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There are two variables $a$ and $b$.
(The first value of variables : $a=1,b=0$)
We will add $1$ to $a$ or $b$ for $n$ times. ($n \in \mathbb N$)
The probability of adding to each variables are : $$P_a = k({a \over a+b})+(1-k)({b \over a+b})$$ $$P_b = k({b \over a+b})+(1-k)({a \over a+b})$$ ($P_a$ : probability of add 1 to a, $P_b$ : probability of add 1 to b)
$k$ is constant number in $(0,1)$.

I want to find $P(a=1)$,$P(a=2)$,...,$P(a=n)$.
I think this discrete probability distribution is similar with binomial distribution. However, each sequences are not independent so I am really confusing what this probability distribution is.

sunghwa Kang
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  • You can get properly sized parentheses that adjust to their content by preceding them with \left and \right. – joriki Sep 01 '18 at 06:24
  • You can view this as an urn model. You start out with one white ball and no black balls. In each step, you draw a ball; with probability $k$ you add a ball of the same colour and with probability $1-k$ you add a ball of the opposite colour. – joriki Sep 01 '18 at 06:28

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