\) possibilities for those
cards, which gives a total of
\(10\cdot4({18\choose5}+{18\choose4}{54\choose1})=6952320\).
\item 2 wilds: There are \(10\choose2\) choices for which wilds we have. We need
3 cards in any one suit to give a flush and there are 4 cards left so there
is no danger of overcounting. Thus, there are 4 choices for a suit and
\({18\choose4}+{18\choose3}{54\choose1}\) choices for the cards in that suit,
which gives a total of \({10\choose2}4({18\choose4}+{18\choose3}{54\choose1})=8482320\).
\item 3 wilds: There are \(10\choose3\) choices for which wilds we have. We need
2 cards in any one suit to give a flush and there are 3 cards left so there
is no danger of overcounting. Thus, there are 4 choices for a suit and
\({18\choose3}+{18\choose2}{54\choose1}\) choices for the cards in that suit,
which gives a total of \({10\choose3}4({18\choose3}+{18\choose2}{54\choose1})=4357440\).
\item 4 wilds: There are \(10\choose4\) choices for which wilds we have, and
any remaining cards will give a flush, so there are a total of
\({10\choose4}{72\choose2}=536760\) possibilities.
\item 5 wilds: There are \(10\choose5\) choices for which wilds we have, and
any remaining set of cards will give a flush, so there are a total of
\({10\choose5}{72\choose1}=18144\) possibilities.
\item 6 wilds: There are \(10\choose6\) possibilities.
\end{itemize}
So, the total is $1924944+6952320+8482320+4357440+536760+18144+210=\mathbf{22272138}$.\\
\hyperref[flush5cardanytop]{Back to table}
\subsection{Straights}
\subsubsection{3 card including a 2}
\label{straight3cardwith2}
We break it down by the length of the straight:
\begin{itemize}
\item 6 card straight: For a 0-5 straight there are $5\cdot9^5$ possibilities,
and for a 1-6 or 2-7 straight there are $9^6$ possibilities for a total of
$5\cdot9^5+2\cdot9^6=1358127$.
\item 5 card straight: For a 0-4 straight there are $5\cdot9^4$ possibilities
and $82-5-9$ possibilities for the other card (subtracting 5 for the cards
used in the straight, and 9 because if the extra card were a 5 that would be a
6 card straight which we counted above). We've overcounted the cases where the
other card is a 0-4, so we need to subtract out all straights of 0-4 where the
card distribution is (2,1,1,1,1). If 0 has 2 cards there are \({5\choose2}9^4\)
of these, and if anywhere from 1-4 has 2 cards there are
\({9\choose2}5\cdot9^3\). So for a 0-4 straight there are
\(5\cdot9^4(82-5-9)-\left({5\choose2}9^4+4{9\choose2}5\cdot9^3\right)\)
possibilities. Using a similar technique, for a 1-5 straight there are
$9^5(82-5-5-9)$ possibilities, but we overcounted the (2,1,1,1,1) distributions
so the true number is \(9^5(82-5-5-9)-5{9\choose2}9^4\), and for a 2-6 straight
there are $9^5(82-5-9-9)$ possibilities, but we overcounted the (2,1,1,1,1)
distributions so the true number is \(9^5(82-5-9-9)-5{9\choose2}9^4\).
This gives us a total of
\(5\cdot9^4(82-5-9)-\left({5\choose2}9^4+4{9\choose2}5\cdot9^3\right)+9^5(82-5-5-9)-\left(5{9\choose2}9^4\right)+9^5(82-5-9-9)\left(5{9\choose2}9^4\right)=6482268\).
\item 4 card straight: We use the inclusion-exclusion technique as described
in the \hyperref[straightflush3cardspef]{3 card straight flush, specific suit} case. For the 0-3 case our universe of cards is all 82 cards except any 4 (to
avoid falling in the 5 card straight case), so there are $82-9=73$ cards in our
universe $U$. We get \({{73}\choose6}-\left({{73-5}\choose6}+3{{73-9}\choose6}\right)+\left(3{{73-14}\choose6}+3{{73-18}\choose6}\right)-\left(3{{73-23}\choose6}+{{73-27}\choose6}\right)+{{73-32}\choose6}=5452920\). For the 1-4 case our universe
is all cards except any 0 or 5, so there are $82-5-9=68$ cards in our universe.
We get \({{68}\choose6}-4{{68-9}\choose6}+6{{68-18}\choose6}-4{{68-27}\choose6}+{{68-36}\choose6}=7488288\). For the 2-5 case our universe is all cards except
any 1 or 6, so there are $82-9-9=64$ cards in our universe, and we get
\({{64}\choose6}-4{{64-9}\choose6}+6{{64-18}\choose6}-4{{64-27}\choose6}+{{64-36}\choose6}=6294186\), for a total of $5452920+7488828+6294186=19235394$.
