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Unformatted text preview: A little review on combinatorics: In general, you can ALWAYS USE THE BCR. However, combinations and permutations apply to very specific situations. However, when applicable, it is probably better to use a combination or permutation. Example: How many ways is it possible to get 2 pairs (and at best 2 pairs)? Using Combinations: There are “13 choose 2” ways to pick the 2 face values of the cards that you want pairs of. There are “4 choose 2” ways to get the suits of your pair. In addition, there are “11 choose 1 ways” to pick the face value of the last card of your hand that is not the same as one of the first 2 cards. Also, there are “4 choose 1” ways that you can pick the suit for this final card. Therefore, the number of ways to get 2 pairs and only 2 pairs is: 2 ⎛13 ⎞ ⎛ 4 ⎞ ⎛11⎞ ⎛ 4 ⎞
⎜ ⎟ * ⎜ ⎟ * ⎜ ⎟ * ⎜ ⎟ . ⎜ 2 ⎟ ⎜ 2 ⎟ ⎜1 ⎟ ⎜1 ⎟
⎝ ⎠⎝⎠ ⎝ ⎠⎝⎠ Now, think about attempting this with the BCR. More than likely, you would end up with the following: 13*4*3*12*4*3*11*4. This is double counting the number of ways suits can occur (4*3 since it is “ordering” the suits). Also, there is a double count in the 2 cards you get for pairs (13*2 instead of “13 choose 2”). CLARIFICATION: The reason we are using “13 choose 2” instead of 13*12 is that we are not including the order in which they occur in your hand. Also, there is no essential difference between the first face value that you choose for the first pair and the second face value you choose for the second pair. Little Difference: Recall the number of ways that you can get a full house is “13 choose 1” * “4 choose 3” * “12 choose 1” * “4 choose 2”. The reason you don’t “lump” the two cards together into a “13 choose 2” instead of the “13 choose 1” * “12 choose 1” is because YOU ARE SPECIFYING WHICH ONE IS A 3 OF A KIND AND WHICH ONE IS A PAIR. Similarly, a pair (and only a pair) has the following number of ways possible: (13 choose 1)* (4 choose 2) * (12 choose 3)*(4 choose 1)^3. You have to specify how many face values you can have for your pair, how many ways you can choose the suits for that face value, how many ways you can choose 3 remaining face values that are not a pair and then the # of suits possible for those singles. Since the singles are not distinguishable we “lump” them together in the “12 choose 3” and do not count them individually like “12 choose 1” * “11 choose 1” * “10 choose 1” because this imposes that “order matters” i.e. it matters which single you get first, second, and third. A BCR example: A lock combination problem: Here combination does not mean “n choose r” or anything similar. It means a combination to the lock or password or whatever. This is not a combination problem. Suppose we have 5 spaces available for the password. We can use numbers (0 thru 9) or letters (NOT case sensitive). a) How many possible combinations are there? There are 36^5 possible combinations. There are 36 possible choices per “blank” and there are 5 blanks. b) If you cannot have a 9 on the first number, how many possible combinations are there? There are 35*36^5 combinations. c) If you cannot have a 9 on the first number, what is the probability that all 5 blanks are odd numbers? The probability is (4/35)*(5/36)^5. A good example of a permutation: Use the above 5 space combination problem. How many possible ways are there to make a “combination” if you cannot repeat the same “number or letter” twice. The answer is (36 permute 5) . Ok, so meow we are going to get into the new subject: Think about the problem of how many ways there are to get your hand in spades or hearts. The answer is “52 choose 13”. Now, how many ways are there for your team to get your hands? The answer is “52 choose 13,13”. How many ways are there to deal out all of the cards in spades or hearts? The answer is “52 choose 13,13,13,13”. These are multinomial coefficients. These are useful when we are grouping samples into specific groups (by player, etc.) or because of an inherent quality (by color of value, etc.). A different example: We have the word D‐A‐L‐D‐E‐R‐F‐A‐R‐G. How many ways are there to arrange these letters? How many ways are there to “sample” 5 letters from this word? What is the probability that the 1st letter could be the same as the second? What is the probability that the 1st letter could be the same as the last? The Olympics: Suppose we are examining the 100m dash. There are 6 teams left in the final 8 spots. There are only duplicates, so there are 4 teams with 1 participant and 2 teams with 2 participants. How many ways are there for the contestants to finish if we are only interested in their countries (not the participants names)? Previous example: If we are only interested in the medals, how many ways are there for this to occur if we are only interested in the countries? ...
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This note was uploaded on 02/26/2012 for the course STAT 225 taught by Professor Martin during the Fall '08 term at Purdue UniversityWest Lafayette.
 Fall '08
 MARTIN
 Probability

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