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.