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<TABLE BORDER="2" CELLPADDING="5" BGCOLOR="#AACCAA"><TD>
<FONT SIZE="+2"><B>Odds & Probabilities in Texas Hold'em</B></FONT>
</TABLE>
<FONT SIZE="-2">motivated by e-mail from <A HREF="javascript:OpenWin('http://www.pokercalculatorreport.com/')"><B>Marty S.</B></A></FONT>
<BR><TABLE BORDER=2 CELLPADDING=5><TD>
In Texas Hold'em, you're dealt two cards.
<BR>Subsequently, the dealer places three cards face up on the table: the <B>Flop</B>.
<BR>Then he places another face up: the <B>Turn</B>.
<BR>Then, a final card, face up: the <B>River</B>.
<BR>Between each event, players make their bets.
<BR>The objective is to achieve the <A HREF="javascript:OpenWin3('poker-hands.gif')"><B>best 5-card poker hand</B></A>, using your two cards and the 5 cards played face up.
</TD></TABLE>
<P><HR><P>
<TABLE BORDER="2" CELLPADDING="5" BGCOLOR="#AACCAA"><TD><FONT SIZE="+1">
<B>the Flop</B></TD></TABLE>
<TABLE><TD>Suppose you're holding the following pair in your hand:
<B>5</B><FONT SIZE="+1" COLOR="red">e&</FONT>, <B>5</B><FONT SIZE="+1" COLOR="black">c&</FONT>
<BR>What are your chances of getting another <B>5</B>, to make it 3-of-a-kind?
<BR>The cards that can improve your hand are called <I>outs</I> ... and, in this case, your outs (to improve to 3-of-a-kind) are <B>5</B><FONT SIZE="+1" COLOR="red">f&</FONT> and <B>5</B><FONT SIZE="+1" COLOR="black">`&</FONT>.
<P>There are 52 cards in the deck and 2 are in your hand, leaving <FONT COLOR="blue"><B>50</B></FONT> cards unseen.
<BR>The probability of getting one of your <FONT COLOR="green"><B>2</B></FONT> "outs" is <FONT COLOR="green"><B>2</B></FONT> in <FONT COLOR="blue"><B>50</B></FONT>.
... and that's 4% <FONT COLOR="#555555" SIZE="-1">(since 100 * <FONT COLOR="green"><B>2</B></FONT> / <FONT COLOR="blue"><B>50</B></FONT> = 4%)</FONT>.
<P><FONT COLOR="#000077">>So it's just (number of outs) / (number of cards unseen) ?</FONT>
<BR>Yes, if there are <FONT COLOR="blue"><B>50</B></FONT> cards unseen.
<BR>However, after <B>the Flop</B> there'll be another 3 cards on the table, so the number of cards unseen is <FONT COLOR="blue"><B>47</B></FONT>.
<BR>Suppose the flop doesn't have a <B>5</B><FONT SIZE="+1" COLOR="red">f&</FONT> or a <B>5</B><FONT SIZE="+1" COLOR="black">`&</FONT>.
<BR>Then you need to calculate the probability of getting
one of these <FONT COLOR="green"><B>2</B></FONT> outs from the <FONT COLOR="blue"><B>47</B></FONT> unseen cards, and ...
<P><FONT COLOR="#000077">>I'd say the chances are 2 out of 47. Right?</FONT>
<BR>Right, and that's about <FONT COLOR="green"><B></B></FONT> <FONT COLOR="magenta"><B>4.25</B></FONT>%.
<P>Some like to note the "odds" rather than the probability.
<BR>If there are X ways to win and Y ways to lose, then the odds are Y-to-X or Y/X to 1.
<P><FONT COLOR="#000077">>That's confusing!</FONT>
<BR>It's just saying for every Y ways to win there are X ways to lose ... and you normally say it as: <I>"The odds are 7-to-1"</I>.
<P><FONT COLOR="#000077">>And that'd mean that for every 7 ways to win there's 1 way to lose. </FONT>
<BR>Yes. Of <B>8</B> possible ways events can occur, <B>7</B> will be losses and <B>1</B> will be a win.
