You play a coin flipping game with 5 coins. On round 1 you flip all of them. On round 2, you pick up all the ones that came up tails (leaving all the heads alone) and flip them again. You continue to do this until all the coins are heads. For example:
Round 1: H T T H T
Round 2: - H T - H
Round 3: - - T - -
Round 4: - - T - -
Round 5: - - H - -
Done in 5 Rounds.
What is the expected number of rounds you'll need to finish the game?
What is the probability you will finish the game in 3 rounds or less?
Please see my next comment - there is an error in this reasoning.
Let f(n) be the expected number of rounds to finish the game when starting with n coins. We seek f(5). Note that
f(n)=1 + (p(0,n)*f(n)+p(1,n)*f(n-1)+...+p(n-1,n)*f(1)+p(n,n)*f(0)),
where p(i,j) is the probability of getting i heads out of j tosses. Using p(i,j) = (j choose i)/(2^j), and f(0)=0, we can recursively compute f(1),f(2),...
I probably made a mistake (doing it by hand), but here's what I got:
f(1)=2
f(2)=8/3
f(3)=22/7
f(4)=368/105
f(5)=2270/651
Edited on October 13, 2004, 5:30 pm
Solution to First Question (high chance of error)