\item 3 card straight: We use inclusion-exclusion again. For a 0-2 straight our
universe is all 82 cards except any 3 (to avoid falling in the 4 card straight
case), so there are $82-9=73$ cards in our universe $U$. So we get
\({{73}\choose6}-\left({{73-5}\choose6}+2{{73-9}\choose6}\right)+\left(2{{73-14}\choose6}+{{73-18}\choose6}\right)-{{73-23}\choose6}=14042295\). For a 1-3
straight our universe is everything except 0's and 4's, and there are $82-5-9=68$ of these. So we get \({{68}\choose6}-3{{68-9}\choose6}+3{{68-18}\choose6}-{{68-27}\choose6}=17456634\). For a 2-4 straight our universe is everything except
1's and 5's, and there are $82-9-9=64$ of these. So we get
\({{64}\choose6}-3{{64-9}\choose6}+3{{64-18}\choose6}-{{64-27}\choose6}=13781016\), for a total of $14042295+17456634+13781016=45279945$.
\end{itemize}
So, the total is $1358127+6482268+19235394+45279945=\mathbf{72355734}$. \\
\hyperref[straight3cardwith2top]{Back to table}
\subsubsection{3 card including a 4}
\label{straight3cardwith4}
We break it down by the length of the straight:
\begin{itemize}
\item 6 card straight: For a 0-5 or 4-9 straight there are $5\cdot9^5$
possibilities, and for a 1-6, 2-7, or 3-8 straight there are $9^6$ possibilities
for a total of $2\cdot5\cdot9^5+3\cdot9^6=2184813$.
\item 5 card straight: For a 0-4 straight there are $5\cdot9^4$ possibilities
and $82-5-9$ possibilities for the other card (subtracting 5 for the cards
used in the straight, and 9 because if the extra card were a 5 that would be a
6 card straight which we counted above). We've overcounted the cases where the
other card is a 0-4, so we need to subtract out all straights of 0-4 where the
card distribution is (2,1,1,1,1). If 0 has 2 cards there are \({5\choose2}9^4\)
of these, and if anywhere from 1-4 has 2 cards there are
\({9\choose2}5\cdot9^3\). So for a 0-4 straight there are
\(5\cdot9^4(82-5-9)-\left({5\choose2}9^4+4{9\choose2}5\cdot9^3\right)\)
possibilities. Using a similar technique, for a 1-5 (or 4-8) straight there are
$9^5(82-5-5-9)$ possibilities, but we overcounted the (2,1,1,1,1) distributions
so the true number is \(9^5(82-5-5-9)-5{9\choose2}9^4\), and for a 2-6 (or 3-7)
straight there are $9^5(82-5-9-9)$ possibilities, but we overcounted the
(2,1,1,1,1) distributions so the true number is
\(9^5(82-5-9-9)-5{9\choose2}9^4\). This gives us a total of
\(5\cdot9^4(82-5-9)-\left({5\choose2}9^4+4{9\choose2}5\cdot9^3\right)+2\left(9^5(82-5-5-9)-\left(5{9\choose2}9^4\right)\right)+2\left(9^5(82-5-9-9)\left(5{9\choose2}9^4\right)\right)=11324286\).
\item 4 card straight: We use the inclusion-exclusion technique as described
in the \hyperref[straightflush3cardspef]{3 card straight flush, specific suit} case. For the 1-4 case our universe is all cards except any 0 or 5, so there
are $82-5-9=68$ cards in our universe.
We get \({{68}\choose6}-4{{68-9}\choose6}+6{{68-18}\choose6}-4{{68-27}\choose6}+{{68-36}\choose6}=7488288\). For the 2-5 (and 3-6 and 4-7) case our universe
is all cards except any 1 or 6, so there are $82-9-9=64$ cards in our universe,
and we get
\({{64}\choose6}-4{{64-9}\choose6}+6{{64-18}\choose6}-4{{64-27}\choose6}+{{64-36}\choose6}=6294186\), for a total of $7488828+3\cdot6294186=26370846$.
\item 3 card straight: We use inclusion-exclusion again. For a 2-4 (and 3-5
and 4-6) straight our universe is everything except 1's and 5's, and there are
$82-9-9=64$ of these. So we get
\({{64}\choose6}-3{{64-9}\choose6}+3{{64-18}\choose6}-{{64-27}\choose6}=13781016\), for a total of $3\cdot13781016=41343048$.
\end{itemize}
So, the total is $2184813+11324286+26370846+41343048=\mathbf{81222993}$. \\
\hyperref[straight3cardwith4top]{Back to table}
\subsubsection{3 card including a 5}
\label{straight3cardwith5}
By symmetry this is the same number as
\hyperref[straight3cardwith4]{3 card straight including a 4}, which is
{\bf 81222993}.\\
\hyperref[straight3cardwith5top]{Back to table}
\subsubsection{3 card including a 7}
\label{straight3cardwith7}
By symmetry this is the same number as
\hyperref[straight3cardwith2]{3 card straight including a 2}, which is
{\bf 72355734}.\\
\hyperref[straight3cardwith7top]{Back to table}
\subsubsection{4 card including a 3}
\label{straight4cardwith3}
We break it down into the length of the straight:
\begin{itemize}
\item 6 card straight: For a 0-5 straight there are $5\cdot9^5$ possibilities,
and for a 1-6, 2-7, or 3-8 straight there are $9^6$ possibilities for a total
of \(5\cdot9^5+3\cdot9^6=1889568\).
