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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:

$P(H) + P(T) = 1$

$P(H) = 2P(T)$

$1 = 2P(T) + P(T)$

$P(T) = 1/3$

$2P(T) = P(H) = 2/3$

Parcly Taxel
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  • Yes, if " the probability of heads in any given toss is twice that of tails" then th probability of heads is 2/3 and the probability of tails is 1/3. From that, the probability of two tails is (1/3)(1/3)= 1/9. The probability of one head and one tail, in either order, is 2(2/3)(1/3)= 4/9. The probability of two heads (no tails) is (2/3)(2/3)= 4/9. (Check that 1/9+ 4/9+ 4/9= 9/9= 1.) If you gain £4 when you get two tails, £1 when you get one tail, and lose £A when you get no tails, then you average winnings would be 4(1/9)+ 1(4/9)- A(4/9)= (8- 4A)/9. Solve (8- 4A)/9= 0 for A. – user247327 Nov 02 '18 at 11:21

1 Answers1

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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.

Parcly Taxel
  • 103,344