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I try to visualize a bernoulli chain. With variables p q and s.

  • p probability for success
  • q probability for failure
  • s length of the chain

For fixed p q and s the chain is easiliy calcultated. However s in my case is determined by a roll of n dice which each have d sides:

  • n = number of dice rolled
  • d = number of sided of the dice (3, 4, 6, 8, 10, 12, 20)

So lets take this example:

  • p = 0.25
  • q = 0.75
  • n = 10
  • d = 3

So s can be anywhere from 10 to 30. I want to cacluclate a probability vecotor with

[No success, exactly 1 success, exactly 2 Successes..., exactly 30 successes]

I think I have to calculate the probability of each possible s, multiplie it with the actual bernoulli formula. And add the probabilities for i.e. 5 successes from each possible outcome of s. Is this correct?

I really struggle to get the probabilities of each sum. I get it for 2 or 3 dice I could just go through all possible results and well add it but for 20 dice it breaks my computer ;)

Dennis Ich
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    I can't understand what you are saying. Please clarify what the procedure is. How do you decide the first roll (which here seems to be ten three sided dice)? Then it seems you roll something a number of times that is the sum from the first roll. How do you decide what that something is? What constitutes success? If we can't understand you, we can't help. – Ross Millikan May 31 '17 at 21:40
  • I agree with Ross Millikan: as it is very very unclear, beginning by your first sentence, unless you suppress all things that are unimportant and concentrate on a more detailed example, I fear that nobody will answer you. – Jean Marie May 31 '17 at 22:04
  • I clarified it and hope thats better understandable. Thanks for the comments. I think the answere given is what I need or atleast near to it. – Dennis Ich Jun 01 '17 at 06:16

1 Answers1

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There is a stars and bars approach...

Lets start with an example:

Suppose you roll 5 six-sided dice and you want to now the probability that they sum to 20.

You can visualize this as 20 pips getting divided into 5 bins, with no more than 6 pips going into any one bin, and every bin receiving at least one pip.

20 pips over 5 bins is with every bin getting at least 1 is ${20-1 \choose 5-1}$

Now we need to excluding the possibilities that there are more than 6 allocated to any 1 bin.

$-{5\choose 1}{13 \choose 4}$

We have over-corrected, and must count the cases with more than 6 going to 2 bins.

$+{5\choose 2}{7 \choose 4}$

${19 \choose 4} -{5\choose 1}{13 \choose 4} + {5\choose 2}{7 \choose 4} = 651$

and there are $6^5$ ways to roll $5$ six sided dice.

$\frac {651}{6^5}$

How to generalize this...

How many ways to roll $m, n$-sided dice, with sum equal to $y.$

Let $k = \lfloor \frac {y-m}n \rfloor$

$\sum_\limits{i=0}^{k} (-1)^i{m\choose i}{y-1 - in\choose m-1}$

Doug M
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