One-player game using coin with head probability $p$. Player tosses a coin and increment score if head, else decrement. Find probability of score $2$.

Question

Consider a game by a single player using a coin with head probability $p\; (0 < p < 1)$. The player starts the game with her initial score $0$, and repeats a trial (called round) in which she tosses a coin and increases her score by one if head, otherwise decreases her score by one. The game is over with her win once her score gets to $2$, while the game is over with her defeat once her score gets to $—2$. Answer the following questions.

1. For each $k=1, 2, 3$, answer the probability that she wins the game exactly at the $2k$-th round and the probability that she looses the game exactly at the $2k$-th round, respectively.
2. Answer the probability that she wins the game.
3. Answer the expectation and the variance of the number of times that she wins when she repeats the game $100$ times.
4. Let $p = 0.6$. Establish with reason whether or not the probability is higher than $1/10$ that the number of times she wins is less than $50$ when she repeats the game $100$ times.

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Here is my attempt.

Can someone please check if the approach is correct, especially for 2 and 4 part?

Answer 1

The probability of winning (resp. losing) at the $2k^{\mathsf{th}}$ round is is $2^{k-1}p^2(p(1-p))^{k-1}$ (resp. $2^{k-1}(1-p)^2(p(1-p))^{k-1}$). This follows from a win (resp. loss) at the $2k^{\mathsf{th}}$ round consisting of two heads (resp. tails) and $k$ heads/tails pairs.

Letting $\tau_i$ be the probability of winning given that the current score is $i$, for $i\in\{-2,-1,0,1,2\}$, we have the recurrence $\tau_i= p\tau_{i+1} + (1-p)\tau_{i-1}$, and with the boundary conditions $\tau_2=1, \tau_{-2}=0$, it follows that $$ \tau_0 = \frac{p^2}{1-2 p+2p^2}. $$ If the game is repeated $n$ times, then the distribution of the number of wins is binomially distributed with parameters $n$ and $\tau_0$, and hence has mean $\frac{np^2}{1-2 p+2p^2}$ and variance $n\left(\frac{p(1-p) }{1-2p+2p^2}\right)^2$.

If $p=3/5$ then the probability of winning less than 50 times when the game is played $100$ times is \begin{align} &\sum _{k=0}^{49} \left(\frac{\left(\frac{3}{5}\right)^2}{\frac{2}{5} 3 \left(\frac{3}{5}-1\right)+1}\right)^k \left(1-\frac{\left(\frac{3}{5}\right)^2}{\frac{2}{5} 3 \left(\frac{3}{5}-1\right)+1}\right)^{100-k} \binom{100}{k}\\ =& \tiny\frac{3733426269496958096470429118425665418368735163407238068404578920487073841498634546664688980706863475589120}{190718085458920964116236375748835779710674959067303165370168392262012207679844273858329666379998629245551661077}\\ &\approx0.0000195756\\ &\ll\frac1{10}. \end{align}