<BR>That means your winning probability is <B>1</B> / <B>8</B> or 12.5%.
</TD><TD><CENTER>
<TABLE BORDER=1 CELLPADDING=5>
<TD><CENTER>Your Hand</TD><TD><CENTER>the Flop</TD><TD><CENTER>Outs</TD><TD><CENTER>Probability</TD><TR>
<TD><CENTER>5<FONT SIZE="+1" COLOR="red">e&</FONT>, 5<FONT SIZE="+1" COLOR="black">c&</FONT></TD>
<TD><CENTER>A<FONT SIZE="+1" COLOR="black">`&</FONT>, 3<FONT SIZE="+1" COLOR="black">c&</FONT>, J<FONT SIZE="+1" COLOR="red">f&</FONT></TD>
<TD><CENTER>5<FONT SIZE="+1" COLOR="red">f&</FONT>, 5<FONT SIZE="+1" COLOR="black">`&</FONT><BR><FONT SIZE="-1">to make 3-of-a-kind</TD>
<TD><CENTER>2/47 = 4.25%</TD><TR>
<TD><CENTER>5<FONT SIZE="+1" COLOR="red">e&</FONT>, 5<FONT SIZE="+1" COLOR="black">c&</FONT></TD>
<TD><CENTER>A<FONT SIZE="+1" COLOR="black">`&</FONT>, 5<FONT SIZE="+1" COLOR="red">f&</FONT>, J<FONT SIZE="+1" COLOR="red">f&</FONT></TD>
<TD><CENTER>5<FONT SIZE="+1" COLOR="black">`&</FONT><BR><FONT SIZE="-1">to make 3-of-a-kind</TD>
<TD><CENTER>1/47 = 2.13%</TD><TR>
<TD><CENTER>5<FONT SIZE="+1" COLOR="red">e&</FONT>, 7<FONT SIZE="+1" COLOR="black">c&</FONT></TD>
<TD><CENTER>6<FONT SIZE="+1" COLOR="black">`&</FONT>, 9<FONT SIZE="+1" COLOR="black">c&</FONT>, A<FONT SIZE="+1" COLOR="red">f&</FONT></TD>
<TD><CENTER>8<FONT SIZE="+1" COLOR="black">c&</FONT>, 8<FONT SIZE="+1" COLOR="red">f&</FONT>, 8<FONT SIZE="+1" COLOR="red">e&</FONT>, 8<FONT SIZE="+1" COLOR="black">`&</FONT><BR><FONT SIZE="-1">to make the inside straight</TD>
<TD><CENTER>4/47 = 8.51%</TD><TR>
<TD><CENTER>J<FONT SIZE="+1" COLOR="black">c&</FONT>, K<FONT SIZE="+1" COLOR="red">f&</FONT></TD>
<TD><CENTER>J<FONT SIZE="+1" COLOR="red">f&</FONT>, 3<FONT SIZE="+1" COLOR="black">c&</FONT>, 5<FONT SIZE="+1" COLOR="black">`&</FONT></TD>
<TD><CENTER>J<FONT SIZE="+1" COLOR="red">e&</FONT>, J<FONT SIZE="+1" COLOR="black">`&</FONT>, K<FONT SIZE="+1" COLOR="black">c&</FONT>, K<FONT SIZE="+1" COLOR="red">e&</FONT>, K<FONT SIZE="+1" COLOR="black">`&</FONT>
<BR><FONT SIZE="-1">to improve your hand</TD>
<TD><CENTER>5/47 = 10.6%</TD><TR>
</TABLE>
<B>Table 1:</B> Some probability calculations<BR>based upon needing <B><U>one</U></B> out to improve your hand
</TD></TABLE>
<TABLE WIDTH=100%><TD>
<FONT COLOR="#000077">>So if odds are <B>Y / X</B>, then the probability of winning is <B>X / (X+Y)</B> .</FONT>
<BR>Uh ... yes, but if somebody says the odds are 7-to-1 it's easier just to calculate the probability as 1/(7+1) = 0.125.
