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Hi,

I have a probability question for all the maths whizs out there.

Given the following probabilities based on a D6:

Turn 2 (4+) - 50.00%
Turn 3 (3+) - 83.50% cumulative (includes the probability of Turn 2)
Turn 4 (2+) - 97.25% cumulative
Turn 5 (2+) - 99.54% cumulative
Turn 6 (2+) - 99.92% cumulative

So event occurs on 4+ Turn 2, 3+ Turn 3 and 2+ Turns 4-6 then there is 99.92% chance of it occuring by Turn 6.

Effectively think of it in terms of Reserves arriving. If I have 9 units I am interested in knowing what the chances are that eight units will arrive by T2, T3, T4, T5, T6 etc....the probability 7 units will arrive T2, T3, T4 etc

I'm looking to work out the distribution pattern for a given number of events.

If I have 9 potential instances of the event occuring then what is the chance of it occuring 9 times on Turn 2 [I calculate it as 0.20%0, 9 times by Turn 3 [my maths says 19.73%], 9 times by Turn 4 [   77.80%], 9 times by Turn 5 [   95.95%] and 9 times by Turn 6 [  99.31%].

So effectively I want to complete this matrix:

Turn 2 3 4 5 6

9 times 0.20% 19.73% 77.80% 95.95% 99.31%

8 times

7 times

6 times

5 times

4 times

3 times

2 times

1 time

0 times 0.20% 0.00% 0.00% 0.00% 0.00%

So can you help me out. It is obviously not a normal distribution past Turn 2.

Any help appreciated

Pete

Take your 1295/1296 number. Call it x.

Matrix is:

(1-x)^9 * x^0 * 9C9
(1-x)^8 * x^1 * 9C8
(1-x)^7 * x^2 * 9C7
(1-x)^6 * x^3 * 9C6
(1-x)^5 * x^4 * 9C5
(1-x)^4 * x^5 * 9C4
(1-x)^3 * x^6 * 9C3
(1-x)^2 * x^7 * 9C2
(1-x)^1 * x^8 * 9C1
(1-x)^0 * x^9 * 9C0

Now take your 215/216. Call it x. Do the math above the same way.

Sorry I'm not sure I understand your explanation.

Just to make sure we are talking the same thing, what I'm after is the following.

I have nine units off table and what to know the probability distribution of them arriving.

On Turn 2 it is a normal distribution because it is 4+ to arrive. For 9 to turn up it would be a 0.20% and the same for 0.....want to know the distribution for 1 through 8.

For me it becomes more difficult when we go to Turn 3 as the distribution becomes assymmetrical. Keen to know what the actual distribution is for 1-8 units to turn up by Turn 3.

Then the same for Turns 4-6.

Please be gentle with the maths...enthusiastic amateur

Pete

Take your 1/2 number. Call it x.

Matrix is:

(1-x)^9 * x^0 * 9C9
(1-x)^8 * x^1 * 9C8
(1-x)^7 * x^2 * 9C7
(1-x)^6 * x^3 * 9C6
(1-x)^5 * x^4 * 9C5
(1-x)^4 * x^5 * 9C4
(1-x)^3 * x^6 * 9C3
(1-x)^2 * x^7 * 9C2
(1-x)^1 * x^8 * 9C1
(1-x)^0 * x^9 * 9C0

Cuz, yeah....that's more clearly explained.....

Sounds like you probably want to look up Binomial distribution.

The other poster was showing you how to calcuate the chance of each event (0 to 9 reserves arriving). All you need to do is plug in the probability of it happening for each turn - you already seem to have worked that out (I didn't check your figures).

So the chance of 8 arriving by turn 3 =

8 successes = .835^8 = 0.2363
1 failure = .165^1 = 0.165
9C8 = 9

0.2363 * 0.165 * 9 = 0.3509

I'm no math guru but I think you can chug this out using the binomial probability function (looks like that is what olinari is proposing but I'm not sure). If I have time I'll see what I can do (but honestly, olinair probably has the easier way to do it). What I'm not sure of is just how accurate this would be (as I said, not a guru and its been a long time since I've had to do probability).

http://faculty.vassar.edu/lowry/binomialX.html

So for say 7 of 9 units coming in by turn 3.
n=9
k=7
p=5/6
q=1/6 (only 1/6 chance of failing both rolls)

Excel can help you with doing this multiple times. It has a function binomdist that you can use. Use the numbers above and set cumulative to false if you want to know the exact chance of 7 success or true to see chance of at most 7 success.
Hope that helps.

EDIT statement about setting binomdist true as it was wrong.

Thanks Guys, that's great.

I'll have a go crunching them this weekend.

Pete