coin tosses that change their possibilities according to the previous toss

Question

Let's say I have a coin that when tossed has a probability of 50% getting heads, and 50% getting tails. But for the next toss, the side that "won" gets half the chances that the side that "lost" had, so for example, if heads "won", in the next toss the probability will be 75% for heads and 25% for tails. According to this, how do I calculate the possibilities of things like: In ten iterations, what is the possibility that heads will be picked exactly in 7 of those (or even at least in 7 of those)? And how do I calculate a more general case where the probability in the start is not 50-50, but can be 70-30, 20-80, 5-95, and so on?

Answer 1

Code a Heads with $1$ and a Tails with $0$. Let $X_1,X_2,\ldots$ be (dependent) Bernoulli random variables modeling the tosses. Say that initially $P(X_1=1)=p$. The conditional distribution of $X_n$ given the past is as follows:

$$X_n|(X_{n-1}=e_{n-1},\ldots ,X_1=e_1) \sim \text{Bernoulli}\Big(\frac 1{2^{n-1}}(p+\sum_{k=1}^{n-1}2^{k-1}e_k)\Big),$$ and this is easily shown via induction.

This essentially gives you the joint distribution of $(X_1,\ldots,X_n)$ which is enough to compute any of the coveted probabilities.

For instance, by the multiplication rule, $$\begin{aligned} P(\sum_{k=1}^{10} X_k = 7) = \sum_{\substack{(e_1,\ldots,e_{10})\in \{0,1\}^{10}\\e_1+\ldots+e_{10}=7}} P(X_n=e_{10}|X_{9}=e_{9},\ldots ,X_1=e_1)\times\ldots\times P(X_1=e_1) \end{aligned} $$ and this can be computed in closed form, or implemented on a computer.