<BR>Normally, you'd multiply by 100 changing it to 12.5%.
<P>In our example, of the <FONT COLOR="blue"><B>47</B></FONT> unseen cards, X = <FONT COLOR="green"><B>2</B></FONT> is the number of cards that'll give you your 3-of-a-kind.
<BR>The number of cards that'll "miss you" is then <FONT COLOR="blue"><B>47</B></FONT> - <FONT COLOR="green"><B>2</B></FONT> = <B>45</B>.
Then <B>Y</B> = <B>45</B>.
<BR>The odds are then <B>Y</B>-to-<B>X</B> or <B>45</B> to <FONT COLOR="green"><B>2</B></FONT> or <FONT COLOR="magenta"><B>22.5</B></FONT> to 1.
</TD><TD>
<TABLE BORDER="2" CELLPADDING="5" BGCOLOR="#DDFFDD"><TD>
If <B>P</B> is the probability of getting a winning card <FONT COLOR="#555555" SIZE="-1">(<U>Example</U>: P = 12.5%)</FONT>
<BR>and <B>Z</B> gives the odds <FONT COLOR="#555555" SIZE="-1">(<U>Example</U>: Z = 7 meaning 7-to-1 odds)</FONT>
<BR>then <B>P</B> = 100/(1+<B>Z</B>) <FONT COLOR="#555555" SIZE="-1">(<U>Example</U>: P = 100/(1+7) = 12.5%)</FONT>
<BR>and <B>Z</B> = 100/<B>P</B> - 1 <FONT COLOR="#555555" SIZE="-1">(<U>Example</U>: Z = 100/P - 1 = 7)</FONT>
</TABLE>
</TD></TABLE>
<FONT COLOR="#000077">>Last time you got the probability as <FONT COLOR="magenta"><B>4.25</B></FONT>%.</FONT>
<BR>Yes, and that's 1/(1+odds) = 1/(1+<FONT COLOR="magenta"><B>22.5</B></FONT>) = 0.0425 or
<FONT COLOR="#555555" SIZE="-1">(multiplying by 100, to get a percentage)</FONT> <FONT COLOR="magenta"><B>4.25</B></FONT>%.
<P><HR><P>
<TABLE BORDER="2" CELLPADDING="5" BGCOLOR="#AACCAA"><TD><FONT SIZE="+1">
<B>the Turn and the River</B></TD></TABLE>
<P>After <B>the Flop</B>, there are three cards face up on the table. That means there are now <FONT COLOR="blue"><B>47</B></FONT> cards unseen.
<BR>After <B>the Turn</B>, there are four cards face up on the table. That means there are now <FONT COLOR="blue"><B>46</B></FONT> cards unseen.
<BR>After <B>the River</B>, there are five cards face up on the table. That means there are now <FONT COLOR="blue"><B>45</B></FONT> cards unseen.
<BR><FONT COLOR="#555555" SIZE="-1">(That's 52, less the 2 in your Hand, less the cards face up, on the table.)</FONT>
<P>Suppose you hold <B>5</B><FONT SIZE="+1" COLOR="red">e&</FONT>, <B>5</B><FONT SIZE="+1" COLOR="black">c&</FONT> and <B>the Flop</B> doesn't have either
the <B>5</B><FONT SIZE="+1" COLOR="red">f&</FONT> or the <B>5</B><FONT SIZE="+1" COLOR="black">`&</FONT>.
<BR>Suppose you want to calculate the probability of getting one of the <FONT COLOR="green"><B>2</B></FONT> cards that'll improve your hand on the next <U>two</U> cards laid face up.
