Expert: Paul Klarreich - 12/29/2007

Question
I’m trying to calculate a distribution table that shows the distribution of how many times in a row an outcome occurs over a given number of events.

Here is the problem. I’d like to see the answer with an explanation of the process.

Each event can have one of four possible outcomes at various probabilities:

Outcome A = p(.3157895)
Outcome B = p(.3157895)
Outcome C = p(.3157895)
Outcome D = p(.0526315)

I want to know over a long period of time, say 1,000 events, how will the outcomes be distributed. In other words, how many times will each outcome show up alone, twice consecutively, 3 times, 4 times….10 times.

A table like so, showing the probabilities:

    1  2  3  4   5  6  7    8  9  10
 A                     

 B                              
 C                              
 D                              

I would think that the probability of outcome A occurring twice in a row would be:
A2p = (.3157895) (.3157895)
And the probability of A happening 3 times in a row is:
A3p = (.3157895) (.3157895) (3.57895)

However that doesn’t work because when I do that for (A1, A2, A3, …., A10), then add up the probabilities I should get a total of .3157895, but I don’t.

Thanks  

Answer
Questioner:   Vance
Category:  Advanced Math
Private:  No
 
Subject:  probabilities
Question:  I’m trying to calculate a distribution table that shows the distribution of how many times in a row an outcome occurs over a given number of events.

Here is the problem. I’d like to see the answer with an explanation of the process.

Each event can have one of four possible outcomes at various probabilities:

Outcome A = p(.3157895)
Outcome B = p(.3157895)
Outcome C = p(.3157895)
Outcome D = p(.0526315)

I want to know over a long period of time, say 1,000 events, how will the outcomes be distributed. In other words, how many times will each outcome show up alone, twice consecutively, 3 times, 4 times….10 times.

A table like so, showing the probabilities:

   1  2  3  4 5  6  7 8  9  10
A

B
C
D

I would think that the probability of outcome A occurring twice in a row would be:
A2p = (.3157895) (.3157895)
And the probability of A happening 3 times in a row is:
A3p = (.3157895) (.3157895) (3.57895)

However that doesn’t work because when I do that for (A1, A2, A3, …., A10), then add up the probabilities I should get a total of .3157895, but I don’t.

Thanks
.........................................
Hi, Vance,

This is a lot of calculation, so I am just going to outline one of the computations and leave it there.

What is the probability of obtaining A exactly twice in a row at the n-th trial?

p(not A at n-1) * p(A at n) * p(A at n+1) * p(not A at n+2)

That is good for n = 2 up to n = 1000 - 2.
Now add those up for n = 2 to 998 and add in the probability at:

n = 1:  p(A at 1) * p(A at 2) * p(not A at 3)
n = 999:p(not A at 998) * p(A at 999) * p(A at 1000)

That's 999 probabilities, which you divide by 999 to get a total p.

IF that makes any sense, you can repeat it for the others.