\item 5 card straight: For a 0-4 straight there are $5(9^4)$ possibilities for
the straight and $82-5-9$ possibilities for the other card (we subtract 5 for the 5 cards used in the straight, and 9 because if the extra card were a 5 that
would be a 6 card straight which we counted above). However, we've overcounted
the cases where the other card is a 0-4. So we need to subtract out all
straights of 0-4 where the card distribution is (2,1,1,1,1). If 0 has 2 cards
there are \({5\choose2}9^4\) of these, and if anywhere from 1-4 has 2 cards
there are \(5{9\choose2}9^3\). So for a 0-4 straight there are \(5\cdot9^4(82-5-9)-({5\choose2}9^4+4{9\choose2}5\cdot9^3)\) possibilities. Using a similar
technique, for a 1-5 straight there are $9^5(82-5-5-9)$ possibilities, but we overcounted the (2,1,1,1,1) distributions so the true number is \(9^5(82-5-5-9)-5{9\choose2}9^4\). For a 2-5 straight (and similarly a 3-6 straight) there
are \(9^5(82-5-9-9)-5{9\choose2}9^4\) possibilities. This gives us a total of
\(5\cdot9^4(82-5-9)-\left({5\choose2}9^4+4{9\choose2}5\cdot9^3\right)+9^5(82-5-5-9)-5{9\choose2}9^4+2\left(9^5(82-5-9-9)-5{9\choose2}9^4\right)=8785179\).
\item 4 card straight: We break it down again by the type of straight:
\begin{itemize}
\item 0-3: There are $5(9^3)$ possibilities for the straight and
\({82-4-9}\choose2\) possibilities for the other two cards (excluding the cards
we already chose and any 4's since that would make it a 5 card straight).
However, again we've overcounted cases where the extra cards are 0-3. The
distributions of cards we have to exclude are (2,1,1,1), (3,1,1,1), and
(2,2,1,1). For (2,1,1,1), if 0 has 2 cards there are \({5\choose2}9^3\) possibilities and if anywhere from 1-3 has 2 cards there are \({9\choose2}5\cdot9^2\)
possibilities. In either case, there are $82-5-9\cdot4$ possibilities for the
extra card, since the extra card can't be anywhere from 0-4, to avoid changing the distribution or making it a 5 card straight. For (3,1,1,1), if 0 has 3 cards
there are \({5\choose3}9^3\) possibilities and otherwise there are
\({9\choose3}5\cdot9^2\) possibilities. For (2,2,1,1), if 0 has 2 cards there
are \({5\choose2}3{9\choose2}9^2\) possibilities and otherwise there are
\({9\choose2}{9\choose2}5\cdot9\) possibilities. Note that we have to subtract
2 times the number of (3,1,1,1) overcounting (since we've counted these
$3\cdot1\cdot1\cdot1$ times and we want to only count them once) and 3 times
the number of (2,2,1,1) overcounting. This gives us a total of
\(
\begin{array}{lll}
5\cdot9^3{{82-4-9}\choose2} & - ({5\choose2}9^3+3{9\choose2}5\cdot9^2)(82-5-9\cdot4) & \\
& - 2({5\choose3}9^3+3{9\choose3}5\cdot0^2) & \\
& - 3({5\choose2}3{9\choose2}9^2+{3\choose2}{9\choose2}{9\choose2}5\cdot9 & = 5452920
\end{array}
\)
\item 1-4: We use a similar technique as the 0-3 case. There are $9^4$
possibilities for the straight and \({{82-4-5-9}\choose2}\) possibilities for the other cards (excluding the cards we chose and 0's and 5's to avoid 5 card
straights). For the (2,1,1,1) distribution there are
\(4{9\choose2}9^3\) possibilities and $82-5-9\cdot5$ possibilities for the extra
card (avoiding any numbers from 0-5). For the (3,1,1,1) distribution there are
\(4{9\choose3}9^3\) possibilities and for the (2,2,1,1) distribution there are
\({4\choose2}{9\choose2}{9\choose2}9^2\) possibilities. This gives us a total
of
\(
\begin{array}{lll}
9^4{{82-4-5-9}\choose2} & - (4{9\choose2}9^3)(82-5-9\cdot5) & \\
& - 2(4{9\choose3}9^3) & \\
& - 3({4\choose2}{9\choose2}{9\choose2}9^2 & = 7488288
\end{array}
\)
\item 2-5 (and similarly 3-6): We use a similar technique as the 0-3 case.
