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I was thinking of a game where a player pays one dollar to toss a coin. If the result is a head they get their 1 dollar bet plus an additional 1 dollar. If the outcome is a tails they lose their 1 dollar bet and get nothing.

If n = 1 the fair price will be 0.5 dollars to play. But what happens if we want to make it n tosses? It becomes complicated because at each iterative step the player has to decide whether the current pay off is greater or less then the expected outcome of tossing another coin.

I simulated this once and got a fair price to be 0.76 dollars for n = 20 but I just cannot recall what my logic was.

Any suggestions will be appreciated.

The players pay off is the number of heads in n tosses.

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define "fair price" in "fair price will be 0.5" –  Roam Feb 15 at 7:36
    
if you were going to sell this opportunity to a unlimited number of punters if you charge 50c for each person to play in the long run you will break even ie. you would expect not to make money and not to lose money –  Seeking Alpha Feb 15 at 7:38
    
@SeekingAlpha How did you get 50c? If $n=1$, you lose one dollar with probability $\frac{1}{2}$ and win one dollar with probability $\frac{1}{2}$. –  JiK Feb 15 at 8:48
    
@JiK, $\frac 12 \cdot 1+\frac 12\cdot0=\$1=50 c$. –  Ragnar Feb 15 at 8:52
    
What do you earn after two tosses? Does one earn $1$ per head or only when all tosses are head? –  Ragnar Feb 15 at 8:54

1 Answer 1

The fair price for $n$ tosses is $n$ dollars.

JiK has shown for a single toss the fair price is one dollar. For $n$ tosses if the player pays a dollar for each toss then the expected profit is $$E[\mathrm{gain}]= {\left( { n \over 2} \right)} (+1) \ + \ {\left( { n \over 2} \right) } (-1)= 0,$$ where ${n \over 2}$ is the expected number of heads and the expected number of tails. So the fair price is $n.$

Intuitively, this makes sense. All tosses are independent, so if the fair price for a single toss is one dollar, then the fair price for $n$ tosses should be $n$ dollars.

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