# What do you think the best job/work is?

#### ldaian

Correct. 4 1 in 4 chances do not equal 1 100% chance, however, there is a .00390625% chance that you find 4 of the items in a 2 hour search... So there is always that.

#### Talamare

##### Well-Known Member
Calculate the chance of my finding 3 items that had 8% chance

Or tell me how to calculate it #### Oogie Da Bruce

Calculate the chance of my finding 3 items that had 8% chance

Or tell me how to calculate it It's been ages since I had to dust off my abacus. It's a probability function that roughly translates into:

2 hours of work 8% chance:

0 items = 71.64%
1 item = 24.95%
2 items = 3.25%
3 items = 0.19%
4 items = 0.004%

Don't ask me to explain how it works, just trust the math. #### Flood

Correct. 4 1 in 4 chances do not equal 1 100% chance, however, there is a .00390625% chance that you find 4 of the items in a 2 hour search... So there is always that.
It's been ages since I had to dust off my abacus. It's a probability function that roughly translates into:

2 hours of work 8% chance:

0 items = 71.64%
1 item = 24.95%
2 items = 3.25%
3 items = 0.19%
4 items = 0.004%

Don't ask me to explain how it works, just trust the math. ill trust one of you but not both of you it seems hard to beleive that both using the correct math, its easier to find four 8% items rather than four 25% items

yours seems more reasonable oogie

#### Talamare

##### Well-Known Member
Dammit did no one pay attention during math class?
I know I didnt I was busy sleeping!
How is it that the whole internet cant figure out a probability problem?!

*looks thru a bunch of math sites*

They said use

(0.25)^4 and (0.08)^4
.00390625 and 0.00004096

But this is chance to find 4 in a row, not chance to find 3 in 4 tries

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#### Dayks

OK if i remember correctly its done by:
For 1 out of 1
1x0.25
For 2 out of 2
0.25x0.25
For 3 out of 3
0.0625x0.25
For 3 out of 4
0.015625x0.75
=0.01171875x100
=1.171875%
but i have a bad memory so i may be wrong
so you just divide the chance by 100 every time and then you just times it by the chance of the next one (so in this case .25 to receive, .75 not to.

#### Talamare

##### Well-Known Member
Ask my friend on msn and this is what he gave me

at 25% chance
1 item in 4 tries = 105.46875%
2 item in 4 tries = 35.15625%
3 item in 4 tries = 11.71875%
4 item in 4 tries = 3.90625%

at 8% chance
1 item in 4 tries = 62.29504%
2 item in 4 tries = 5.41696%
3 item in 4 tries = 0.47104%
4 item in 4 tries = 0.04096%

#### Flood

i didnt pay attention or work and somehow walked out of highschool with a C in maths, granted i couldnt of done much much better in every subject but i didnt bother

#### Talamare

##### Well-Known Member
i didnt pay attention or work and somehow walked out of highschool with a C in maths, granted i couldnt of done much much better in every subject but i didnt bother
In my math class more often then not the students were correcting the teacher, so much so that sometimes some students would take over the class, oh and in my English class, the teacher didnt speak English so we had to have one of the spanish kids translate it -_- Gotta love public schools

#### Flood

In my math class more often then not the students were correcting the teacher, so much so that sometimes some students would take over the class, oh and in my English class, the teacher didnt speak English so we had to have one of the spanish kids translate it -_- Gotta love public schools
lol had one with english as his second language, that was his excuse whenever he got something wrong

there was this lad in my computer course in college who was insanely bright, knew most things before the teacher started teaching them, he helped the teachers, wrote better programs than them, helped the students more than the teacher, and walked out with lower qualifications than quite a few of the students..

#### Dayks

ha, i didnt do a single homework for my maths and barely any classwork and got an A
Unfortunately i'm now doing it at A level and I've got the same teacher that just cant teach me, and of course its an awful lot harder now.

#### Morzanka

Alright, here it is how to calculate it:

You have x% chance to find a product in 30 mins

In 2 hours:
0 products: 100 * (1 - (x/100))^4
1 product: 100 * 4 * (1 - (x/100))^3 * (x/100)
2 products: 100 * 6 * (1 - (x/100))^2 * (x/100)^2
3 products: 100 * 4 * (1 - (x/100)) * (x/100)^3
4 products: 100 * (x/100)^4

So if you have 25% chance:

0 p: 31,6%
1 p: 42,2%
2 p: 21,7%
3 p: 4,7%
4 p: 0,4%

And to get the chance of finding at least one, you should add everything but the chance of 0 p.

#### Oogie Da Bruce

If searching for 2 hours there are 4 occurances where you have X% of finding an object. There are a fixed amount of potential outcomes where F=failing to find find the object & S=suceeding in finding the object. The distribution of those potential outcomes are as follows:

FFFF=0
FFFS=1
FFSF=1
FFSS=2
FSFF=1
FSFS=2
FSSF=2
FSSS=3
SFFF=1
SFFS=2
SFSF=2
SFSS=3
SSFF=2
SSFS=3
SSSF=3
SSSS=4

Therefore for any given 2 hours of work the potential for finding objects is:

0 objects = 1 instance
1 object = 4 instances
2 objects = 6 instances
3 objects = 4 instances
4 objects = 1 instance

Now apply the probabilty of finding an object to the instances.