There are $9^4$
possibilities for the straight and \({{82-4-9-9}\choose2}\) possibilities for the other cards (excluding the cards we chose and 1's and 6's to avoid 5 card
straights). For the (2,1,1,1) distribution there are
\(4{9\choose2}9^3\) possibilities and $82-9\cdot6$ possibilities for the extra
card (avoiding any numbers from 1-6). For the (3,1,1,1) distribution there are
\(4{9\choose3}9^3\) possibilities and for the (2,2,1,1) distribution there are
\({4\choose2}{9\choose2}{9\choose2}9^2\) possibilities. This gives us a total
of
\(
\begin{array}{lll}
9^4{{82-4-9-9}\choose2} & - (4{9\choose2}9^3)(82-9\cdot6) & \\
& - 2(4{9\choose3}9^3) & \\
& - 3({4\choose2}{9\choose2}{9\choose2}9^2 & = 6294186
\end{array}
\)
\end{itemize}
\end{itemize}
So, the total is $1889568+8785179+5452920+7488288+2\cdot6294186=\mathbf{36204327}$.\\
\hyperref[straight4cardwith3top]{Back to table}
\subsubsection{4 card including a 6}
\label{straight4cardwith6}
By symmetry this is the same number as
\hyperref[straight4cardwith3]{4 card straight including a 3},
which is {\bf 36204327}.\\
\hyperref[straight4cardwith6top]{Back to table}
\subsubsection{5 card}
\label{straight5card}
We break it down into the length of the straight:
\begin{itemize}
\item 6 card straight: To make a 0-5 straight, there are $5\cdot9^5$
possibilities (and the same number for a 4-9 straight), and for 1-6 there are
$9^6$ for a total of $2(5\cdot 9^5)+3(9^6)=2184813$.
\item 5 card straight: For a 0-4 straight, there are $5\cdot9^4$ ways. There
are $82-5-9$ ways to choose the next card (to avoid ones we've already chosen
and to avoid making a 6 card straight). However, this double counts cases with
a (2,1,1,1,1) distribution of cards over 0-4, so we have to subtract those -
there are \({5\choose2}9^4\) where we have 2 0's and \(4({9\choose2}5\cdot9^3)\) where we have 2 of something else. We use a similar technique to count
the rest of the straights, giving us a total of
\begin{eqnarray*}
2[(5\cdot9^4)(82-5-9)-({5\choose2}9^4+4({9\choose2}5\cdot9^3))] & & \\
+2(9^5(82-5-9-5)-5{9\choose2}9^4) & &\\
+2(9^5(82-5-9-9)-5{9\choose2}9^4) & = & 12964536
\end{eqnarray*}
\end{itemize}
So, the total is $2184813+12964536=\mathbf{15149349}$.\\
\hyperref[straight5cardtop]{Back to table}
\subsection{Straight Flushes}
\subsubsection{3 card, specific suit}
\label{straightflush3cardspef}
Since we're dealing in a specific suit (let's say G(reen)), there are 3 cards
of that suit (including 1 wild) for each number except 0 and 9, and there are
2 cards for those. We break it down by the length of the straight flush:
\begin{itemize}
\item 6 card straight flush: For a 0-5 or 4-9 straight there are $2\cdot3^5$
possibilities and for a 1-6, 2-7, or 3-8 straight there are $3^6$ possibilities
for a total of \(2(2\cdot3^5)+3(3^6)=3159\).
\item 5 card straight flush: For a 0-4 straight, there are $2\cdot3^4$
ways to choose the straight, and $82-5-3$ ways to choose the next card (to avoid
cards we've already chosen and to avoid making a 6 card straight). However,
this double counts cases with a (2,1,1,1,1) distribution of cards over 0-4,
so we have to subtract those - there are \({2\choose2}3^4\) where we have 2 0's
and \(4{3\choose2}2\cdot3^3\) where we have 2 of something else. We use a
similar technique to count the rest of the straights, giving us a total of
\begin{eqnarray*}
2\left[(2\cdot3^4)(82-5-3)-\left({2\choose2}3^4+4{3\choose2}2\cdot3^3\right)\right] & & \\
+2\left[3^5(82-5-2-3)-5{3\choose2}3^4\right] & & \\
+2\left[3^5(82-5-3-3)-5{3\choose2}3^4\right] & = & 87156
\end{eqnarray*}
\item 4 card straight flush: We could use a similar technique to the 5 card
case (like we do in the
\hyperref[straight4cardwith3]{4 card straight including a 3} case). However,
let's try something different and use the \href{http://en.wikipedia.org/wiki/Inclusion-exclusion_principle}{Inclusion-exclusion principle}.