<P><FONT COLOR="#000077">>Huh?</FONT>
<BR>The probability of getting either the <B>5</B><FONT SIZE="+1" COLOR="red">f&</FONT> or the <B>5</B><FONT SIZE="+1" COLOR="black">`&</FONT>, immediately after the flop
<FONT COLOR="#555555" SIZE="-1">(when there are <FONT COLOR="blue"><B>47</B></FONT> cards unseen)</FONT>
is <FONT COLOR="magenta"><B>4.25</B></FONT>% as we've noted before in Table 1. <FONT COLOR="#555555" SIZE="-1">(That's 2/47.)</FONT>
<BR>That's the probability of getting one of your outs on <B>the Turn</B> <U>and</U> <B>the River</B>.
<BR>If there were just <U>one</U> card to go, you'd expect the probability to be half as much.
<P><FONT COLOR="#000077">>That's <I>still</I> confusing!</FONT>
<BR>Okay, we'll do it this way:
<P><HR><P>
<TABLE BORDER="2" CELLPADDING="5" BGCOLOR="#AACCAA"><TD><FONT SIZE="+1">
<B>Combinatorics</B></TD></TABLE>
<P>
<UL>
<LI>If you have 52 cards and you ask: <I>"How many ways can I be dealt 2 cards?"</I>
<LI>The answer is: 52!/(2! 50!) = (52)(51)/(2)(1) = 1326, denoted by <SUB>52</SUB>C<SUB>2</SUB> <FONT COLOR="#555555" SIZE="-1">... read as <I>52-Choose-2</I></FONT>.
<BR><FONT COLOR="#555555" SIZE="-1">Note that 52! = (52)(51)(50)...(3)(2)(1), the product of all integers from 1 to 52.</FONT>
<LI>In general, if you have N cards and you ask: <I>"How many ways can I be dealt M cards"</I>, the answer is <SUB>N</SUB>C<SUB>M</SUB>.
</UL>
<FONT COLOR="#000077">>So there are 1326 possible 2-card hands dealt in Texas Hold'em?</FONT>
<BR>Yes. In fact, there are 52 possibilities for the first card and <FONT COLOR="#555555" SIZE="-1">(having dealt that card)</FONT> 51 for the second so that makes (52)(51) = 2652.
<BR>However, among these 2652 possibilities we'd find <FONT COLOR="red"><B>2</B></FONT> identical hands: [<B>5</B><FONT SIZE="+1" COLOR="red">f&</FONT>, <B>5</B><FONT SIZE="+1" COLOR="black">`&</FONT>]
as well as [<B>5</B><FONT SIZE="+1" COLOR="black">`&</FONT>, <B>5</B><FONT SIZE="+1" COLOR="red">f&</FONT>].
<BR>Since one doesn't distinguish between these <FONT COLOR="red"><B>2</B></FONT> hands, we divide 2652 by <FONT COLOR="red"><B>2</B></FONT> getting our 1326 ... and that's <SUB>52</SUB>C<SUB>2</SUB>.
<P>Suppose we ask: <I>"How many ways can I deal a pair?"</I>
<UL>
<LI>There are 1326 ways to be dealt two cards, how many of these are pairs?
<LI>There are four 2s in the deck, 2<FONT SIZE="+1" COLOR="red">e&</FONT>, 2<FONT SIZE="+1" COLOR="red">f&</FONT>, 2<FONT SIZE="+1" COLOR="black">c&</FONT>, 2<FONT SIZE="+1" COLOR="black">`&</FONT>.
<LI>We want to choose 2 of them. The number of ways is <SUB>4</SUB>C<SUB>2</SUB> = (4)(3)/2 = 6.
<LI>For all 13 cards, from the 2 to the A, that means 13*6 = 78 possible pairs.
<LI>So, of the 1326 possible 2-card hands, 78 will be a pair.
<LI>The probability of being dealt a pair is then: 78/1326 = 0.0588 or 5.88%.
<BR><FONT COLOR="#555555" SIZE="-1">The odds are then: 100/probability - 1 = 100/5.88 - 1 = 16, hence 16-to-1.</FONT>
</UL>
<FONT COLOR="#000077">>Isn't there some magic (and simple!) formula?</FONT>
<UL>
<LI>You calculate the total number of ways that something can happen <FONT COLOR="#555555" SIZE="-1">(like 1326 ways to deal 2 cards)</FONT>.
