It's been ages since I had to dust off my abacus. It's a probability function that roughly translates into:Calculate the chance of my finding 3 items that had 8% chance

Or tell me how to calculate it

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% itemsIt'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.

yours seems more reasonable oogie

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

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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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.

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%

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 wrongIn 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

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..

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.

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.

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:

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

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!!!

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

Thats excelent...

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

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

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:

**(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:

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:

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:

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:

Or by moving the parentheses:

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 >_<

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

The chance of winning

The chance of winning

The chance of winning

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

Now, let's analyze this to see

For each

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)

- 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)

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)

As it turns out, this progression is part of something called

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.

- 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.

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:

-----------------------------------------------------------

The size of it scares me >_<

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