For the 0-3 case, our universe of cards is all 82 cards except for any Green
4 (to avoid falling in the 5 card straight flush case), so there are $82-3=79$
cards in our universe $U$. Let $A_0$ be the set of hands that don't have a
Green 0, $A_1$
be the set of hands that don't have a Green 1, up to $A_3$ being the set of
hands that don't have a Green 3. Then what we want is the set of hands that
satisfy none of $A_0, \ldots A_4$, or \(\left|\bigcap \overline{A_i}\right|\),
which by de Morgan's laws is \(\left|\overline{\bigcup A_i}\right|=\left|U\right|-\left|\bigcup A_i\right|\). Now, $\left|A_0\right|$ is the
number of hands that don't have a Green 0, which is \({{79-2}\choose6}\), and the other $\left|A_i\right|$'s are \({{79-3}\choose6}\). $\left|A_0\cap A_i\right|$ is the number of hands that don't have a Green 0 or a Green $i$, which is
\({{79-2-3}\choose6}\), and the other $\left|A_i\cap A_j\right|$'s are
\({{79-3-3}\choose6}\). We continue on in this manner to find that the number
of 4 card straight flushes of numbers 0-3 is
\({{79}\choose6}-\left[{{79-2}\choose6}+3{{79-3}\choose6}\right]+\left[3{{79-5}\choose6}+3{{79-6}\choose6}\right]-\left[3{{79-8}\choose6}+{{79-9}\choose6}\right]+{{79-11}\choose6}=136161\) (and by symmetry this is the number of 4 card
straight flushes 6-9). We use a similar technique (letting our universe of
cards be all 82 cards except Green 0's and Green 5's) to calculate the number
of 4 card straight flushes of numbers 1-4 (and 5-8 by symmetry) is
\({{77}\choose6}-4{{77-3}\choose6}+6{{77-6}\choose6}-4{{77-9}\choose6}+{{77-12}\choose6}=190134\). Again, letting our universe be all 82 cards except Green 1's
and Green 6's, the number of 4 card straight flushes of numbers 2-5 (and 3-6
and 4-7 by symmetry) is
\({{76}\choose6}-4{{76-3}\choose6}+6{{76-6}\choose6}-4{{76-9}\choose6}+{{76-12}\choose6}=184626\), which gives us a total of $2\cdot136161+2\cdot190134+3\cdot184626=1206468$.
\item 3 card straight flush: We use a similar technique as above, although we
have to consider the case where we have two non-adjacent 3 card straight flushes
separately. Notice that under this technique, the expressions get simpler the
fewer cards we have, unlike the technique of subtracting off distributions we
overcounted which gets more complicated the fewer cards we have.
For 0-2, our universe of cards is 82-3=79, so we get
\({{79}\choose6}-\left[{{79-2}\choose6}+2{{79-3}\choose6}\right]+\left[2{{79-5}\choose6}+{{79-6}\choose6}\right]-{{79-8}\choose6}=1144050\), although we have to
subtract the two non-adjacent 3 card straight flushes which is
\((2\cdot3^2)(3\cdot3^3+2\cdot3^2)=1782\). (these are the same as 7-9 by symmetry) For 1-3 (and by symmetry 6-8) our universe of cards is 82-5=77, so we get
\({{77}\choose6}-3{{77-3}\choose6}+3{{77-6}\choose6}-{{77-9}\choose6}=1545075\),
then subtracting \(3^3(2\cdot3^3+2\cdot3^2)=1944\). For 2-4 (and 5-7) our
universe of cards is 82-6=76, so we get
\({{76}\choose6}-3{{76-3}\choose6}+3{{76-6}\choose6}-{{76-9}\choose6}=1479843\),
then subtracting \(3^3(3^3+2\cdot3^2)=1215\). For 3-5 (and 4-6) our
universe of cards is 82-6=76, so we get
\({{76}\choose6}-3{{76-3}\choose6}+3{{76-6}\choose6}-{{76-9}\choose6}=1479843\),
then subtracting \(3^3(2\cdot3^2)=486\).
Finally, we have to add in the case where there are two non-adjacent 3 card
straights. If the first straight is 0-2, the other straight could be anywhere
from 4-6 to 7-9, which gives $2\cdot3^2(3\cdot3^3+2\cdot3^2)$. Adding up
all such cases gives an additional
$2\cdot3^2(3\cdot3^3+2\cdot3^2)+3^3(2\cdot3^3+2\cdot3^2)+3^3(3^3+2\cdot3^2)+3^3(2\cdot3^2)=5427$ possibilities.
This gives a total of
$2(1144050-1782)+2(1545075-1944)+2(1479843-1215)+2(1479843-486)+5427=11292195$.