<LI>From these, you identify the number of ways that something <U>special</U> can happen <FONT COLOR="#555555" SIZE="-1">(like 78 ways to get a pair)</FONT>.
<LI>You divide the latter by the former <FONT COLOR="#555555" SIZE="-1">(like 78/1326 to get 5.88% probability of something <U>special</U> happening)</FONT>.
</UL>
Okay, let's do something more complicated:
<UL>
<LI>Suppose there are 4 outs and you want 1 of these 4 outs to improve your hand.
<LI>After the <B>Flop</B>, there are 47 card unseen.
<LI>The number of possible combinations of 2 cards for the <B>Turn</B> and the <B>River</B> is <SUB>47</SUB>C<SUB>2</SUB>.
<LI>How many of these will have 1 of the 4 outs?
<LI>To calculate this, we remove the 4 outs, leaving 47 - 4 = 43 cards.
<LI>There are <SUB>43</SUB>C<SUB>2</SUB> ways to deal 2 cards from these 43 cards ... and NONE will have any of the 4 outs <FONT COLOR="#555555" SIZE="-1">(because we've removed them)</FONT>.
<LI>So how many of the 2 cards (from the <B>Turn</B> and the <B>River</B>) WILL have 1 of the 4 outs?
<LI>It's <SUB>47</SUB>C<SUB>2</SUB> - <SUB>43</SUB>C<SUB>2</SUB> = (total 2-card deals) - (those that do NOT have an out card).
<LI>The probability is then: [ <SUB>47</SUB>C<SUB>2</SUB> - <SUB>43</SUB>C<SUB>2</SUB>] / <SUB>47</SUB>C<SUB>2</SUB> = (number with at least 1 of the outs) / (total 2-card deals).
<LI>That's a probability of 1 - <SUB>43</SUB>C<SUB>2</SUB> / <SUB>47</SUB>C<SUB>2</SUB> = 1 - 0.835 = 0.165 or 16.5%.
</UL>
<TABLE><TD>
<FONT COLOR="#000077">>I think there's a magic formula there.</FONT>
<BR>Yes, there are a couple and they go like this:
<TABLE BORDER="2" CELLPADDING="5" BGCOLOR="#DDFFDD" ><TD>
<B>After the Flop:</B>
<BR>There are 47 cards unseen and there are <B>n</B> outs which will improve your hand <BR><FONT COLOR="#555555" SIZE="-1"><U>Example</U>: n = 3 outs</FONT>.
<BR>From the next 2 cards dealt (<B>Turn</B> + <B>River</B>), you want 1 out of the n outs <BR><FONT COLOR="#555555" SIZE="-1"><U>Example</U>: 1 out of n = 3 outs</FONT>.
<BR>The probability of getting 1 of the n outs in the next 2 cards is:
<BR><B>Probability</B> = 1 - <SUB>(47-n)</SUB>C<SUB>2</SUB> / <SUB>47</SUB>C<SUB>2</SUB>
<BR><FONT COLOR="#555555" SIZE="-1"><U>Example</U>: 1 - <SUB>44</SUB>C<SUB>2</SUB> / <SUB>47</SUB>C<SUB>2</SUB> = 0.125 or 12.5%</FONT>.
</TABLE>
<TABLE BORDER="2" CELLPADDING="5" BGCOLOR="#DDFFDD"><TD>
<B>After the Flop + Turn:</B>
<BR>There are 46 cards unseen and there are <B>n</B> outs which will improve your hand <BR><FONT COLOR="#555555" SIZE="-1"><U>Example</U>: n = 3 outs</FONT>.
<BR>From the next card dealt (the <B>River</B>), you want 1 out of the n outs <BR><FONT COLOR="#555555" SIZE="-1"><U>Example</U>: 1 out of n = 3 outs</FONT>.
