Coin Flip Game
January 31st, 2014 at 1:47:42 PM
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Say you were offered a coin flip game where the pot starts at $1 and if you win the pot doubles and if you win again the pot doubles again and it doubles each time you win. And you win the amount of the pot that existed the last time you win.
For example the pot goes from $1,$2,$4,$8,$16,$32 etc.
ie if you win 3 times and then lose you win $4
if if you win 4 times and then lose you win $8.
There is no limit to how much the pot will get to.
How much are you willing to pay to have a go at this opportunity?
For example the pot goes from $1,$2,$4,$8,$16,$32 etc.
ie if you win 3 times and then lose you win $4
if if you win 4 times and then lose you win $8.
There is no limit to how much the pot will get to.
How much are you willing to pay to have a go at this opportunity?
January 31st, 2014 at 1:49:54 PM
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$undefined
January 31st, 2014 at 2:01:46 PM
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Quote: AceTwoSay you were offered a coin flip game where the pot starts at $1 and if you win the pot doubles and if you win again the pot doubles again and it doubles each time you win. And you win the amount of the pot that existed the last time you win.
For example the pot goes from $1,$2,$4,$8,$16,$32 etc.
ie if you win 3 times and then lose you win $4
if if you win 4 times and then lose you win $8.
There is no limit to how much the pot will get to.
How much are you willing to pay to have a go at this opportunity?
you can buy this for your website for cheap and it connects to many payment processors. just google "coin flip game" but there are so many out there, not many people draw interest.
http://www.vietsol.com/products/vsflip.php
8 more years till retirement.
January 31st, 2014 at 3:11:10 PM
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Quote: AceTwoSay you were offered a coin flip game where the pot starts at $1 and if you win the pot doubles and if you win again the pot doubles again and it doubles each time you win. And you win the amount of the pot that existed the last time you win.
For example the pot goes from $1,$2,$4,$8,$16,$32 etc.
ie if you win 3 times and then lose you win $4
if if you win 4 times and then lose you win $8.
There is no limit to how much the pot will get to.
How much are you willing to pay to have a go at this opportunity?
If you paid $2 to play then, if my calculations are correct, this would be a break even game for you.
Imagine playing the game 8 times:-
You win 4 and you lose 4. The outcomes are:-
HHH
HHT
HTH
THH
TTT
TTH
THT
HTT
If you win with 'Heads' then you win $1, $1, $2 and $4 for the winning outcomes above.
So, if the 4 losing outcomes total $8 then you are at break even.
January 31st, 2014 at 5:54:36 PM
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Most of the cash in my wallet (~$100).
January 31st, 2014 at 6:11:58 PM
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Say you always calls Heads.
Using Geoff's plans
HHH (1 perm) You're still in the game with the pot starting at £8
HHT (1 perm) Losing on the third call - so it doubled to 2, then doubled to 4, then you won the pot
HTx (2 perms) Losing on the second call - so it doubled to 2, then you won the pot
Txx (4 perms) Losing on the first call - so you just won £1
8 results paid (4*1)+(2*2)+(1*4)+1*"still in game"
Thus the average result X = (4+4+4+8*X)/8 = 12/8 + X
There is no solution to X = 1.5+X !
I think the mistake is assuming that "HHH" is only worth £8 when in fact it is the same as starting the game from the beginning but with 8 times as much in the pot.
Another way of looking at it is
You win £1 if you lose on the first round: P=1/2 so contributes £1*.5 = 50p to the expected value.
You win £2 if you win ONE and lose on the 2nd round : P=1/4 so that also contributes £2*(1/4) = 50p to the expected value.
You win £4 if you win TWO and lose on the 3rd round : P=1/8 so that also contributes £4*(1/8) = 50p to the expected value.
This series continues for ever, hence the expected value is infinite.