\end{itemize}
So, the total is $3159+87156+1206468+11292195=\mathbf{12588978}$. \\
\hyperref[straightflush3cardspeftop]{Back to table}
\subsubsection{3 card, any suit}
\label{straightflush3cardany}
You can perhaps do some sort of inclusion-exclusion argument where your sets
are the hands that have 3 cards in any particular suit, but it gets very
complicated very fast - open to suggestions.\\
\hyperref[straightflush3cardanytop]{Back to table}
\subsubsection{4 card, any suit}
\label{straightflush4cardany}
You can perhaps do some sort of inclusion-exclusion argument where your sets
are the hands that have 4 cards in any particular suit, but it gets very
complicated very fast - open to suggestions.\\
\hyperref[straightflush4cardanytop]{Back to table}
\subsection{Poker Hands}
\subsubsection{2 pair}
\label{twopair}
We can make two pair by all the possibilities under \hyperref[fullhouse]{Full House}, plus the following:
\begin{itemize}
\item 2-2-2 split (i.e. three pairs): If we use the numbers 0 and 9, there are
\({5\choose2}{5\choose2}\) for the choices of the cards of number 0 and 9, 8
choices for the other number, and \({9\choose2}\) choices for the cards of that
number. If we use only one of 0 and 9, there are 2 choices for which one to use,
\({5\choose2}\) choices for the cards of that number, \({8\choose2}\) choices
for the other two numbers, and \({9\choose2}\) choices for the cards in each number. If we use neither 0 nor 9, there are \(8\choose3\) choices for the numbers
and \({9\choose2}\) for the cards in each number. This gives a total of
\({5\choose2}{5\choose2}\cdot8{9\choose2}+2{5\choose2}{8\choose2}{9\choose2}{9\choose2}+{8\choose3}{9\choose2}{9\choose2}{9\choose2}=3367296\).
\item 2-2 split with no other pair: If we use 0 and 9, there are \({5\choose2}{5\choose2}\) possibilities for those cards. For the other two cards, they can't
be 0 or 9, so there are \({{82-5-5}\choose2}\) choices, except \(8{9\choose2}\) of these lead to another pair, so we subtract them. Similarly, if we use one of
0 and 9, there are \(2{5\choose2}8{9\choose2}({{82-5-9}\choose2}-(7{9\choose2}+{5\choose2}))\) choices and if we use neither 0 nor 9 there are
\({8\choose2}{9\choose2}{9\choose2}({{82-9-9}\choose2}-(6{9\choose2}+2{5\choose2}))\) for a total of \({5\choose2}{5\choose2}({{82-5-5}\choose2}-8{9\choose2})+
2{5\choose2}8{9\choose2}({{82-5-9}\choose2}-(7{9\choose2}+{5\choose2}))+
{8\choose2}{9\choose2}{9\choose2}({{82-9-9}\choose2}-(6{9\choose2}+2{5\choose2}))=76431600\).
\end{itemize}
So, the total is $12646280+3367296+76431600=\mathbf{92445176}$.\\
\hyperref[twopairtop]{Back to table}
\subsubsection{3 of a kind}
\label{kind3}
We break it down by the number of cards of the same number:
\begin{itemize}
\item 6 of a kind: Since there are only 5 0's and 9's, we must be a 1-8, so
there are 8 choices for the number and \({9\choose6}\) ways to choose the cards
of that number for a total of \(8{9\choose6}=672\).
\item 5 of a kind: If the number is 0 or 9, there are \({5\choose5}\) ways to
choose the cards of that number and $82-5$ ways to choose the other card.
Otherwise, there are \({9\choose5}\) ways to choose the cards of that number
and $82-9$ (to avoid picking the same number again) to choose the other card.
This gives a total of \(2{5\choose5}(82-5)+8{9\choose5}(82-9)=73738\).
\item 4 of a kind: If the number is 0 or 9, there are \({5\choose4}\) ways to
choose the cards of that number and \({82-5}\choose2\) ways to choose the other
card. Otherwise, there are \({9\choose4}\) ways to choose the cards of that
number and \({82-9}\choose2\) ways to choose the other card. This gives a
total of \(2{5\choose4}{{82-5}\choose2}+8{9\choose4}{{82-9}\choose2}=2678284\).
\item 3 of a kind and 3 of another kind: (we break this out separately to avoid
overcounting) - we count this in the \hyperref[fullhouse]{Full house} section,
and the result is 211108.
\item 3 of a kind (and not 3 of another kind): If the number is 0 or 9, there
are \(5\choose3\) ways to choose the cards of that number and
\({{82-5}\choose3}\) ways to choose the other cards, but we have to subtract
off the number of ways to choose another 3 of a kind, which is
\({5\choose3}+8{9\choose3}\). Otherwise, there are \(9\choose3\) ways to choose
the cards of that number and \({{82-9}\choose3}\) ways to choose the other
cards, but again we subtract \(2{5\choose3}+7{9\choose3}\), which is the number
of ways to choose another 3 of a kind. This gives us a total of
\(2{5\choose3}\left({{82-5}\choose3}-\left({5\choose3}+8{9\choose3}\right)\right)+8{9\choose3}\left({{82-9}\choose3}-\left(2{5\choose3}+7{9\choose3}\right)\right)=42836496\).
\end{itemize}
So, the total is $672+73738+2678284+211108+42836496=\mathbf{45800298}$.\\
\hyperref[kind3top]{Back to table}
\subsubsection{Full house}
\label{fullhouse}
We break it down by the distribution of cards in the full house:
\begin{itemize}
\item 4 of one number and 2 of another: If the number of the 4 cards is 0 or 9,
there are 2 choices for the number, \(5\choose4\) ways to choose the cards in that number and
\(8{9\choose2}+{5\choose2}\) ways to choose the cards in the other number.
