Thread: Combinatorics Question of the Day

1. Originally Posted by mistuhbull
but I almost got it, which is a different possibility than getting it, or not being close to getting it.
No. Read the OP. There are exactly three possibilities in this scenario:
1. You don't use a extraroll
2. You use a extraroll and don't get an item
3. You use a extraroll and get an item

It's completely irrelevant if you "almost" got an item, the possibilities don't change with the probabilities (unless the chance is 0 or 100%).

2. Originally Posted by Stede
Why only D? Why aren't we considering the win / no-win for not using a bonus roll? If we do, then the outcome space has 4 discrete possibilities, and we're looking at 4^10.

In this hypothetical scenario, since each raider is an individual, I suppose we'd consider the order unique - and so, 4^10 is accurate.
What's the fourth possibility?

I count only 3:

1. No bonus roll
2. Bonus roll + no item
3. Bonus roll + item

3. Originally Posted by mistuhbull
but there is an almost. If I rolled a 49 but the win roll was 50, then I almost won. I didn't win, but I almost did
No, rolling a 49 is no different than rolling a 1 or any other number below 50, in this context. Both are non-win scenarios and constitute the same outcome ie possibility.

4. 59049 in 10 man and 847288609443 in 25 man
or 3 to the power of the difficulty.
Rolls are not important as each person does not affect the others chances.(there is a calculable odd that everyone rolls and gets loot witch is 0.000017)
It is a system of 10 elements each with 3 different states(not roll,roll and lose,roll and win).The solution to this is in a mathematical formula.

5. Originally Posted by Laurabelle
What's the fourth possibility?

I count only 3:

1. No bonus roll
2. Bonus roll + no item
3. Bonus roll + item
I think Stede is trying to say that without using a bonus roll, you can win or lose? I don't really understand this, as you can't win or lose a roll that never happened.

6. Originally Posted by oltavp1
59049 in 10 man and 847288609443 in 25 man
or 3 to the power of the difficulty.
Rolls are not important as each person does not affect the others chances.(there is a calculable odd that everyone rolls and gets loot witch is 0.000017)
It is a system of 10 elements each with 3 different states(not roll,roll and lose,roll and win).The solution to this is in a mathematical formula.
This is exactly correct.

7. Originally Posted by Static Transit
You can't win or lose a game you don't play, my friend.
Well, let's see. Here's the full space.

(Initial Roll / Bonus Roll)
Win / na
Lose / na
Win / Win
Win / Lose
Lose / Win
Lose / Lose

It's unclear exactly what the OP is after, but it seems like we're looking at those that actually used their bonus roll. That means #1 & #2 are out. Then, depending on whether we're looking at everything that's left or just at the scenarios with a win, we have either four possibilities left or we exclude #6. Then we consider there's 10 raiders, and since we're talking about bonus rolls, we can assume independence.

8. More formally, we can construct a bijection from the set of all distributions of (not roll = A, roll and lose = B, roll and win = C) over a 3 letter alphabet {A,B,C}. Forming 10 letter words for every possible state (the ith letter in the string would correspond to the ith raider's state). Since we can easily enumerate the later, the former holds the same count by the definition bijection.

9. Originally Posted by Static Transit
I think Stede is trying to say that without using a bonus roll, you can win or lose? I don't really understand this, as you can't win or lose a roll that never happened.
I guess I didn't follow that we're only considering the bonus roll and we're not considering the the intial roll at all. That's a pretty straightforward question - straightforward enough that I didn't think anyone would really ask it.

10. Originally Posted by Stede
Well, let's see. Here's the full space.

(Initial Roll / Bonus Roll)
Win / na
Lose / na
Win / Win
Win / Lose
Lose / Win
Lose / Lose

It's unclear exactly what the OP is after, but it seems like we're looking at those that actually used their bonus roll. That means #1 & #2 are out. Then, depending on whether we're looking at everything that's left or just at the scenarios with a win, we have either four possibilities left or we exclude #6. Then we consider there's 10 raiders, and since we're talking about bonus rolls, we can assume independence.
Stede, I thought the question was pretty clear. I did not include the normal rolls, only bonus in the question.

---------- Post added 2013-02-13 at 11:15 AM ----------

OK, anyways this experiment worked out. Look for other Combinatorics Questions in the future.

11. Originally Posted by Laurabelle
Stede, I thought the question was pretty clear. I did not include the normal rolls, only bonus in the question.

---------- Post added 2013-02-13 at 11:15 AM ----------

OK, anyways this experiment worked out. Look for other Combinatorics Questions in the future.
Looking back on it, it was clear. I just didn't think you'd start a thread about something so rudimentary and slap a label on it that has more syllables than the question has calculation steps.

Side note that the independence of bonus rolls is important. The calculation for the normal roll would be different.

12. Originally Posted by Stede
Looking back on it, it was clear. I just didn't think you'd start a thread about something so rudimentary and slap a label on it that has more syllables than the question has calculation steps.
Well it was just meant to be an elementary counting problem from the Combinatorics branch of Math (Probability & Statistics may also cover this area). Though, you'd be surprised how even something so basic can puzzle very smart people.

13. Originally Posted by oltavp1
59049 in 10 man and 847288609443 in 25 man
or 3 to the power of the difficulty.
Rolls are not important as each person does not affect the others chances.(there is a calculable odd that everyone rolls and gets loot witch is 0.000017)
It is a system of 10 elements each with 3 different states(not roll,roll and lose,roll and win).The solution to this is in a mathematical formula.
This was my thought initially with one difference - we assume *some* of the raiders will use the coin. Doesn't that translate into "not the entire group"?

14. Originally Posted by Ranjit
This was my thought initially with one difference - we assume *some* of the raiders will use the coin. Doesn't that translate into "not the entire group"?
May be an ambiguity with the English language in the problem statement, but I had intended that the possibility that all 10 raiders use a bonus roll as a valid option.

15. Originally Posted by Laurabelle
Well it was just meant to be an elementary counting problem from the Combinatorics branch of Math (Probability & Statistics may also cover this area). Though, you'd be surprised how even something so basic can puzzle very smart people.
Generally, my rule is that, after I spend 5-10 minutes working through something, I go back and make sure I'm reading the question right. Generally. Look forward to more of these.

16. Well english is not my native either, but i assumed "any number of" is the correct term so was suspecting it might be a tricky question and the catch is in the wording

17. Originally Posted by Laurabelle
May be an ambiguity with the English language in the problem statement, but I had intended that the possibility that all 10 raiders use a bonus roll as a valid option.
Even if it wasn't so.You just had to subtract the two states(all roll/no roll).Problem solved

18. Originally Posted by oltavp1
Even if it wasn't so.You just had to subtract the two states(all roll/no roll).Problem solved
Also, the state of "all/no rollers won" since *some* of the rollers won, right?

Anywho, we got the answer, this was nice, please keep them coming.

19. Originally Posted by Ranjit
Also, the state of "all rollers won" since *some* of the rollers won, right?

Anywho, we got the answer, this was nice, please keep them coming.
Yes,as long as you remove the "no one won" state as well

Ps.The fact that i'm studying maths is kinda helping here

20. Edited it right when you were posting i guess. And when it comes to your PS, i guess everyone here studied some sort of maths to be even remotely interested in this

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