ÿþ<HTML> <TITLE>Poker Odds</TITLE> <HEAD> <SCRIPT LANGUAGE ="JavaScript"> <!-- function OpenWin(url) { window.open(url, "DisplayWindow", "location=1, scrollbars=1, menubar=1, status=1, toolbar=1, resizable=1, width=780, height=400"); } function OpenWin2(url) { window.open(url, "DisplayWindow", "width=200, height=200"); } function OpenWin3(url) { window.open(url, "DisplayWindow", "width=300, height=800"); } <!-- //--> </SCRIPT> <!-- <A HREF="javascript:OpenWin('URL')"><B>NAME</B></A> --> </HEAD> <BODY BGCOLOR="#DDDDDD"> <TABLE BORDER="1" BGCOLOR="white" CELLPADDING="5" WIDTH="100%"><TD> <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: &nbsp; <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 &nbsp; &nbsp; <FONT COLOR="#555555" SIZE="-1">(<U>Example</U>: P = 12.5%)</FONT> <BR>and <B>Z</B> gives the odds &nbsp; &nbsp; <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>) &nbsp; &nbsp; <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 &nbsp; &nbsp; <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> &#8776; 4.3 n &nbsp; &nbsp; <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? &nbsp; <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. &nbsp; <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. <P><P><IMG SRC="in_progress.gif"> </TD></TABLE> <P><A HREF="gummy_stuff.htm"><IMG SRC="gummy_link.gif" ALIGN="bottom"></A> <A HREF="Excel"><IMG SRC="notice.gif"></A> <!-- <FONT SIZE="+1" COLOR="red">e&</FONT> <FONT SIZE="+1" COLOR="red">f&</FONT> <FONT SIZE="+1" COLOR="black">c&</FONT> <FONT SIZE="+1" COLOR="black">`&</FONT> <SUB>N</SUB>C<SUB>M</SUB> --> </BODY></HTML> <FONT COLOR="#000077"></FONT> <FONT COLOR="#555555" SIZE="-1"></FONT> <FONT COLOR="green"><B></B></FONT> <FONT COLOR="blue"><B></B></FONT> </BODY></HTML>