Infinity Game

Probability Level 2

You're invited to play a special coin-flipping game with a fairly simple rule: heads earn money; tails end party.

If you flip head at the first try, you'll be rewarded 1 dollar. If you flip another head at the second try, you'll be rewarded 2 more dollars. If you flip another consecutive head at the third try, you'll be rewarded 3 more dollars, and so on.

As long as the coin continuously turns up head, you can keep playing as many times as you like, but whenever it's tail face, the game is over immediately. (Therefore, it's also possible for you to gain nothing if you flip tail at your first try.)

Assuming the coin is unbiased, what is the expected earnings (in dollars) from this game?


The answer is 2.

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Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

Since the question indicates that the game would end if the tail face occurs, we can come up with the following scenarios (T for Tail & H for Head):

T earns 0 dollars.

HT earns 1 dollar.

HHT earns 1 + 2 = 3 dollars.

HHHT earns 1 + 2 + 3 = 6 dollars.

Therefore, the probability distribution of a discrete random variable X X , where X X denotes a number of successful consecutive heads, is as follows:

P ( X = 0 ) = 1 2 P(X = 0) = \dfrac{1}{2} P ( X = 1 ) = ( 1 2 ) 2 P(X = 1) = (\dfrac{1}{2})^2 P ( X = 2 ) = ( 1 2 ) 3 P(X = 2) = (\dfrac{1}{2})^3 P ( X = 3 ) = ( 1 2 ) 4 P(X = 3) = (\dfrac{1}{2})^4 ⋮ \vdots P ( X = n ) = ( 1 2 ) n + 1 P(X = n) = (\dfrac{1}{2})^{n+1}

The expected earn E E from the accumulative heads = n ( n + 1 ) 2 \dfrac{n(n+1)}{2} .

Therefore, the expected earn from this game = ∑ i = 0 ∞ P E \sum_{i=0}^\infty PE

= ∑ i = 0 ∞ ( n ( n + 1 ) 2 ) ( ( 1 2 ) n + 1 ) \sum_{i=0}^\infty (\dfrac{n(n+1)}{2})((\dfrac{1}{2})^{n+1})

= 0 × 1 2 + 1 × ( 1 2 ) 2 + ( 1 + 2 ) × ( 1 2 ) 3 + ( 1 + 2 + 3 ) × ( 1 2 ) 4 + . . . 0\times \dfrac{1}{2} + 1\times (\dfrac{1}{2})^2 + (1+2)\times (\dfrac{1}{2})^3 + (1+2 + 3)\times (\dfrac{1}{2})^4 + ...

= [ 1 × ( 1 2 ) 2 + ( 2 ) × ( 1 2 ) 3 + ( 3 ) × ( 1 2 ) 4 + . . . ] + [ ( 1 ) × ( 1 2 ) 3 + ( 2 ) × ( 1 2 ) 4 + . . . ] + [ ( 1 ) × ( 1 2 ) 4 + . . . ] [1\times (\dfrac{1}{2})^2 + (2)\times (\dfrac{1}{2})^3 + (3)\times (\dfrac{1}{2})^4 + ...] + [(1)\times (\dfrac{1}{2})^3 + (2)\times (\dfrac{1}{2})^4 + ...] + [(1)\times (\dfrac{1}{2})^4 + ...]

Suppose S = 1 × ( 1 2 ) 2 + ( 2 ) × ( 1 2 ) 3 + ( 3 ) × ( 1 2 ) 4 + . . . S = {1\times (\dfrac{1}{2})^2 + (2)\times (\dfrac{1}{2})^3 + (3)\times (\dfrac{1}{2})^4 + ...} .

Then, by using geometric progression sum, ∑ i = 0 ∞ P E = S ( 1 + 1 2 + 1 4 + . . . ) = 2 S \sum_{i=0}^\infty PE = S(1 + \dfrac{1}{2} + \dfrac{1}{4} + ...) = 2S .

Now S = 1 × ( 1 2 ) 2 + ( 2 ) × ( 1 2 ) 3 + ( 3 ) × ( 1 2 ) 4 + . . . S = {1\times (\dfrac{1}{2})^2 + (2)\times (\dfrac{1}{2})^3 + (3)\times (\dfrac{1}{2})^4 + ...}

= [ 1 × ( 1 2 ) 2 + ( 1 ) × ( 1 2 ) 3 + ( 1 ) × ( 1 2 ) 4 + . . . ] + [ 1 × ( 1 2 ) 3 + ( 1 ) × ( 1 2 ) 4 + . . . ] + [ 1 × ( 1 2 ) 4 . . . ] [1\times (\dfrac{1}{2})^2 + (1)\times (\dfrac{1}{2})^3 + (1)\times (\dfrac{1}{2})^4 + ...] + [1\times (\dfrac{1}{2})^3 + (1)\times (\dfrac{1}{2})^4 + ...] + [1\times (\dfrac{1}{2})^4...] .

Let T = 1 × ( 1 2 ) 2 + ( 1 ) × ( 1 2 ) 3 + ( 1 ) × ( 1 2 ) 4 + . . . = 1 2 T = {1\times (\dfrac{1}{2})^2 + (1)\times (\dfrac{1}{2})^3 + (1)\times (\dfrac{1}{2})^4 + ...} = \dfrac{1}{2} .

Then S = T ( 1 + 1 2 + 1 4 + . . . ) = ( 1 2 ) × 2 = 1 S = T(1 + \dfrac{1}{2} + \dfrac{1}{4} + ...) = (\dfrac{1}{2})\times 2 = 1 .

Thus, the expected earn = 2 S = 2 2S = \boxed{2} .

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