For a 2 hours seach:

Where S=probability of finding an object
And F=probability of failing (or 1-S)

0 items = FFFF = F^4
1 item = 4*FFFS = 4*F^3*S
2 items = 6*FFSS = 6*F^2*F^2
3 items = 4*FSSS = 4*F*S^3
4 items = SSSS = S^4

Example:

S = 25%
F = 1-S = 75%

0 items = 0.75^4 = 31.64%
1 item = 4*0.75^3*0.25 = 42.19%
2 items = 6*0.75^2*0.25^2 = 21.09%
3 items = 4*0.75*0.25^3 = 4.69%
4 items = 0.25^4 = 0.39%

Or for my 8% example

S=8%
F=92%

0 = 0.92^4 = 71.64%
1 = 4*0.92^3*0.08 = 24.92%
2 = 6*0.92^2*0.08^2 = 3.25%
3 = 4*0.92*0.08^3 = 0.19%
4 = 0.92^4 = 0.004%

This is basically the same math that Morzanka uses, I had already began proving the equation, so I thought I'd finish it.

Damn it!!! I told you to just trust the math!!! And Damn my compulsion!!!

#### Talamare

##### Well-Known Member
Ive learned more today then 4 years of high school math

#### Morzanka

Ive learned more today then 4 years of high school math
Then this game has some use after all... #### gueedo

Thats excelent...

#### Jercat

Oogie, I'm glad you explained it. Yay, knowledge!

#### Scottish Jim

But .... for instance, I'm tending hogs and while its a 20% to find Ham but the LUCK is shown as 0% ... I've tried off and on for three days now and can't get a ham ... and can't get that 0% higher.

Isn't that 0% telling me I have NO chance to find the Ham?

#### Talamare

##### Well-Known Member
Luck doesnt affect products

In a "ask" website (Note I asked before you guys put your responses) I just now got a response, and I figured it rude to just let it rot, by "Psi_Phi_Org"

First, we can look at the "brute force" approach, just listing out all of the win/lose scenarios and calculating the chance for each. I don't recommend this for the general case, but it can be helpful to see how the formulas we develop later work.

Option Number Game 1 Game 2 Game 3 Game 4 Chances
0 L L L L 0.75×0.75×0.75×0.75 = 0.31640625
1 L L L W 0.75×0.75×0.75×0.25 = 0.10546875
2 L L W L 0.75×0.75×0.25×0.75 = 0.10546875
3 L L W W 0.75×0.75×0.25×0.25 = 0.03515625
4 L W L L 0.75×0.25×0.75×0.75 = 0.10546875
5 L W L W 0.75×0.25×0.75×0.25 = 0.03515625
6 L W W L 0.75×0.25×0.25×0.75 = 0.03515625
7 L W W W 0.75×0.25×0.25×0.25 = 0.01171875
8 W L L L 0.25×0.75×0.75×0.75 = 0.10546875
9 W L L W 0.25×0.75×0.75×0.25 = 0.03515625
10 W L W L 0.25×0.75×0.25×0.75 = 0.03515625
11 W L W W 0.25×0.75×0.25×0.25 = 0.01171875
12 W W L L 0.25×0.25×0.75×0.75 = 0.03515625
13 W W L W 0.25×0.25×0.75×0.25 = 0.01171875
14 W W W L 0.25×0.25×0.25×0.75 = 0.01171875
15 W W W W 0.25×0.25×0.25×0.25 = 0.00390625

Now to correlate these to your specific scenarios:

The chance of winning once in four games is the sum of the chances of winning options 1, 2, 4, or 8. That's 0.10546875×4 = 0.421875 = 42.1875%.

The chance of winning twice in four games is the sum of options 3, 5, 6, 9, 10, or 12. That's 0.03515625×6 = 0.2109375 = 21.09375%.

The chance of winning thrice in four games is the sum of options 7, 11, 13, or 14. That's 0.01171875×4 = 0.046875 = 4.6875%.

The chance of winning all four games is option 15. That's 0.00390625 = 0.390625%.

And though you didn't ask for it specifically, the odds of winning none of the games is option 0. That's 0.31640625 = 31.640625%.

Now, let's analyze this to see how this all works out.

For each individual option that had a certain number of wins, the chances were identical. Therefore, to find the total chance for a certain number of wins in all posiitons, we had to take the odds for that number of wins in any position, and multiply by the number of possible ways that could happen. So we need to find formulas for each half of this.

Let's define a couple variables to keep things kind of generic. We'll start by saying P is the percentage for a win in a single game, and N is the total number of games.