<BR>The probability of getting 1 of the n outs in the next card is:
<BR><B>Probability</B> = 1 - <SUB>(46-n)</SUB>C<SUB>1</SUB> / <SUB>46</SUB>C<SUB>1</SUB>
<BR><FONT COLOR="#555555" SIZE="-1"><U>Example</U>: 1 - <SUB>43</SUB>C<SUB>1</SUB> / <SUB>46</SUB>C<SUB>2</SUB> = 0.065 or 6.5%</FONT>.
</TABLE>
</TD><TD><CENTER>
<TABLE BORDER=1><TD><CENTER><FONT SIZE="-1">Outs</TD><TD><CENTER><FONT SIZE="-1">2 cards to go</TD><TD><CENTER><FONT SIZE="-1">1 card to go</TD><TR><TD><CENTER><FONT SIZE="-1">1</TD><TD><CENTER><FONT SIZE="-1">4.3%</TD><TD><CENTER><FONT SIZE="-1">2.2%</TD><TR><TD><CENTER><FONT SIZE="-1">2</TD><TD><CENTER><FONT SIZE="-1">8.4%</TD><TD><CENTER><FONT SIZE="-1">4.3%</TD><TR><TD><CENTER><FONT SIZE="-1">3</TD><TD><CENTER><FONT SIZE="-1">12.5%</TD><TD><CENTER><FONT SIZE="-1">6.5%</TD><TR><TD><CENTER><FONT SIZE="-1">4</TD><TD><CENTER><FONT SIZE="-1">16.5%</TD><TD><CENTER><FONT SIZE="-1">8.7%</TD><TR><TD><CENTER><FONT SIZE="-1">5</TD><TD><CENTER><FONT SIZE="-1">20.4%</TD><TD><CENTER><FONT SIZE="-1">10.9%</TD><TR><TD><CENTER><FONT SIZE="-1">6</TD><TD><CENTER><FONT SIZE="-1">24.1%</TD><TD><CENTER><FONT SIZE="-1">13.0%</TD><TR><TD><CENTER><FONT SIZE="-1">7</TD><TD><CENTER><FONT SIZE="-1">27.8%</TD><TD><CENTER><FONT SIZE="-1">15.2%</TD><TR><TD><CENTER><FONT SIZE="-1">8</TD><TD><CENTER><FONT SIZE="-1">31.5%</TD><TD><CENTER><FONT SIZE="-1">17.4%</TD><TR><TD><CENTER><FONT SIZE="-1">9</TD><TD><CENTER><FONT SIZE="-1">35.0%</TD><TD><CENTER><FONT SIZE="-1">19.6%</TD><TR><TD><CENTER><FONT SIZE="-1">10</TD><TD><CENTER><FONT SIZE="-1">38.4%</TD><TD><CENTER><FONT SIZE="-1">21.7%</TD><TR><TD><CENTER><FONT SIZE="-1">11</TD><TD><CENTER><FONT SIZE="-1">41.7%</TD><TD><CENTER><FONT SIZE="-1">23.9%</TD><TR><TD><CENTER><FONT SIZE="-1">12</TD><TD><CENTER><FONT SIZE="-1">45.0%</TD><TD><CENTER><FONT SIZE="-1">26.1%</TD><TR><TD><CENTER><FONT SIZE="-1">13</TD><TD><CENTER><FONT SIZE="-1">48.1%</TD><TD><CENTER><FONT SIZE="-1">28.3%</TD><TR><TD><CENTER><FONT SIZE="-1">14</TD><TD><CENTER><FONT SIZE="-1">51.2%</TD><TD><CENTER><FONT SIZE="-1">30.4%</TD><TR><TD><CENTER><FONT SIZE="-1">15</TD><TD><CENTER><FONT SIZE="-1">54.1%</TD><TD><CENTER><FONT SIZE="-1">32.6%</TD><TR><TD><CENTER><FONT SIZE="-1">16</TD><TD><CENTER><FONT SIZE="-1">57.0%</TD><TD><CENTER><FONT SIZE="-1">34.8%</TD><TR><TD><CENTER><FONT SIZE="-1">17</TD><TD><CENTER><FONT SIZE="-1">59.8%</TD><TD><CENTER><FONT SIZE="-1">37.0%</TD><TR><TD><CENTER><FONT SIZE="-1">18</TD><TD><CENTER><FONT SIZE="-1">62.4%</TD><TD><CENTER><FONT SIZE="-1">39.1%</TD><TR><TD><CENTER><FONT SIZE="-1">19</TD><TD><CENTER><FONT SIZE="-1">65.0%</TD><TD><CENTER><FONT SIZE="-1">41.3%</TD><TR><TD><CENTER><FONT SIZE="-1">20</TD><TD><CENTER><FONT SIZE="-1">67.5%</TD><TD><CENTER><FONT SIZE="-1">43.5%</TD></TABLE>
<B>Table 2:</B></TD></TABLE>
<TABLE><TD>
Now look carefully at Table 2. Notice that the probabilities, when there are 2 cards to go, is roughly 4x the outs.