If the number of the 4 cards is not 0 or 9, there are 8 choices for the number,
\(9\choose4\) ways to choose the cards in that number and \(7{9\choose2}+2{5\choose2}\) ways to choose the cards in the other number. This gives a total of
\(2{5\choose4}(8{9\choose2}+{5\choose2})+8{9\choose4}(7{9\choose2}+2{5\choose2})=277156\).
\item 3 of one number and 3 of another: If the numbers are 0 and 9, there are
\({5\choose3}{5\choose3}\) choices. If one of the numbers is 0 or 9, there are
\(2{5\choose3}\cdot8{9\choose3}\) choices. If neither number is 0 nor 9, there
are \({8\choose2}\) ways to choose the numbers, so there are \({8\choose2}{9\choose3}{9\choose3}\) choices. All together, this gives \({5\choose3}{5\choose3}+2{5\choose3}\cdot8{9\choose3}+{8\choose2}{9\choose3}{9\choose3}=211108\).
\item 3 of one number and 2 of another: If the number of the 3 cards is 0 or 9,
there are 2 choices for the number, \({5\choose3}\) choices for the cards in that number, and \(8{9\choose2}(82-5-9)+{5\choose2}(82-5-5)\) for the other 3 cards. If the number of the 3 cards is not 0 or 9, there are 8 choices for the number,
\({9\choose3}\) choices for the cards in that number, and \(7{9\choose2}(82-9-9)+2{5\choose2}(82-9-5)\) choices for the other 3 cards. All together, this gives
\(2{5\choose3}(8{9\choose2}(82-5-9)+{5\choose2}(82-5-5))+8{9\choose3}(7{9\choose2}(82-9-9)+2{5\choose2}(82-9-5))=12158016\).
\end{itemize}
So, the total is $277156+211108+12158016=\mathbf{12646280}$.\\
\hyperref[fullhousetop]{Back to table}
\subsection{Miscellaneous}
\subsubsection{5 odd cards}
\label{fiveodd}
We break it down by the number of odd cards:
\begin{itemize}
\item 6 odd cards: There are 41 odd cards, so there are \({41\choose6}=4496388\) possibilities.
\item 5 odd cards: There are also 41 even cards, so there are \({41\choose5}41=30725318\) possibilities.
\end{itemize}
This gives a total of $4496388+30725318=\mathbf{35221706}$.\\
\hyperref[fiveoddtop]{Back to table}
\subsubsection{5 even cards}
\label{fiveeven}
Since there are 41 even cards and 41 odd cards, by symmetry this is the same number as \hyperref[fiveodd]{5 odd cards}, which is {\bf 35221706}.\\
\hyperref[fiveeventop]{Back to table}
\subsubsection{3 cards totalling $\leq$ 4}
\label{total4}
We break it down by the lowest two cards in the hand:
\begin{itemize}
\item 0 and 0: Firstly, we count how many hands have 0 and 0 as their lowest
cards. There are \({82}\choose6\) total hands, \({82-5}\choose6\) hands that have
no 0 cards, and \(5{{82-5}\choose5}\) hands with only 1 0, so the number of hands
that have 0 and 0 as their lowest cards is \({{82}\choose6}-\left({{82-5}\choose6}+5{{82-5}\choose5}\right)\). Of these, the only hands that don't have 3 cards
totalling $\leq 4$ are those that only have 5's or higher for the remaining
cards, and there are \({5\choose2}{{82-5-9\cdot4}\choose4}\) of these, so our
total is \({{82}\choose6}-\left({{82-5}\choose6}+5{{82-5}\choose5}\right)-{5\choose2}{{82-5-9\cdot4}\choose4}=13266257\).
\item 0 and 1: There are 5 choices for which 0 we have, and for the remaining
5 cards there are \({82-5}\choose5\), and \({82-5-9}\choose5\) have no 1's,
so the number of these hands is \(5\left({{82-5}\choose5}-{{82-5-9}\choose5}\right)\). Of these, the only hands that don't have 3 cards totalling $\leq 4$ are
those that only have 4's or higher for the remaining cards, and there are
\(5\cdot9{{82-5-9\cdot3}\choose4}\) of these, so our total is
\(5\left({{82-5}\choose5}-{{82-5-9}\choose5}\right)-5\cdot9{{82-5-9\cdot3}\choose4}=36304935\).
\item 0 and 2: There are 5 choices for which 0 we have, and for the remaining
5 cards (which have no 1's) there are \({82-5-9}\choose5\), and
\({82-5-9\cdot2}\choose5\) have no 1's,
so the number of these hands is \(5\left({{82-5-9}\choose5}-{{82-5-9\cdot2}\choose5}\right)\).