Using Geoff's plans
HHH (1 perm) You're still in the game with the pot starting at £8
HHT (1 perm) Losing on the third call - so it doubled to 2, then doubled to 4, then you won the pot
HTx (2 perms) Losing on the second call - so it doubled to 2, then you won the pot
Txx (4 perms) Losing on the first call - so you just won £1
8 results paid (4*1)+(2*2)+(1*4)+1*"still in game"
Thus the average result X = (4+4+4+8*X)/8 = 12/8 + X
There is no solution to X = 1.5+X !
I think the mistake is assuming that "HHH" is only worth £8 when in fact it is the same as starting the game from the beginning but with 8 times as much in the pot.
Another way of looking at it is
You win £1 if you lose on the first round: P=1/2 so contributes £1*.5 = 50p to the expected value.
You win £2 if you win ONE and lose on the 2nd round : P=1/4 so that also contributes £2*(1/4) = 50p to the expected value.
You win £4 if you win TWO and lose on the 3rd round : P=1/8 so that also contributes £4*(1/8) = 50p to the expected value.
This series continues for ever, hence the expected value is infinite.
January 31st, 2014 at 6:29:32 PM
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The EV of the game is infinite.
0*1/2+1*1/4+2*1/8+4*1/16... = 0+1/4+1/4+1/4+... = +infinity
However, the diminishing returns of the value of money means that the utility of the game need not be infinite as well. Hence, I am willing to lose a decent amount of money, but not so much as I would really worry about losing on a one time basis.
Another interesting question is what would you pay if you could play the game as many times as you wanted? However, you need to bring sufficent cash with you to cover each time you will play. Winnings can be used to get further games, but if you go bust, you must quit and can not play again. What would you pay to play and how much cash would you bring?
0*1/2+1*1/4+2*1/8+4*1/16... = 0+1/4+1/4+1/4+... = +infinity
However, the diminishing returns of the value of money means that the utility of the game need not be infinite as well. Hence, I am willing to lose a decent amount of money, but not so much as I would really worry about losing on a one time basis.
Another interesting question is what would you pay if you could play the game as many times as you wanted? However, you need to bring sufficent cash with you to cover each time you will play. Winnings can be used to get further games, but if you go bust, you must quit and can not play again. What would you pay to play and how much cash would you bring?
January 31st, 2014 at 7:29:52 PM
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I stand by my original answer.
February 1st, 2014 at 6:08:52 AM
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You wouldn't even pay less than $1 ?
February 3rd, 2014 at 11:38:12 AM
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The mathematical answer in terms of Ev is Infinite. Each round gives a $0.25 of addititional EV.
But most people intuitively (meaning that they do not understand the math) would not pay a lot of money to have the opportunity to play such game.
Even though the Ev is infinite and that at any payment of cost you have positive Ev, the utility considerations come into effect.
And of course the practical considerations of the other side honoring the bet.
After 11 wins the Pot becomes around $1.000
After 21 wins the Pot becomes around $1 million
After 31 wins the pot becomes around $1 billion
After 41 wins the pot becomes around $1 trillion
So if the other side could pay you max $1 trillion the Ev = 41 x $0.25 = $10.
So even though the Ev for the game as proposed is Infinite, practical considerations make it a small number.
This problem is called the St. Petersburg paradox.
http://en.wikipedia.org/wiki/St._Petersburg_paradox
But most people intuitively (meaning that they do not understand the math) would not pay a lot of money to have the opportunity to play such game.
Even though the Ev is infinite and that at any payment of cost you have positive Ev, the utility considerations come into effect.
And of course the practical considerations of the other side honoring the bet.
After 11 wins the Pot becomes around $1.000
After 21 wins the Pot becomes around $1 million
After 31 wins the pot becomes around $1 billion
After 41 wins the pot becomes around $1 trillion
So if the other side could pay you max $1 trillion the Ev = 41 x $0.25 = $10.
So even though the Ev for the game as proposed is Infinite, practical considerations make it a small number.
This problem is called the St. Petersburg paradox.
http://en.wikipedia.org/wiki/St._Petersburg_paradox
