k-man's kasino

[bonanova]

To celebrate the Grand Opening of his new casino, k-man offered the first 1000 patrons the opportunity to win some pocket change by playing a game that carried the catchy phrase Flip while you're a-Head. Each patron paid $100 to flip a coin multiple times, winning $100 for every H (heads) that appeared, and stopping (without penalty) at the first appearnace of a T (tails.)

What was k-man's expected cost for this gesture of good will?

 

A coveted bonanova gold star for an answer from the The Book.

[k-man]

Hah! Who would know that better than k-man?

 

there is no such thing as non-profit casino 

[gavinksong]

E(n) = expected number of coin tosses before a tails appears

E(n) = ∑ n * 2^{-n}

E(n) = 2 * E(n) - E(n) = 1 + ∑ 2^{-n} = 1 + 1 = 2

 

My guess is k-man loses $100, 000.

[gavinksong]

It turns out k-man has been using a two-faced coin all along!

AHAHAHAHAHAHAHA

[bonanova]

E(n) = expected number of coin tosses before a tails appears

E(n) = ∑ n * 2^{-n}

E(n) = 2 * E(n) - E(n) = 1 + ∑ 2^{-n} = 1 + 1 = 2

 

My guess is k-man loses $100, 000.

Does this account for all 1000 patrons?

[gavinksong]

 

E(n) = expected number of coin tosses before a tails appears

E(n) = ∑ n * 2^{-n}

E(n) = 2 * E(n) - E(n) = 1 + ∑ 2^{-n} = 1 + 1 = 2

 

My guess is k-man loses $100, 000.

Does this account for all 1000 patrons?

 

 

Ohhhhhh, wait. I missed the part where it says that each patron pays $100.

 

k-man doesn't expect to lose any money.

[k-man]

E(n) = expected number of coin tosses before a tails appears

E(n) = ∑ n * 2^{-n}

E(n) = 2 * E(n) - E(n) = 1 + ∑ 2^{-n} = 1 + 1 = 2

 

My guess is k-man loses $100, 000.

 

Your E(n) is correct and is 2, but if the tails appears on the second toss then the player breaks even, so expected gain/loss from each game is $0.

 

Outcome, Probability, Player's net gain

 

T, 1/2, -$100

HT, 1/4, $0

HHT, 1/8, $100

HHHT, 1/16, $200

HHHHT, 1/32, $300

......

 

The probability weighted net gain series can be expressed as 100(n-2)/2ⁿ, for n in (1 ... infinity). The series converges and the sum of the series is 0.

 

 

P.S. I see that you already recognized it.

Edited February 10, 2015 by k-man

[dgreening]

K-Man is right

 

 

The sequence converges to $0, but even after including the effects of 93
heads in a row the Kasino is still slightly ahead

[bonanova]

All correct conclusions. k-man was first, and his post was thoughtful enough to signal the answer without spoiling the puzzle. I'll mark it solved after giving some time to come up with a words-only solution, which Erdős would very likely agree belongs in The Book, and earn the gold star.

 

Specifically, neither the enumeration of cases nor an infinite sum, both of which give the correct answer, are needed. It will also show that dgreening's last comment does not quite apply to k-man's expected cost, which is exact.

[gavinksong]

I'll have a go.

 

Essentially, k-man earns $100 each time a tails is flipped, and he loses $100 each time a heads is flipped. Incidentally, the game stops once 1000 tails have been flipped, but that is irrelevant. Since both events have an equal chance of occurring, k-man neither expects to gain or lose any money.

Edited February 11, 2015 by gavinksong

[k-man]

I'll have a go.

 

Essentially, k-man earns $100 each time a tails is flipped, and he loses $100 each time a heads is flipped. Incidentally, the game stops once 1000 tails have been flipped, but that is irrelevant. Since both events have an equal chance of occurring, k-man neither expects to gain or lose any money.

 

Right on!

[bonanova]

Solved by k-man.

 

for gavinksong

 

Simply appeals to the fact that the coin is fair. Thus k-man's expected cost is the expected number of heads, which is the number of tails (1000), minus the number of bettors (also 1000.)