Anything Math
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27-02-2017, 12:21 AM
RE: Anything Math
I can explain what the IRR means now, if anyone wants me to Blush

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27-02-2017, 11:30 PM
RE: Anything Math
BEAM math quiz

https://www.nytimes.com/interactive/2017...-quiz.html

“I am quite sure now that often, very often, in matters concerning religion and politics a man’s reasoning powers are not above the monkey’s.”~Mark Twain
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27-02-2017, 11:33 PM
RE: Anything Math
(26-02-2017 01:45 PM)morondog Wrote:  
(22-02-2017 12:28 AM)Full Circle Wrote:  The IRR is the interest rate, also called the discount rate, that is required to bring the net present value (NPV) to zero. That is, the interest rate that would result in the present value of the capital investment, or cash outflow, being equal to the value of the total returns over time, or cash inflow.

That's about as clear as a shit sandwich. Jesus. Someone needs to rewrite that.

What do you need to calculate? This IRR thing? The page you linked has instructions for how to get Excel to do it automagically. I vote do that if all you want is the IRR. If you want something in-depth, for what purpose are you calculating it?

I used the automagic button. Bowing

% return on investment when additional dollars are added throughout the year.

“I am quite sure now that often, very often, in matters concerning religion and politics a man’s reasoning powers are not above the monkey’s.”~Mark Twain
“Ocean: A body of water occupying about two-thirds of a world made for man - who has no gills.”~ Ambrose Bierce
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28-02-2017, 12:37 AM
RE: Anything Math
I have a little challenge I made up Smile Please put solutions in spoilers! Probably quite easy for those well versed in probability.

I buy booster packs for a card game. Each pack contains 10 cards. The cards can be common, uncommon or rare. A pack always contains a guaranteed rare card, and two guaranteed uncommon cards. The remaining 7 cards can each be of any rarity. For each one, 95% of the time it will be a common, 4% of the time an uncommon and 1% a rare.

I open a booster pack, and randomly select a card from it to look at. It's a rare!

What is the probability that there's at least one more rare in the pack?

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28-02-2017, 11:46 PM
RE: Anything Math
Bump Big Grin

I'll give out a hint to get you started if you want.

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01-03-2017, 12:10 AM
RE: Anything Math
OK I had a go.

Since a rare card has been selected, that means the remaining 9 cards consist of two uncommon cards and 7 other cards satisfying the probability conditions. We are looking at the probability that there is at least one more rare card in the set of 10. That will be the complement of the probability that there are *no* more rare cards. There are 7 slots for a rare card, and each slot has 0.99 probability not to be rare. So 0.99^7 is the probability that there is no rare card in the remaining 9 cards. Therefore 1-0.99^7 = 0.068 is the probability that there is at least one more rare card.

This is assuming that rare cards are not uncommon cards. If rare cards are a subset of uncommon cards then I don't think there's enough information to solve the problem.

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(06-02-2014 03:47 PM)Momsurroundedbyboys Wrote:  And I'm giving myself a conclusion again from all the facepalming.
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01-03-2017, 12:41 AM
RE: Anything Math
(01-03-2017 12:10 AM)morondog Wrote:  OK I had a go.

Since a rare card has been selected, that means the remaining 9 cards consist of two uncommon cards and 7 other cards satisfying the probability conditions. We are looking at the probability that there is at least one more rare card in the set of 10. That will be the complement of the probability that there are *no* more rare cards. There are 7 slots for a rare card, and each slot has 0.99 probability not to be rare. So 0.99^7 is the probability that there is no rare card in the remaining 9 cards. Therefore 1-0.99^7 = 0.068 is the probability that there is at least one more rare card.

This is assuming that rare cards are not uncommon cards. If rare cards are a subset of uncommon cards then I don't think there's enough information to solve the problem.

Certainly along the right lines, you're very close! Yes, rares are not uncommons.

You've assumed that the rare that has been chosen is the guaranteed rare, and not one of the other 7 cards which happened to be a rare Smile

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01-03-2017, 01:13 AM
RE: Anything Math
(01-03-2017 12:41 AM)Robvalue Wrote:  
(01-03-2017 12:10 AM)morondog Wrote:  OK I had a go.

Certainly along the right lines, you're very close! Yes, rares are not uncommons.

You've assumed that the rare that has been chosen is the guaranteed rare, and not one of the other 7 cards which happened to be a rare Smile

I think it's OK to make that assumption. If a pack has two rare cards in it, does it matter which one was put in as the guaranteed one and which was the randomly chosen one? So again, when I pick a card out of the pack and it turns out to be rare, does it matter how it was chosen to be *in* the pack?

But OK. Let's see if I can calculate a difference. First I will assume that it was the guaranteed one that I chose. Then I get a probability of 0.068 that there's one more rare. If it was one of the random ones, then having chosen it there's a probability of 1 that there is one more rare, since there is a guaranteed rare in the pack.

I'm not sure how to combine these two observations, but my gut says multiply them, which gives me the same answer I already got. Otherwise, how would you combine them? These conditional probability things drive me nuts.

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(06-02-2014 03:47 PM)Momsurroundedbyboys Wrote:  And I'm giving myself a conclusion again from all the facepalming.
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01-03-2017, 01:49 AM
RE: Anything Math
(01-03-2017 01:13 AM)morondog Wrote:  
(01-03-2017 12:41 AM)Robvalue Wrote:  Certainly along the right lines, you're very close! Yes, rares are not uncommons.

You've assumed that the rare that has been chosen is the guaranteed rare, and not one of the other 7 cards which happened to be a rare Smile

I think it's OK to make that assumption. If a pack has two rare cards in it, does it matter which one was put in as the guaranteed one and which was the randomly chosen one? So again, when I pick a card out of the pack and it turns out to be rare, does it matter how it was chosen to be *in* the pack?

But OK. Let's see if I can calculate a difference. First I will assume that it was the guaranteed one that I chose. Then I get a probability of 0.068 that there's one more rare. If it was one of the random ones, then having chosen it there's a probability of 1 that there is one more rare, since there is a guaranteed rare in the pack.

I'm not sure how to combine these two observations, but my gut says multiply them, which gives me the same answer I already got. Otherwise, how would you combine them? These conditional probability things drive me nuts.

Yep it does matter, because as you say, if you happened to have picked out one of the 7 cards and it was a rare, the probability becomes 1. So this is a different scenario which needs to be taken into account. It is then about combining the two, yep Smile You're almost there!

The next step is to calculate these probabilities, and then see if you can figure out where to go from there. Imagine we're going back to before we picked out the rare card:

Probability of finding a rare by getting lucky with a pick from one of the 7 undefined cards.

Total probability of finding a rare to begin with.

It's about using Bayes theorem, but it can be thought out using a probability tree.

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01-03-2017, 02:05 AM
RE: Anything Math
Hmpf. OK I'll think about this thing some more. Most I touch probability with these days is random number generation.

We'll love you just the way you are
If you're perfect -- Alanis Morissette
(06-02-2014 03:47 PM)Momsurroundedbyboys Wrote:  And I'm giving myself a conclusion again from all the facepalming.
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