<BR>That's because of the following (when there are n outs):
<UL>
<LI>Probability = 1 - <SUB>(47-n)</SUB>C<SUB>2</SUB> / <SUB>47</SUB>C<SUB>2</SUB> = 1 - [(47-n)(46-n)]/[(47)(46)] = 1 - (1-n/47)(1-n/46) = n (1/47 + 1/46) - n<SUP>2</SUP>/(47)(46)
<LI>To get a percentage, multiply by 100 and get:
<BR> Probability = n (100/47 + 100/46) + 0.046 n<SUP>2</SUP> ≈ 4.3 n <FONT COLOR="#555555" SIZE="-1">... when n (hence n<SUP>2</SUP>) isn't too large.</FONT>
</UL>
</TD><TD><IMG SRC="poker-1.gif"></TD></TABLE>
<FONT COLOR="#000077">>Earlier, you got 2/47 and now you get ...</FONT>
<BR>And now I get 1 - <SUB>46</SUB>C<SUB>2</SUB> / <SUB>47</SUB>C<SUB>2</SUB> = 1 - 45/47 = 2/47.
<BR>Neat, eh? <IMG SRC="smiley14.gif">
<P><FONT COLOR="#000077">>When you say the dealer has 47 cards, you're assumuing I'm the only player, right?</FONT>
<BR>Good point.
<P><HR><P>
<TABLE BORDER="2" CELLPADDING="5" BGCOLOR="#AACCAA"><TD><FONT SIZE="+1">
<B>Other players</B></TD></TABLE>
<P>If the dealer deals 2 cards to you, he'll have 50 left.
<BR>However, before <B>the Flop</B>, he'll deal 2 cards to each of the your opponents.
<BR>So, if there are m opponents, how many ways can 2 cards be dealt to each?
<BR>There are <SUB>50</SUB>C<SUB>2</SUB> ways to deal the first 2 and, for <I>each</I> of these, there are <SUB>48</SUB>C<SUB>2</SUB> to deal the next 2, then <SUB>46</SUB>C<SUB>2</SUB> <I>etc. etc.</I>
<BR>So, for m = 3 opponents, that'll be (<SUB>50</SUB>C<SUB>2</SUB>)(<SUB>48</SUB>C<SUB>2</SUB>)(<SUB>46</SUB>C<SUB>2</SUB>) = 1,430,163,000.
<P><FONT COLOR="#000077">>Mamma mia!</FONT>
<BR>Hold on!
<BR>Suppose the hands dealt to the 3 other players are:
<BR>[J<FONT SIZE="+1" COLOR="red">e&</FONT>, 9<FONT SIZE="+1" COLOR="black">`&</FONT>] [6<FONT SIZE="+1" COLOR="black">c&</FONT>, 9<FONT SIZE="+1" COLOR="black">`&</FONT>]
[2<FONT SIZE="+1" COLOR="red">e&</FONT>, A<FONT SIZE="+1" COLOR="red">f&</FONT>].
