Dice: Full HouseDate: 6/28/96 at 22:44:23 From: Anonymous Subject: Dice Full House Probability I want to know the probability of rolling a full house with 5 six sided dice. For example, a successful roll of five dice would be 11222. I believe the answer is 300/(6 to the 5th power). There are thirty ways that you can roll a full house. 11 with 222,333,444,555,666 22 with 111,333,444,555,666 33 ... 44 ... 55 ... 66 ... therefore the number of successful rolls is 5*6=30 Here is where I am a little unsure of my solution. I believe that there is a difference between rolling 23332 and 22333 This means that there are more possible correct answers than 30. Is the answer the number of permutations of 5 things taken 5 at a time * the 30 possible Or is the answer the number of combinations 5 things taken 5 at a time * the 30 possible ? Each die is an individual event. By the way, 11111 and 11114 are not a successful possibilities. Thank you for your time. Sincerely James Eastman Date: 6/29/96 at 11:11:50 From: Doctor Anthony Subject: Re: Dice Full House Probability As you have correctly noted, you could choose the pair in 6 ways and then the threesome in 5 ways, giving 6*5 = 30 ways of choosing the two numbers to make up the full house. Now the number of different sequences in any one of these full houses is the number of ways of arranging 5 things, 2 being alike of one kind and 3 being alike of a different kind = 5!/(2!.3!) which is also the same as 5_C_2 = 10. So having the two numbers chosen in 30 ways, these can be arranged in 10 ways within each group of 5, giving a total of 30*10 = 300 ways. Now the unrestricted way of choosing the 5 numbers is 6^5, and so the probability of a full house is: P(Full House) = 300/6^5 = 300/7776 = 25/648 = 0.03858 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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