Lecture 29 21 127 Concepts of Math 11 09 2012 Counting Examples Poker Hands In these examples we will come to realize why poker hands consisting of 5 cards each are ranked the way they are Why does three of a kind beat two pair Why does a flush beat a straight We will answer these questions by showing how many of the possible 5 card hands fall into each of these categories Note In each of these examples we will be thorough about describing where we use ROS and ROP pointing out a partition and describing a multi step process This is the level of rigor that we expect on homework submissions Since we also realize this might be your first time working with these kinds of counting problems we will preface each formal argument with an intuitive explanation of how the argument will go In each of these explanantions you ll notice that we are essentially asking ourselves What kinds of questions would I need to ask of someone who has such a hand in mind What are the determining characteristics of such a hand How many ways could they answer our probing questions We recommend that you approach problems in this way as well Think of a specific object that you are trying to count and identify its properties that would uniquely determine it then figure out how many options there are for each of those properties One Pair Intuition To construct a One Pair hand we need to select two cards of the same rank and then we need to select three other cards that are all different ranks and that are not of the same rank as the paired cards To identify these cards we will choose their ranks and then their suits Claim There are 3 13 4 12 4 1 2 3 1 One Pair hands Proof We will construct One Pair hands via a six step process 1 Select a rank that will appear in our hand twice There are 13 1 ways to do this 2 For that rank selected in Step 1 select the suits of the two cards that will appear in our hand There are 42 ways to do this 3 Of the remaining 12 unchosen ranks i e not the rank from Step 1 select 3 ranks to appear in our hand There are 12 ways to do this 3 4 For the lowest of those 3 ranks chosen in Step 3 select a suit There are 41 ways to do this 5 For the middle of those 3 ranks chosen in Step 3 select a suit There are 41 ways to do this 6 For the highest of those 3 ranks chosen in Step 3 select a suit There are 41 ways to do this 3 By ROP and simplifying 41 41 41 41 we have shown the expression above is correct 1 Summary Notice the subtlety of steps 4 6 above We could have just lumped them into one step and said For each of those 3 chosen ranks select one suit There are 41 options in each case so there 3 are 41 options overall by ROP We thought it might be clearer to show how we are really doing the same step three times in a row To do this though we had to find some way to order the 3 ranks we had chosen in Step 3 Alternative Solution We could also approach this problem in a slightly different way We could determine a One Pair hand by asking which 4 ranks appear then asking which one of those 4 is repeated twice determining the 2 suits for that rank and then determining the 1 suit for each of the other 3 ranks Claim There are 3 13 4 4 4 4 1 2 1 One Pair hands Proof We will construct One Pair hands via a six step process 1 Select 4 ranks that will appear in our hand There are 13 4 ways to do this 2 Of the 4 ranks selected in Step 1 select one of them Two cards of that rank will appear in our hand There are 41 ways to do this 3 For that rank chosen in Step 2 select 2 suits These will appear in our hand There are 42 ways to do this 4 For the lowest of those 3 ranks not chosen in Step 2 select a suit There are 41 ways to do this 5 For the middle of those 3 ranks not chosen in Step 2 select a suit There are 41 ways to do this 6 For the highest of those 3 ranks not chosen in Step 2 select a suit There are 41 ways to do this By ROP and simplifying 4 1 4 4 1 1 4 3 1 we have shown the expression above is correct Summary Again we could have lumped Steps 4 6 into one by saying For each of the three ranks 3 not chosen in Step 2 select a suit In each case there are 41 options so by ROP there are 41 total options Two Pair Intuition To construct a Two Pair hand we need to identify two ranks that are doubled and then a third rank to appear once For each of those ranks we then need to assign suits There is no worry about making a straight or a flush inadvertently since we have repeated ranks Claim There are 2 13 4 11 4 2 2 1 1 Two Pair hands 2 Proof We will construct Two Pair hands by a four step process 1 Select two ranks that will be doubled in our hand There are 13 2 ways to do this 2 For each of the two ranks chosen in Step 2 assign two suits 2 By ROP there are 42 42 42 ways to do this 3 From the 11 unchosen ranks i e those not selected in Step 1 choose one more to appear There are 11 1 ways to do this 4 Assign a suit to that rank chosen in Step 3 There are 41 ways to do this By ROP we have proven the claim Straight Intuition The ranks of the cards in a straight are uniquely determined by the starting rank the lowest card of the hand If I told you I had a 5 card straight starting with 7 you d know immediately I have a 789TJ straight Since we can have a straight like A2345 or one like 23456 all the way up to TJQKA Note There is no going around the corner in a straight like QKA23 this means we have 10 possible lowest ranks in a straight Thus there are ten types of straight and after that we just need to assign the suits so that they aren t all the same in which case we d have a straight flush Claim There are 5 10 4 4 10 45 4 1 1 1 5 card hands that aere straights Proof We will describe 5 card hands that are straights by a two step process 1 Select one …