How Many Ways Can You Roll Six Dice with Repeating Numbers?

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The discussion revolves around calculating the number of ways to roll six dice with varying conditions of repeated numbers. Participants suggest starting with simpler cases, such as rolling two dice with the same number, to build understanding. The concept of negation is introduced to simplify the problem, focusing on outcomes where numbers are different. Various strategies, including tabling possibilities and using combinatorial formulas, are discussed to tackle cases of at least two, three, four, and five identical numbers. The conversation highlights the complexity of probability calculations and the importance of breaking down the problem into manageable parts.
  • #61
Shouldn't the probability that all numbers are different plus exaclt 2 are alike and the rest different plus exactly three are alike plus...up to they are all the same be no greater than one?
 
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  • #62
ArcanaNoir said:
Shouldn't the probability that all numbers are different plus exaclt 2 are alike and the rest different plus exactly three are alike plus...up to they are all the same be no greater than one?

Correct, they should sum up to exactly one (or less than one depending on what you calculate exactly).
 
  • #63
Okay, I think I have this problem all done. I've solved it probably more than one way by now, and I got consistent answers. *runs away*
 
  • #64
ArcanaNoir said:

Homework Statement



How many ways can you roll six dice so that at least 2 numbers are the same? At least 3? At least 4? At least 5?

Homework Equations





The Attempt at a Solution



:cry: I've used every equation in the chapter and filled page after page with numbers all moved around. I'm drowning! Please help. I hate probability.

{at least two the same} = E2 + E3 + E4 + E5 + E6, where Ej = {exactly j the same} (and "+" denotes set union). E2 = {2 1s, others all different}+{2 2s, others all different}+... Clearly, all these have the same number of elements, so the number of elements in E2 is 6*|{2 1s, all others different}| (|{.}| = set cardinality). Now think of 6 bins and 6 balls tossed at random into them; we want the number of outcomes in which bin 1 has 2 balls and the others all have 0 or 1 ball each. This the type of thing for which the _multinomial_ distribution was designed.

RGV
 
  • #65
ArcanaNoir said:
*runs away*

Hey! Come back! :smile:
There's more...
 

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