Probability of Rolling a 6 on One DieDate: 30 Apr 1995 02:51:31 -0400 From: Anonymous Subject: Dice The following questions came up during a game of Risk... First, we know that the odds of rolling a six with one die are 1 in six. What are the odds of rolling a six in two, three, four (etc.) rolls? It can't be 1/6 + 1/6 + ... or else you would have a 100% chance of a six in 6 rolls. How do the probabilities add up? Secondly, my friend says that if you roll several low numbers first, your chances are then better of getting a high roll. While he admits that this makes no logical sense probability-wise, he swears that there has been research done to back it up. Ever hear of anything like this? Thanks in advance, cantalupo@aol.com Date: 1 May 1995 15:00:35 GMT From: Daniel Eisenbud Subject: Re: Dice The odds of rolling a six with each of two consecutive rolls is 1/6 * 1/6 = 1/36. This makes sense, if you realize that when you roll the second die, you will only have rolled a 6 on the first one out of six times. So there's only a 1/6 chance that if you do roll a six on the second die, it will be your second six. The probability of rolling n sixes in a row is in fact (1/6)^n, for similar reasons. >Secondly, my friend says that if you roll several low numbers first, your >chances are then better of getting a high roll. While he admits that this >makes no logical sense probability-wise, he swears that there has been >research done to back it up. Ever hear of anything like this? No. The chance of rolling a six is always 1/6, no matter what you roll before it, no matter what. -Dan "Dr. Math" Eisenbud Date: 11 Feb 2000 13:26:22 GMT From: Dr. TWE Subject: Re: dice Hi! I think Dan misunderstood your first question. Are you asking what is the probability of rolling _at least_ one six on two, three, four (etc.) rolls? Let's examine this roll by roll. In one roll, the probability of rolling a 6 is 1/6. For two rolls, there is a 1/6 probability of rolling a six on the first roll. If this occurs, we've satisfied our condition. There is a 5/6 probability that the first roll is not a 6. In that case, we need to see if the second roll is a 6. The probability of the second roll being a 6 is 1/6, so our overall probability is 1/6 + (5/6)*(1/6) = 11/36. Why did I multiply the second 1/6 by 5/6? Because I only need to consider the 5/6 of the time that the first roll wasn't a 6. As you can see the probability is slightly less than 2/6. For three rolls, there is a 1/6 probability of rolling a six on the first roll. There is a 5/6 probability that the first roll is not a 6. In that case, we need to see if the second roll is a 6. The probability of the second roll being a 6 is 1/6, giving us a probability of 11/36. There is a 25/36 probability that neither of the first two rolls was a 6. In that case, we need to see if the third roll is a 6. The probability of the third roll being a 6 is 1/6, giving us a probability of 1/6 + (5/6)*(1/6) + (25/36)*(1/6) = 91/216. Again, this is less than 3/6. The general formula for rolling at least one 6 in n rolls is 1 - (5/6)^n. One final note: the probability of rolling _exactly_ one 6 in n rolls is different, and a little more difficult to compute. - Doctor TWE, The Math Forum http://mathforum.org/dr.math/ |
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