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osanseviero
 one year ago
What is the probability that the total after rolling 4 fair dice is 21?
And of 22?
http://projects.iq.harvard.edu/files/stat110/files/strategic_practice_and_homework_1.pdf
osanseviero
 one year ago
What is the probability that the total after rolling 4 fair dice is 21? And of 22? http://projects.iq.harvard.edu/files/stat110/files/strategic_practice_and_homework_1.pdf

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osanseviero
 one year ago
Best ResponseYou've already chosen the best response.1I don't understand the answer given

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1If you roll 4 fair dice, how many total possible outcomes are there?

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.11296, right? (6*6*6*6)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1You have 6 possible outcomes with each roll. For each of the 6 outcomes of the first die, there are 6 possible outcomes of the second die. Then there are 6 possible outcomes of the third die and another 6 possible outcomes of the 4th die. The total number of outcomes is 6 * 6 * 6 * 6 = 6^4 = 1296

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Ok, you got that correct. Now we need to see how many of the outcomes add to 21.

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.1Yep...that's where I get confused. I know we can use (6,6,6,3) (6,5,5,5) and (6,6,5,4). My problem is knowing how many of each one there are

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Just using three 6's and one 3, there are 4 ways: 3,6,6,6 6,3,6,6 6,6,3,6 6,6,6,3

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.15,5,5,6 5,6,5,5 5,5,6,5 5,5,5,6

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1I don't think we need to count all the ways. The question is asking what is more likely to happen a sum of 21 or a sum of 22 using 4 rolls of a die. You don't need to know each one. You just need to compare the results.

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.1Yep. But if you check the answer they explain using permutations. " To get a 21, the outcome must be a permutation of (6, 6, 6, 3) (4 possibilities), (6, 5, 5, 5) (4 possibilities), or (6, 6, 5, 4) (4!/2 = 12 possibilities). To get a 22, the outcome must be a permutation of (6, 6, 6, 4) (4 possibilities), or (6, 6, 5, 5) (4!/22 = 6 possibilities). "

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.1I would like to understand why it is 4!/2...or how to do this in a less exhaustive way (so I can do this with big numbers)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.14! = 4 * 3 * 2 * 1

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.1Yep...but from where does the 4! come for this example?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Use the counting principle. In how many way can you arrange 1,2,3,4? There are 4 possibilities for the first position, then 3 possibilities for the second position, then 2 possibilities for the third position then 1 for the last position. That gives 4 * 3 2 * 1 That means the digits 1,2,3,4 (without repetition) can make 4! different combinations. Now what about the letters A,A,B,C? Again you can do the counting principle and get 4 * 3 * 2 * 1, but since there are two A's, you must divide by 2 to eliminate groupings that look the same. If we call one A "A1" and the other A "A2" A1A2BC is indistinguishable from A2A1BC. That's why we divide by 2.

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.1Ok...so lets go with the example of 3,6,6,6. There are 4 possibilities for every position. So that would be 4! = 24. Then how did that decrease to 4? I'm thinking in: (n!)/(r!*(nr)!) So that would be 4! / 3! (using r=3), but I'm not sure

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Since there are three 6's, you must divide by 3! = 6 4!/3! = 24/6 = 4

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.1Oh so 4 + 4 + 4!/2!. You divide by the repeated ones

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.1So now for the (5,5,6,6) it is the same 4!/(2 * 2!) Thanks!

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1When you roll 1 die, there is an equal probability of landing any of the 6 outcomes. When you roll 2 dice, the minimum you can get is 2 and the maximum is 12. (2 + 12)/2 = 7 The most likely outcome is 7. The least likely are 2 and 12. As you go down from 7 to 2, each outcome is less likely. As you go up from 8 to 12, each outcome is less likely. With 3 dice, the outcomes are from 3 to 18. The average is (3 + 18)/2 = 10.5. The most likely outcomes are equally 10 and 11. Outcomes from 9 to 3 and outcomes from 12 to 18 become progressively less likely. When 4 dice are rolled, the outcomes range from 4 to 24. The average is (4 + 24)/2 = 14. Outcomes from 13 to 4 become progressively less likely, as do outcomes from 15 to 24. Since 22 is more than 21, 22 is less likely than 21.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Take a look at this: http://alumnus.caltech.edu/~leif/FRP/probability.html

osanseviero
 one year ago
Best ResponseYou've already chosen the best response.1Thanks a lot, much better now
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