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Assuming following situation: I have a 10 sided dice (1-10). I´m allowed to roll the dice 10 times. For the first roll the proability to hit each number is the same 10%. For each roll consecutive roll the chance to roll a 5 is increased by 2%. Now i want to calculate the cumulative probability to hit the 5 exactly two times in 10 tries. I know how to calculate it if the probability to hit the 5 per roll stays constant, but how do i do this if the probability changes with each roll? For the first roll the chance to roll the 5 is 10/100, for the second roll the chance to hit the 5 is 12/100 and so on...

LordCatGod
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  • Welcome to math SE. have a look at mathjax to improve your mathematical expression. – Alain Remillard Jan 09 '20 at 19:35
  • You want the probability to have exactly two $5$ in $10$ rolls or the probability to have at least two $5$ in $10$ rolls? – Alain Remillard Jan 09 '20 at 19:37
  • @AlainRemillard I want to hit the 5 exactly two times. Sorry it wasn´t clear in my initial post, edited it. – LordCatGod Jan 09 '20 at 19:39
  • There are $45$ possibilities to have two $5$ with $10$ rolls, and I wonder if there is a better way than brute force to find the probability you are looking for. – Alain Remillard Jan 09 '20 at 19:44
  • @AlainRemillard First, thanks for your answer but excuse me i´m a bit lost. What do you mean with 45 possibilities or better saying how is this expressed in a percentage chance? – LordCatGod Jan 09 '20 at 19:47
  • Sorry I wasn't clear. The two $5$ could be on rolls $1$ and $2$ (first possibility), or rolls $1$ and $3$ (second possibility), $\ldots$, or on rolls $9$ and $10$ (forty-fifth possibility). There are $45$ different outcomes where you have two $5$. And brute force means evaluating each probability independently, then adding them up. – Alain Remillard Jan 09 '20 at 19:51
  • @AlainRemillard Ah understood. Thanks, now i understand what you meant with brute force! – LordCatGod Jan 09 '20 at 19:54
  • I did it with Excel and got $0.304759153\approx30.5%$ – Alain Remillard Jan 09 '20 at 19:58

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I hope there is a simplier answer than this. I decided to create a speadsheet to solve the problem by brute force.

There are $45$ different outcomes where there is two $5$ on $10$ rolls. Since the probability of getting a $5$ change with each roll of the die, $$P(\text{having $5$ on roll $i$}) = 0.1+0.02*(i-1)$$ I created a spreadsheet to evaluate the probability.

The number are

  • the first row is the number of the roll
  • the second row is the probability to have a $5$ on that roll
  • the first column is just the number of the possibility, it is not use in the computation
  • The second and third row are the two rolls where the $5$ appear. E.g. possibility $30$ (row $32$), the $5$ where on roll 4 and roll 10.
  • The cells D3 to M47 is the probability associated with the roll. For cell D3, the formula is =SI(OU(D$1=$C3;D$1=$B3);D$2;1-D$2) (my version of Excel is in french, formula might be slightly different for an english version)
  • Column N is the product of the probabilities of the 10 rolls
  • Finally, the green cell (N2) is the sum of all $45$ possibilities.

The answer is $0.304759152681615\approx 30.5\%$

enter image description here

Alain Remillard
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