TITLE: A coin is weighted, such that the probability of heads in any given toss is twice that of tails.
QUESTION [5 upvotes]: A coin is weighted, such that the probability of heads in any given toss is twice that of tails. A player tosses two such coins. The player wins £4 if 2 tails occur and £1 if 1 tail occurs. The player should lose less than what amount of money if no tails occur for the game to be favourable to the player? 
What I have done so far:
$P(H) + P(T) = 1$
$P(H) = 2P(T)$
$1 = 2P(T) + P(T)$
$P(T) = 1/3$
$2P(T) = P(H) = 2/3$

REPLY [3 votes]: Let's calculate the expected value:
$$\frac19\cdot4+2\cdot\frac29\cdot1+\frac49\cdot x>0$$
where the three terms refer to the events of 2, 1 and 0 tails respectively. Then
$$\frac49+\frac49+\frac49\cdot x>0$$
$$\frac49\cdot x>-\frac89$$
$$x>-2$$
Thus the player should lose less than £2 for the game to be favourable to them.