<BR>When we counted up to 1,430,163,000, that number included all of the rearrangements of these 3 hands.
<P><FONT COLOR="#000077">>Huh?</FONT>
<BR>It counted
[J<FONT SIZE="+1" COLOR="red">e&</FONT>, 9<FONT SIZE="+1" COLOR="black">`&</FONT>] [6<FONT SIZE="+1" COLOR="black">c&</FONT>, 9<FONT SIZE="+1" COLOR="black">`&</FONT>] [2<FONT SIZE="+1" COLOR="red">e&</FONT>, A<FONT SIZE="+1" COLOR="red">f&</FONT>]
and
[6<FONT SIZE="+1" COLOR="black">c&</FONT>, 9<FONT SIZE="+1" COLOR="black">`&</FONT>] [J<FONT SIZE="+1" COLOR="red">e&</FONT>, 9<FONT SIZE="+1" COLOR="black">`&</FONT>] [2<FONT SIZE="+1" COLOR="red">e&</FONT>, A<FONT SIZE="+1" COLOR="red">f&</FONT>]
and
[2<FONT SIZE="+1" COLOR="red">e&</FONT>, A<FONT SIZE="+1" COLOR="red">f&</FONT>] [6<FONT SIZE="+1" COLOR="black">c&</FONT>, 9<FONT SIZE="+1" COLOR="black">`&</FONT>] [J<FONT SIZE="+1" COLOR="red">e&</FONT>, 9<FONT SIZE="+1" COLOR="black">`&</FONT>]
<I>etc. etc.</I> as distinct.
<BR>Clearly, it makes no difference which opponent holds which hand ... so they shouldn't be counted as distinct possibilities.
<BR>Since there are 3*2*1 = 3! = 6 possible rearrangements for each set of 3, the number 1,430,163,000 is 6x too large ... so we divide by 6.
<BR>In fact, for m opponents, we'd divide by m! = 1*2*3*...*m. <FONT COLOR="#555555" SIZE="-1">(In our m=3 example, that's 1*2*3 = 6.)</FONT>
<P>Okay, so we have:
<TABLE BORDER="2" CELLPADDING="5" BGCOLOR="#DDFFDD"><TD>
From 50 cards, the number of different ways to deal <B>m</B> sets of 2 cards <FONT COLOR="#555555" SIZE="-1">(to m opponents)</FONT> is:
<BR><B> (<SUB>50</SUB>C<SUB>2</SUB>) (<SUB>48</SUB>C<SUB>2</SUB>) (<SUB>46</SUB>C<SUB>2</SUB>) ... (<SUB>52-2m</SUB>C<SUB>2</SUB>) / m! </B>
</TD></TABLE>
<FONT COLOR="#000077">>Okay, suppose there are 9 opponents and I'd like to know all the possibilities so that ...</FONT>
<BR>Answer: Over 20,000,000,000,000,000,000,000,000,000.
<P><FONT COLOR="#000077">>Yeah, but if I had a computer I'd just ...</FONT>
<BR>If your computer could calculate the possibilities at the rate of 10,000,000,000,000 per second, it's take over 600 million years.
<BR>In real cases, one wouldn't bother calculating all the possible combinations that would beat <U>your</U> hand ... else you'll be late for supper.
<BR>Instead, one imagines thousands (millions?) of hands being dealt to your opponents and seeing what percentage beat your hand.
<P><FONT COLOR="#000077">>And that's the probability of being beaten by an opponent?</FONT>
<BR>Approximately. It's sorta like polling 1000 people to see which prefer chocolate ice cream to vanilla.
<BR>Then you use that percentage to gauge the entire population of millions.
<BR>Or it's like doing a few thousand Monte Carlo simulations to see how long your portfolio might last if you withdraw x% each year.
<BR>Or it's like ...
<P><FONT COLOR="#000077">><B>Okay!</B> I get it.</FONT>
<BR>Good.
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