Of these, the only hands that don't have 3 cards totalling $\leq 4$ are
those that only have 3's or higher for the remaining cards, and there are
\(5\cdot9{{82-5-9\cdot2}\choose4}\) of these, so our total is
\(5\left({{82-5-9}\choose5}-{{82-5-9\cdot2}\choose5}\right)-5\cdot9{{82-5-9\cdot2}\choose4}=6608040\).
\item 1 and 1: There are \({82-5}\choose6\) hands that have no 0's, but
\({82-5-9}\choose6\) of these hands have no 1's, and \(9{{82-5-9}\choose5}\)
of these hands have exactly 1 1. So, there are
\({{82-5}\choose6}-\left({{82-5-9}\choose6}+9{{82-5-9}\choose5}\right)\) hands
that have 1 and 1 as their lowest cards. Of these, the only hands that
don't have 3 cards totalling $\leq 4$ are those that only have 3's or higher
for the remaining cards, and there are \({9\choose2}{{82-5-9\cdot2}\choose4}\)
of these, so our total is
\({{82-5}\choose6}-\left({{82-5-9}\choose6}+9{{82-5-9}\choose5}\right)-{9\choose2}{{82-5-9\cdot2}\choose4}=17438748\).
\end{itemize}
Therefore, the total number is $13266257+36304935+6608040+17438748=\mathbf{73617980}$.\\
\hyperref[total4top]{Back to table}
\subsubsection{3 cards totalling $\geq$ 23}
\label{total23}
By symmetry this is the same number as
\hyperref[total4]{3 cards totalling $\geq 4$}, which is
{\bf 73617980}.\\
\hyperref[total23top]{Back to table}
\subsubsection{Skip straight}
\label{skipstraight}
A skip straight is a 4 card straight where there's a 1 number gap between
each card: i.e. 0 2 4 6, 1 3 5 7, etc.\\
We break it down by the length of the skip straight:
\begin{itemize}
\item 5 card skip straight: The possibilities are 0 2 4 6 8 and 1 3 5 7 9. For
0 2 4 6 8, there are $5\cdot9^4$ possibilities for the straight and $82-5$
possibilities for the extra card. However, this double counts cases with
a (2,1,1,1,1) distribution of cards in the straight. If there are 2 0's, there
are \({5\choose2}9^4\) possibilities, and if there are 2 of anything else there
are \({9\choose2}5\cdot9^3\) possibilities. This gives us a total of
\(5\cdot9^4(82-5)-\left({5\choose2}9^4+4{9\choose2}5\cdot9^3\right)=1935495\),
and by symmetry the same total for 1 3 5 7 9.
\item 4 card skip straight: We use the inclusion-exclusion technique as
described in the
\hyperref[straightflush3cardspef]{3 card straight flush, specific suit} case.
For 0 2 4 6 (and similarly 3 5 7 9), our universe of cards is all 82 cards
except any 8 (to avoid falling in the 5 card skip straight case), so there are
$82-9=73$ cards in our universe $U$. We get
\({{73}\choose6}-\left({{73-5}\choose6}+3{{73-9}\choose6}\right)+\left(3{{73-14}\choose6}+3{{73-18}\choose6}\right)-\left(3{{73-23}\choose6}+{{73-27}\choose6}\right)+{{73-32}\choose6}=5452920\).\\
For 1 3 5 7 (and similarly 2 4 6 8), our universe of cards is all 82 cards
except any 9, so there are $82-5=77$ cards in our universe $U$. We get
\({{77}\choose6}-4{{77-9}\choose6}+6{{77-18}\choose6}-4{{77-27}\choose6}+{{77-36}\choose6}=10558836\).
\end{itemize}
So, our total is $2\cdot1935495+2\cdot5452920+2\cdot10558836=\mathbf{35894502}$.\\
\hyperref[skipstraighttop]{Back to table}
\subsubsection{Mixed straight}
\label{mixedstraight}
A mixed straight is a 4 card straight in which each of the cards is a different
suit. I'm open to suggestions on how to count this efficiently.\\
\hyperref[mixedstraighttop]{Back to table}
\section{About this project}
This page was generated from the \href{targetprobs.tex.txt}{\LaTeX source} with
\href{http://www.cse.ohio-state.edu/~gurari/TeX4ht/}{TeX4ht}.
\begin{itemize}
\item \label{calculateprobs} \href{calculateprobs.py.txt}{calculateprobs.py} is the Python script that
calculates the exact number of 6 card hands that satisfy each target card. I
was intending to use it as just a check on my by-hand calculations, but as
the combinatorics got harder and harder I used it as the only calculation for a few target cards.
\item \href{outputgraph.py.txt}{outputgraph.py} generates the pretty
\hyperref[graph]{graph of target cards}. It uses \href{http://matplotlib.sourceforge.net}{matplotlib}.
\item \href{compile.sh.txt}{compile.sh} compiled the source to this HTML page.
\end{itemize}
\end{document}