--Msstaffo 17:51, 26 January 2009 (UTC) I will start by running a simulation of this game and discuss the probabilities of the first and most likely 5 (of the infinite) possible outcomes of a single run. The following table summarizes these first 5 possible outcomes of one turn playing the game.
1: Heads
Payout: $2
2: Tails Heads
Payout: $4
3: Tails Tails Heads
Payout: $8
4: Tails Tails Tails Heads
Payout: $16
5: Tails Tails Tails Tails Heads
Payout: $32
Notice that for these 5 possible outcomes, the run is terminated when the coin lands "Heads up."
Also notice that for each flip of the coin there are two possible outcomes, Heads or Tails. So each time the coin is flipped there is a 50% chance that the game will end.
Lets look at our 5 simulations of possible outcomes for one turn of the game: Attempt 1: There is a 50% chance for the coin to land on Heads, and a 50% chance for the coin to land tails. So this result of a single coin toss landing heads and terminating the run will occur for 50% of the runs. Attempt 2: Once again there is a 50% chance for the coin to land heads and a 50% chance for the coin to land tails for each toss. So this result will occur (.5)(.5)=(.25) or 25% of the turns. Attempt 3: " (.5)(.5)(.5)=.125 or 12.5% of the turns. Attempt 4: " (.5)(.5)(.5)(.5)=.0625 or 6.25% of the turns. Attempt 5: " (.5)(.5)(.5)(.5)(.5)=.03125 or 3.125% of the turns.
The reason each individual toss' possibility to land on a certain side (50%) is multiplied with each subsequent toss is because in order to make it to each next tier, (worded loosely) you have to "luck out" and pass each previous 50/50 chance, so in essence you are "using up" your chance/luck with each toss leaving you less likely to roll a tails and move on to the next toss. In case you noticed the pattern, we can use the experiment we created above to create a type of formula for the possible results of this game:
Let n equal the total number of tosses of the coin up to and including the heads toss that terminates the turn, then
there is a (.5)^n chance for a payout of $2^n.
Looking at this logically (I'll let someone who has done limits more recently explain this in more detail) we see that the chance of flipping the coin n times as n approaches infinite converges to zero, (and the payout diverges to $infinite). Long story short, the higher the payout the lower the chance to obtain it.
Side note: I think that the question asks what is the -most- money you could bet and still win a profit from the game or at least have a statistically probable chance to break even. IE: If I bet $10 am I likely to win over $10 or at least break even? If I bet $100? $1000? Note that the amount of money bet at the beginning has absolutely no effect on the outcome of the coin toss(es). That said, for you realists out there, the most logical move (if you were seeking a profit from this game) would be to bet an amount less than or equal to $2 so as to never do any worse than break even with the worst possible (and most likely) scenario, a single heads toss.
According to the statistics and probabilities we have discussed so far, as the number of turns playing this game approaches infinite, a portion of the results approaching one half of the total turns will yield $2. One quarter of the turns will yield $4, one eighth will yield 8 dollars, etc.
So there is a 50% chance to win 2 dollars, a 25% chance to win 4 dollars, a 12.5% chance to win 8 dollars, a 6.25% chance to win 16 dollars, a 3.125% chance to win 32 dollars, ... , and a (.5)^n chance to win 2^n dollars, PER TURN.
Therefore, in order to sum up our winnings for an infinite amount of turns playing this game, we take
The summation as n approaches infinite of [.5n(2)+.25n(4)+.125n(8)+.0625n(16)+.03125n(32)+...+(.5^n)(n)(2^n)]
which can also be written as
The summation as n approaches infinite of [(1/2)n(2)+(1/4)n(4)+(1/8)n(8)+(1/16)n(16)+(1/32)n(32)+...+((1/2)^n)n(2^n)]
which can also be written as
The summation as subscript K=N approaches infinite of Nsub1+Nsub2+Nsub3+Nsub4+Nsub5+...+NsubK.
So we have "K amount of N's", where K=N, so we have "N amount of N's" or (N)(N) or N^2.
What does this mean?
This means that as we attempt this game an infinite amount of times our winnings, statistically, total up to the square of the number of times that we play the game.
Because some information was omitted from the original description of the game I will state the solution for both possibilities:
Solution for Possibility #1: You only have to pay one time to take as many turns flipping the coin as you would like. So, pay once and play infinite times.
The maximum amount of money that I would pay to play this game and expect to at least break even would be the square of the amount of turns I planned to play the game.
Pay N^2 dollars once then play N turns in order to win N^2 dollars and break even.
Amount paid = N^2 = amount won after playing N turns.
Solution for Possibility #2: You have to pay each time before you begin flipping the coin. So, pay then flip the coin until it lands on heads and terminates the turn, then pay again and flip the coin for your second turn until it lands on heads, then pay again for your third turn, etc, then
The maximum amount of money that I would pay to play this game and expect to at least break even would be equal to the amount of turns I planned to play the game.
Pay N dollars every time I play a turn and then play N turns in order to win N^2 dollars
(N)(N)=N^2
N^2=N^2
Once again, thanks to the original poster for an interesting puzzle. --Msstaffo 17:51, 26 January 2009 (UTC)
