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## Equitable Coin Flip Game

This seems to be a fairly simple probability question but it's stumping me for some reason.

"You flip a fair coin until you get a head and win x dollars, where x is the number of flips it takes to get a head. (e.g. H = win $1; TH = win$2; TTH = win \$3, and so on.) How much should you pay to play this game for it to be equitable (each player's expectation is equal to 0)?"

I'm really not sure if the answer is supposed to be a specific value or a variable, although my guess would be a specific dollar amount. I tried setting up a few equations like (1/2)(x)+(1/2)(L)=0 where x is the dollar amount and L is a lose, but that only gave me L=-x which didn't make any sense to me. I also tried setting it up as a series that produced x=0,-1,-1/2 which didn't seem to make sense either. I feel like I'm making it more complicated than it has to be, but I have no idea what to do.

Anyway, any help would be greatly appreciated. Thanks!

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Quote by royboyz12 					 				 				This seems to be a fairly simple probability question but it's stumping me for some reason.  "You flip a fair coin until you get a head and win x dollars, where x is the number of flips it takes to get a head. (e.g. H = win $1; TH = win$2; TTH = win \$3, and so on.) How much should you pay to play this game for it to be equitable (each player's expectation is equal to 0)?"  I'm really not sure if the answer is supposed to be a specific value or a variable, although my guess would be a specific dollar amount. I tried setting up a few equations like (1/2)(x)+(1/2)(L)=0 where x is the dollar amount and L is a lose, but that only gave me L=-x which didn't make any sense to me. I also tried setting it up as a series that produced x=0,-1,-1/2 which didn't seem to make sense either. I feel like I'm making it more complicated than it has to be, but I have no idea what to do.  Anyway, any help would be greatly appreciated. Thanks!
The number of flips taken to get a head is modelled by a geometric distribution with an obvious choice for its parameter (based on the fact that the coin is fair). The answer should be a specific value..

ok so then if I take the geometric distribution g(x;theta)=theta(1-theta)^(x-1) and plug in theta=(1/2), I get the geometric distribution is = (1/2)^x for x=1,2,3,...  So now where would the equitable part come in? Since its obvious that there isn't a solution to (1/2)^x=0, then would I let x=0, which gives a solution of 1? So would \$1 be the answer?

## Equitable Coin Flip Game

Nevermind...I spent well over an hour over-examining this, woke up this morning and it hit me...Forgot that I could just use the geometric distribution as a function to the find the expectation and then just set that equal to 0

Did you get \$2 as the answer?

No i didn't....I probably did it wrong then.  I set the geometric ditribution (1/2)^x equal to 0 and solved and got 0 as the answer...I figured that this made sense since the only way to have an expectation of 0 was to not play the game. Because when you play the game, the player is guaranteed to win money, but the probability of winning more money decreases over time. I seemed to make sense at the time.  How did you get \$2?

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Quote by royboyz12 					 				 				No i didn't....I probably did it wrong then.  I set the geometric ditribution (1/2)^x equal to 0 and solved and got 0 as the answer...I figured that this made sense since the only way to have an expectation of 0 was to not play the game. Because when you play the game, the player is guaranteed to win money, but the probability of winning more money decreases over time. I seemed to make sense at the time.  How did you get \$2?
You're not guaranteed to win money if it costs you money to play the game. If I pay ten bucks to play, and the flips come up TH, then I'm out 8 dollars. The question is how much should I be paying so that the expected amount that I make each game is zero

If I understand the problem correctly, the player of this game always gets a return of at least $1. To get an equitable price to pay, I used a weighted average. The possible returns are distributed as follows: P($0) = 0  P($1) = 0.5 [H] P($2) = 0.25 [TH] P($3) - 0.125 [TTH] etc. You simply sum these. A 50% chance of winning$1, a 25% chance of winning $2 etc. (.5*1)+(.25*2)+(.125*3)+(.0625*4)+(...) =$2  Each player's (profit) expectation is equal to zero as on average the player paying $2 to play this game will win$2 per game.  If you use your answer of zero, it would mean the (staking) player would win an average of $2 per game. The variable x is not related to the stake. He would be playing a riskless game where he wins money every time he plays. I have taken 'win amount' to mean 'return'. So that a payment of$2 and an outcome of H, would lead to a return of $1 (for a$1 loss overall). If he would literally win $1 (returning$2+\$1), the problem would not be solvable.

Ohhh ok I see what your doing. I was somewhat close when I originally summed (1/2)(win)+(1/2)(lose), but the infinite sum case never occurred to me. I guess I was kinda reading the problem a little wrong. That makes much more sense now. Thanks.

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coin, equitable, probability, statistics

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