Grab & Guess

Probability Level 2

You friend challenges you to play "Grab & Guess" game. As the name suggests, he will grab a handful amount of coins from his money bag, and you'll guess whether the number of coins is odd or even.

Each correct guess will grant you 1 dollar before your friend puts the coins back and resumes the game, and you can continue playing infinitely as long as it's the right guess. On the other hand, the wrong guess will end the game immediately.

If your friend charges you 1 dollar as an entry fee, will this game earn or lose you money? (Assume there's no cheating!)

Earn Lose Neither earn nor lose Not enough information

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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 player fails at guessing, we can come up with the following scenarios (P for Pass & F for Fail):

F earns 0 dollars.

PF earns 1 dollar.

PPF earns 2 dollars.

PPPF earns 3 dollars.

Therefore, the probability distribution of a discrete random variable X X , where X X denotes a number of successful guesses, 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 .

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

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

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

From the geometric progression, we will get: S = 1 4 1 − 1 2 = 1 2 S = \dfrac{\frac{1}{4}}{1-\frac{1}{2}} = \dfrac{1}{2} .

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

From this game, the player's expected to earn 1 dollar, but since the entry fee is already 1 dollar. The player will neither earn or lose money from playing this game.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

I thought, it's 1 dollar entry per guess. Then he will always lose.

Siva Bathula - 4 years, 11 months ago

Hm, but that assumes that each liklihood of each number appearing (or at least each even-odd consecutive pair) is the same. I would argue that "it is more likely to grab a fewer number of coins", and so if he has to grab at least 1 coin, then odds are more likely. Whereas, if he is allowed to grab 0 coins, then even is more likely.

Calvin Lin Staff - 4 years, 11 months ago

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