When there are four games:

• The base odds of 0 wins are (1-P)(1-P)(1-P)(1-P)
• The base odds of 1 win are (P)(1-P)(1-P)(1-P)
• The base odds of 2 wins are (P)(P)(1-P)(1-P)
• The base odds of 3 wins are (P)(P)(P)(1-P)
• The base odds of 4 wins are (P)(P)(P)(P)
To generalize to N games:

• The base odds of 0 wins are (P)^(0)×(1-P)^(N)
• The base odds of 1 win are (P)^(1)×(1-P)^(N-1)
• The base odds of 2 wins are (P)^(2)×(1-P)^(N-2)
...
• The base odds of X wins are (P)^(X)×(1-P)^(N-X)
...
• The base odds of N-2 wins are (P)^(N-2)×(1-P)^(2)
• The base odds of N-1 wins are (P)^(N-1)×(1-P)^(1)
• The base odds of N wins are (P)^(N)×(1-P)^(0)
The key line here is for X wins: (P)^(X)×(1-P)^(N-X).

Now let's figure out how many ways there are to win X number of times...

When there are four games:

• To win 0 times, there is 1 way (lose all four)
• To win 1 time, there are 4 ways (win game #1, or #2, or #3, or #4)
• To win 2 times, there are 6 ways (4 ways to win one game × 3 ways to win a second game, but divided by 2 because winning games #2 and #1 is the same as winning games #1 and #2)
• To win 3 times, there are 4 ways (6 ways to win two games × 2 ways to win a third game, but divided by 3 because winning games 1-2-3 is the sams as 1-3-2 and 3-1-2)
• To win 4 times, there is 1 way (win all four)
You may notice that there is a progression from one to the next: ×4/1, ×3/2, ×2/3, ×1/4. That's because for each additional win, you multiply by how many possibilities there are for the next win, but divide by which the number of possibilities with the same game numbers in a different order.

As it turns out, this progression is part of something called Pascal's Triangle. That's a triangle where each row is one number longer than the row above it, and each number is the sum of the two numbers above it (pretend that there are extra 0's at both ends of the previous row to calculate the numbers on the ends). Take a look at the first few rows:

1​
1 · 1​
1 · 2 · 1​
1 · 3 · 3 · 1​
1 · 4 · 6 · 4 · 1​
1 · 5 · 10·10 · 5 · 1​
1 · 6 · 15 ·20· 15 · 6 · 1​
1 · 7 · 21 · 35·35 · 21 · 7 · 1​
1 · 8 · 28 · 56 ·70· 56 · 28 · 8 · 1​

For any given row in Pascal's Triangle, the same progression holds. For example, on row 7 (counting the row with a single 1 as "row 0"), the progression is ×7/1, ×6/2, ×5/3, ×4/4, ×3/5, ×2/6, ×1/7.

Another way of calculating this is to use a function called factorial, which is the product of all the numbers from the given number to 1. The expression 4! is read aloud as "four factorial", and it means 4×3×2×1. (By definition, 0! = 1, which is the only part of the factorial that may not be obvious.) With this function, the numbers on row 4 can also be written like this:

4!/(0!×4!), 4!/(1!×3!), 4!/(2!×2!), 4!/(3!×1!), 4!/(4!×0!)

For 4 games:

• For 0 wins, there are 4!/(0!×4!) ways that could happen.
That's (4×3×2×1)/((1)×(4×3×2×1)) = 1.
• For 1 win, there are 4!/(1!×3!) ways. That's (4×3×2×1)/((1)×(3×2×1)) = 4.
• For 2 wins, there are 4!/(2!×2!) ways. That's (4×3×2×1)/((2×1)×(2×1)) = 6.
• For 3 wins, there are 4!/(3!×1!) ways. That's (4×3×2×1)/((3×2×1)×(1)) = 4.
• For 4 wins, there are 4!/(4!×0!) ways. That's (4×3×2×1)/((4×3×2×1)×(1)) = 1.
Then for N games:

• For 0 wins, there are N!/(0!×N!) ways that could happen.
• For 1 win, there are N!/(1!×(N-1)!) ways.
• For 2 wins, there are N!/(2!×(N-2)!) ways.
...
• For X wins, there are N!/(X! × (N-X)!) ways.
...
• For N-1 wins, there are N!/((N-1)!×1!) ways.
• For N wins, there are N!/(N!×0!) ways.
Now to combine everything together...

If the chance of winning a game is P, and there are N games, then the chance of winning exactly X of those games is:
((P)^(X) × (1-P)^(N-X)) × (N!/(X! × (N-X)!))​

Or by moving the parentheses:
((P)^(X) × (1-P)^(N-X) × N!)/(X! × (N-X)!)​

Let's run one of our initial questions through, just to make sure it's right. If the chance of winning a game is 25% (P), and there are 4 games (N), then the chance of winning 2 times (X) should be:

((0.25)^(2) × (1-0.25)^(4-2) × 4!)/(2! × (4-2)!)
= (0.0625 × (0.75)^(2) × 4×3×2×1)/((2×1) × 2!)
= (0.0625 × (0.5625) × 24)/(2 × (2×1))
= (0.84375)/(4)
= 0.2109375

That matches what we got by the brute-force method. That confirms that the general-case solution is:

((P)^(X) × (1-P)^(N-X) × N!)/(X! × (N-X)!)

-----------------------------------------------------------
The size of it scares me >_<

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#### Dayks

ah i now get what i did wrong, forgot to times the result by the amount of different ways it could happen