Ok math geeks I need help

[Nifft]

I think its the fact that you go for the chance of *not* getting the results he asked about that is throwing people. It threw me to start with :S

IMHO it's the most intuitive way to think about this particular kind of problem, where you can have multiple independent success events.

Anyway, glad it wasn't an error on my part -- dyslexia strikes me every now and again.

Thanks, -- N

 

[Ferret]

I dunno, to me it sort of twists. It goes from saying the chances of them al being four or above, then it shows the chance of that not happening....I would always assume to looks at the successful outcome(s). I dunno maybe I'm reading it wrong. Either way, good job!

 

[orsal]

I dunno, to me it sort of twists. It goes from saying the chances of them al being four or above, then it shows the chance of that not happening....I would always assume to looks at the successful outcome(s). I dunno maybe I'm reading it wrong. Either way, good job!

The catch is that sometimes, to make a mathematical problem easier, you sometimes need to cleverly restate it in superficially different but logically equivalent terms. The key idea that Nifft was exploiting is that "or" and "and" are duals -- if you replace everything with its complement (i.e. put "not" in front of everything) then they switch places.

NOT (X OR Y) is the same as (NOT X) AND (NOT Y)
NOT (X AND Y) is the same as (NOT X) OR (NOT Y)

Which would you rather compute. "OR" probabilities are only straightforward if the events are mutually exclusive (in which case you can just add probabilities: prob(3 on a d6)=1/6, prob(5 on a d6)=1/6, so prob(3 or 5 on a d6)=2/6. But that's for a single die -- if you're rolling different dice, outcomes won't be mutually exclusive.

In contrast, "AND" probabilities are straightforward to calculate if the events are independent, i.e. knowing what happens with one of them won't affect the probability of the other(s). Events associated with different dice are independent -- if you know the first one came up 2, it won't change the probabilities of the next one. So the probabilities involved in this problem are independent.

Once you see this, you know you'd rather think about "AND" combinations than "OR" combinations. So, Nifft rephrased the question in terms of the opposite events, and got the answer quickly. It's a trick that comes up in any beginning probability theory course.

 

[Delta]

Probability that all four d10s will be above 3: .7 ^ 4 = 0.24 = 24% chance of failure
Probability that both d10s will be above 5: .5 ^ 2 = 0.25 = 25% chance of failure

Nifft's right, but I too need to see the answer in the same terms as the actual initial question. Taking the OP literally when he says "under 4" and "under 6":

- Chance of rolling 4d10 and getting one die under 4: 100-24 = 76%
- Chance of rolling 2d10 and getting one die under 6: 100-25 = 75%

As a side note, one of the most basic college math classes I teach has a segment on introductory probability, and the "complimentary principle" (solving problems via success = 100 - failure) is, oddly, one of the hardest things for them to grasp. That's the day that the whole class looks like they want to run to the side of a boat and puke overboard.

 

[Darklone]

Common problem. I know a guy who worked 50 years as an engineer and still he refuses to understand mathematical proves that assume the opposite and prove it wrong.

 

[Moon-Lancer]

I find this to be *insanely* helpful: http://www.fnordistan.com/smallroller.html

Thanks for the link

 

[DM-Rocco]

I find this to be *insanely* helpful: http://www.fnordistan.com/smallroller.html

That is what I need. I such at math and that has already can in handy, thanks.

 

[Nifft]

So, Nifft rephrased the question in terms of the opposite events, and got the answer quickly. It's a trick that comes up in any beginning probability theory course.

You make me sound smart.

Therefore, I love you.

Cheers, -- N

 

[Nonlethal Force]

Common problem. I know a guy who worked 50 years as an engineer and still he refuses to understand mathematical proves that assume the opposite and prove it wrong.

Well, from a logic standpoint, in mathmatics we have the four "types" of negation:

1. If A, then B.
2. If B, then A.
3. If not A, then not B.
4. If not B, then not A.

1 and 4 always have the same truth value for any hypothesis. 2 and 3 also always have the same truth value. However 1 and 2 do not necessarily have the same truth value. Just giving mathematical validation to your point, so long as you make sure that you do both: assume the opposite and then prove it wrong.

 

[Catavarie]

I just make out charts in excel and count out the number of desired responses. Now i feel like a math idiot.