Expected number of coin flips before a player's total score reaches $3$

Question

there is a simple game using number of coins. initially, player has score 0 and is given three number of coins. then he flips all his coins at once. the number of coins showed head will be added to his score. if player scored 1, he will do the next flip, with only 2 coins, scoring rules are same. if player scored 2, he will do the next flip with only one coin and scoring rules are same. if player scored 3, he won. ... so, a player with score s will flip (3-s) coins, and with number of coins showed head being added to score, until his total score reaches 3. what is expected number of flips in this game?

I think this 'expected number of flips' is confusing me.

let's say his current score is 2. he can score 1 with probability of 0.5. therefore expected number of flips to win his game is 2.

let's say his current score is 1. he can score 1 with probability of 0.5 and score 2 with probability of 0.25. Now it's where I'm stuck. if he wants to score 1+1 and win the game, the probability is 0.5*0.5 - which means 4 more turns are expected to win the game in such scoring combination. if he wants to score 2 and win the game, the probability is 0.25, which means also 4 more turns are expected to win the game by flipping two coins and scoring two heads. I don't really know what else to do here.

Answer 1

The following is a general-purpose approach. At any time, our score (State) is $0$, $1$, $2$, or $3$. Let $a$ be the expected number of additional rounds needed if we are in State $0$, $b$ be the expected number of additional rounds needed if we are in State $1$, and $c$ the expected number of additional rounds needed if we are in State $2$. We only want to find $a$, but $b$ and $c$ will be useful.

Suppose we are in State $0$. Flip. This has a cost of $1$ (round). With probability $\frac{1}{8}$ we get $0$ heads. Then we stay in State $0$, and our expected number of additional rounds needed is $a$. With probability $\dfrac{3}{8}$ we get $1$ head, and enter State $1$, and our expected number of additional rounds becomes $b$. With probability $\dfrac{3}{8}$ we get $2$ heads, and enter State $2$, with expected additional rounds needed equal to $c$. Finally, with probability $\dfrac{1}{8}$ the game is over, no additional rounds needed. We have obtained the equation $$a=1+\frac{a}{8}+\frac{3b}{8}+\frac{3c}{8}.$$

Suppose we are in State $1$. Flip (cost, $1$ round). With probability $\dfrac{1}{4}$, we remain in State $1$, and the expected number of additional rounds needed is $b$. With probability $\dfrac{2}{4}$ we enter State $2$, and the expected number of additional rounds needed is $c$. With probability $\dfrac{1}{4}$ the game is over. So $$b=1+\frac{b}{4}+\frac{2c}{4}.$$

Finally, suppose we are in State $c$. The same reasoning shows that $$c=1+\frac{c}{2}.$$

It is almost obvious that $a$, $b$, and $c$ are finite. Under the assumption that they are, to find $a$, $b$, and $c$ we only need to solve a simple system of $3$ linear equations in $3$ unknowns.