Question

"Online purchasers of a computer game A also often bought two other games L and H.
A recent study has shown that 16% of purchasers of A also bought L; 10% also bought H and that 10% of purchasers of one of H and L also bought the other.
If a purchaser of A is found to have bought H; what is the probability that they also bought L?"

Answer 1

This is all I have for now :(
\[P(L|A) = 0.16\]\[P(H|A) = 0.10\]\[P(H|L) + P(L|H) = 0.10\]

Answer 2

\[P(L|A,H) = \]

Answer 3

\[P(L|A\cap H) = \]I'm a little stuck....

Answer 4

give me a hint :(

Answer 5

I should probably compute the probabilty that someone bought A first

Answer 6

Ahh that hint means I don't have to :-P

Answer 7

The problem is intended to be entirely about purchasers of a computer game A and what other games they might purchase. You can therefore treat purchasers of the game A as the universal set. The 16% and 10% for L and H respectively are intended as independent figures

Answer 8

\[P(H) = 0.16\]\[P(L) = 0.10\]\[P(L|H) + P(H|L) = 0.10\] that looks better

Answer 9

\[P(L|H) = P(L)\]

Answer 10

!!! :(

Answer 11

Let's check that answer, and run a Python simulation :-D

Answer 12

or is it wrong?

Answer 13

Try posting it here; they love problems like these over there http://math.stackexchange.com/

Answer 14

I will go shower and come back

Answer 15

Alright

Answer 16

I wish JamesJ would help.... I'm extremely sure I'm looking at this problem the wrong way, and that it is actually an elementary application of probability laws :-P

Answer 17

Alright I will try the python solution.

Answer 18

Ok, so what you actually have is
P(L|A)=0.16
P(H|A)=0.10
P(H|L)=0.10
P(L|H)=0.10

and you want to know P(L|H,A)

Answer 19

Is P(L|H,A)=P(L|H and A)?

Answer 20

yes

Answer 21

H intersect A :-D

Answer 22

this is tricky. Here's what I've got.

Answer 23

My strategy is to find P(L|HA) as part of a complete probability calculation

P(L|A) = P(L|HA)P(H|A) + P(L|-HA)P(-H|A)

Now we know P(L|A), P(H|A) and P(-H|A). We want P(L|HA) and we could get it except for this problem term of P(L|-HA).

Following so far?

Answer 24

yes

Answer 25

on the python simulation I wrote I ran into 0.01 as the number of people who first bought H and then bought L O.o

Answer 26

So let's expand P(L|-HA)

P(L|-HA) = P(-H|LA)P(L|A)/P(-H|A)
= ( 1 - P(H|LA) ) P(L|A) / (1 - P(H|A)

Now we know every term except P(H|LA)

Answer 27

so one more crank of the Bayes Rule handle:

P(H|LA) = P(L|HA)P(H|A)/P(L|A)

Now P(L|HA) is the quantity we want to find. Put all of this back together and it should be the only variable in the complete probability equation. Solve for it.

Answer 28

P(L|A)=0.16
P(H|A)=0.10
P(H|L)=0.10
P(L|H)=0.10

punching those in the calculator :-D

Answer 29

You guys are trying to make my brain explode by reading that stuff... There should be a warning label

Answer 30

I got 8/5 O.o

Answer 31

Here is my calculation : \[0.16 = x*0.10 + \frac{0.16-x*0.10}{1-0.10}\]

Answer 32

This is the python script I wrote.... I think it's also wrong
http://ideone.com/Dr4xd

Answer 33

in the script, I had 1000000 purchasers of game A.

Answer 34

I see what happens. Actually all the algebra collapses to a the trivial equation, 0 = 0.

Answer 35

lol

Answer 36

Which is perhaps to be expected, applying Bayes Rule twice to the same term.

We need a new strategy

Answer 37

I think the A part is a red herring. What if we treat the universal set just as 'people' and then draw subsets of buyers of H, buyers of L etc. from that

Answer 38

or am I mistaken?

Answer 39

No, buying A does give you information on buying L or H. In the Bayes network of L, H and A, A is at the top of triangle with arrows pointing down to L and H, and L and H have arrows going each way from one to the other

Answer 40

right

Answer 41

is that the right model of the problem? That deciding to buy a game is influenced by a previous decision to buy a different game?

Answer 42

It's actually not helpful to think of this in a Bayes network; I was just using that as an example of the dependencies here. All I want to say in particular is it is NOT true that
P(L|HA) = P(L|H)

Answer 43

right

Answer 44

I don't know. I'm at a dead end right now.

Answer 45

Let me think about it and get back to you. It's worth asking Zarkon as well, as he works in probability and statistics, while I'm just a visitor with a good travel guide.

Answer 46

I'll see if I can post this one at the stackexchange.

Answer 47

yes, and let me know if you get a good answer. Where does this problem come from btw?

Answer 48

Tomas.A just asked me in one of those 'Fan Testimonial' things.

Answer 49

lol

Answer 50

Thomas just sent me this link, that unravels the riddle of this problem:
http://pastebin.com/s4Erdcfi

Answer 51

ah

Answer 52

@Thomas, post your question here.

But look at this, quoting from the teacher's response:
"The problem in Q1 is intended to be entirely about purchasers of a computer game A and what other games they might purchase. You can therefore treat purchasers of the game A as the universal set. The 16% and 10% for L and H respectively are intended as independent figures"

However again a lot of students complained that it's unclear and he said ( worth mentioning that he write it only 15mins before deadline for all assignments):

"There are 2 sets involved
A: those who bought one (at least) of the games
B: those who bought both
B comprises 10% OF A"

Hence in fact
P(H|LA) = P(H|L) , because A is the universal set here, or formally P(A) = 1
= P(H) , because in fact H and L are independent
= 0.10

**********

I put this down as a poorly phrased question.

Answer 53

so the answer to the question is 0.10 :-D phew!