osanseviero
  • osanseviero
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
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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osanseviero
  • osanseviero
I don't understand the answer given
mathstudent55
  • mathstudent55
If you roll 4 fair dice, how many total possible outcomes are there?
osanseviero
  • osanseviero
1296, right? (6*6*6*6)

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mathstudent55
  • mathstudent55
You 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
  • mathstudent55
Ok, you got that correct. Now we need to see how many of the outcomes add to 21.
osanseviero
  • osanseviero
Yep...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
  • mathstudent55
Just 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
  • mathstudent55
5,5,5,6 5,6,5,5 5,5,6,5 5,5,5,6
mathstudent55
  • mathstudent55
I 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
  • osanseviero
Yep. 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
  • osanseviero
I 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
  • mathstudent55
4! = 4 * 3 * 2 * 1
osanseviero
  • osanseviero
Yep...but from where does the 4! come for this example?
mathstudent55
  • mathstudent55
Use 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
  • osanseviero
Ok...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!*(n-r)!) So that would be 4! / 3! (using r=3), but I'm not sure
mathstudent55
  • mathstudent55
Since there are three 6's, you must divide by 3! = 6 4!/3! = 24/6 = 4
osanseviero
  • osanseviero
Oh so 4 + 4 + 4!/2!. You divide by the repeated ones
osanseviero
  • osanseviero
So now for the (5,5,6,6) it is the same 4!/(2 * 2!) Thanks!
mathstudent55
  • mathstudent55
When 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
  • mathstudent55
Take a look at this: http://alumnus.caltech.edu/~leif/FRP/probability.html
osanseviero
  • osanseviero
Thanks a lot, much better now
mathstudent55
  • mathstudent55
You're welcome.

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