Probability and the 'Ways Method'Date: 10/25/2002 at 12:16:52 From: Rae Subject: Investigation of games of chance Suppose we roll one six-sided die. What are the possible outcomes? What is the probabiliy of rolling a 4? If we have two dice how many outcomes are there? With two dice what is the probability of rolling a 5? I would appreciate it if you could lead me in the direction of getting the correct answer. Thanks. Date: 10/25/2002 at 13:35:56 From: Doctor Achilles Subject: Re: Investigation of games of chance Hi Rae, Thanks for writing to Dr. Math. I like to think about these problems in terms of what I call the "ways method" [Note: this is my own name for a common statistical device that other people call other things.] Here's how the "ways method" works: Step 1: List all the possible outcomes. For one die, that is easy to do; there are 6: 1 2 3 4 5 6 Step 2: List the number of equally likely ways that you can get each outcome: For 1, there is one way to get it (roll a 1) For 2, there is one way to get it (roll a 2) For 3, there is one way to get it (roll a 3) For 4, there is one way to get it (roll a 4) For 5, there is one way to get it (roll a 5) For 6, there is one way to get it (roll a 6) Step 3: add up the total number of ways for all outcomes: 1+1+1+1+1+1 = 6 Step 4: the probability of a given outcome is equal to the number of ways you can get it divided by the total number of ways for all outcomes. So, for example, there is one way to get a 5 and there are 6 ways total, so the probability of getting a 5 is equal to 1/6. Let's apply the "ways method" to two dice. Step 1: list the possible outcomes: 2 3 4 5 6 7 8 9 10 11 12 Step 2: list the number of ways to get each outcome: For 2: There is only one way: you have to roll a 1 on the first die and a 1 on the second die. For 3: There are two ways: you can roll a 1 on the first die and a 2 on the second or a 2 on the first die and a 1 on the second. For 4: There are 3 ways: you can roll a 1 on the first die and a 3 on the second, or a 2 on the first die and a 2 on the second, or a 3 on the first die and a 1 on the second. For 5: There are 4 ways: you can roll a 1 on the first die and a 4 on the second, or a 2 on the first and a 3 on the second, or a 3 on the first and a 2 on the second, or a 4 on the first and a 1 on the second. For 6: There are 5 ways: you can roll a 1 on the first and a 5 on the second, or a 2 on the first and a 4 on the second, or a 3 on the first and a 3 on the second, or a 4 on the first and a 2 on the second, or a 5 on the first and a 1 on the second. For 7: There are 6 ways: you can roll a 1 on the first and a 6 on the second, or a 2 on the first and a 5 on the second, or a 3 on the first and a 4 on the second, or a 4 on the first and a 3 on the second, or a 5 on the first and a 2 on the second, or a 6 on the first and a 1 on the second. For 8: There are 5 ways: you can roll a 2 on the first and a 6 on the second, or a 3 on the first and a 5 on the second, or a 4 on the first and a 4 on the second, or a 5 on the first and a 3 on the second, or a 6 on the first and a 2 on the second. For 9: There are 4 ways: you can roll a 3 on the first and a 6 on the second, or a 4 on the first and a 5 on the second, or a 5 on the first and a 4 on the second, or a 6 on the first and a 3 on the second. For 10: There are 3 ways: you can roll a 4 on the first and a 6 on the second, or a 5 on the first and a 5 on the second, or a 6 on the first and a 4 on the second. For 11: There are 2 ways: you can roll a 5 on the first and a 6 on the second or a 6 on the first and a 5 on the second. For 12: There is only 1 way: you have to roll a 6 on both. To summarize: 2 has 1 way 3 has 2 ways 4 has 3 ways 5 has 4 ways 6 has 5 ways 7 has 6 ways 8 has 5 ways 9 has 4 ways 10 has 3 ways 11 has 2 ways 12 has 1 way Step 3: add up the total number of ways: 1+2+3+4+5+6+5+4+3+2+1 = 36 Step 4: to calculate the probability of each outcome, take the number of ways for that outcome divided by the total number of ways (36). 2 has a probability of 1/36 3 has a probability of 2/36 = 1/18 4 has a probability of 3/36 = 1/12 5 has a probability of 4/36 = 1/9 6 has a probability of 5/36 7 has a probability of 6/36 = 1/6 8 has a probability of 5/36 9 has a probability of 1/9 10 has a probability of 1/12 11 has a probability of 1/18 12 has a probability of 1/36 How can we use this method to find the probability with 3 dice? One thing to note: this method works for all probability calculations, but it is not necessarily always the best. It assumes that each "way" is EQUALLY likely (which is true for dice, as long as they aren't weighted). You have to do some complicated adjustments if the "ways" aren't equally likely. Also, the calculations become more difficult if the probabilities change (for example, if events are not independent). For example, with a deck of cards, the chances of getting a King of Spades in one card is 1/52, but if you draw a card and it's not the King of Spades, then your chances of getting the King of Spades in the second trial become 1/51. Again, this method can deal with that, but there may be others that are more efficient. I hope this helps. If you have other questions about this or you're still stuck, please write back. - Doctor Achilles, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/