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Old 2007-10-24, 17:44   #1
davar55
 
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Default Coin Toss Game

A casino opens a new game using a fair (50-50) coin toss.
The player pays (bets) one chip to play. The coin is then tossed
repeatedly until one TAILS occurs. The player gets paid one chip for
each HEADS that occurs before that first TAILS. So for example
if TAILS happens on the first coin toss, the player just loses his
one chip bet; if one HEADS precedes the TAILS, the player comes
out even; if two HEADS precede the TAILS, the player comes out
one chip ahead; and so on.

The question is: is this a fair game, or does the house (or player)
have a monetary advantage (and if so, how much)?

Now in the corner of the casino is the high stakes version of the game,
with different payment rules. Here, the player pays M dollars to play,
but gets paid 2^N dollars if N HEADS occur before the first TAILS.

The question here is: for what value of M is this a fair game for the
player and the house?
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Old 2007-10-24, 19:36   #2
Mini-Geek
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Quote:
Originally Posted by davar55 View Post
A casino opens a new game using a fair (50-50) coin toss.
The player pays (bets) one chip to play. The coin is then tossed
repeatedly until one TAILS occurs. The player gets paid one chip for
each HEADS that occurs before that first TAILS. So for example
if TAILS happens on the first coin toss, the player just loses his
one chip bet; if one HEADS precedes the TAILS, the player comes
out even; if two HEADS precede the TAILS, the player comes out
one chip ahead; and so on.

The question is: is this a fair game, or does the house (or player)
have a monetary advantage (and if so, how much)?

Now in the corner of the casino is the high stakes version of the game,
with different payment rules. Here, the player pays M dollars to play,
but gets paid 2^N dollars if N HEADS occur before the first TAILS.

The question here is: for what value of M is this a fair game for the
player and the house?
I think with the first question, it is a fair game.
With the second, M = 2 seems to get the closest to being fair, but there does seem to be a slight favor in the player's direction.
I wrote a program to calculate the numbers.
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Old 2007-10-24, 22:57   #3
davieddy
 
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Quote:
Originally Posted by Mini-Geek View Post
I think with the first question, it is a fair game.
Expected return is:

1*(1/2)^2+2*(1/2)^3+3*(1/2)^4+........
This is the sum of GPs:
(1/2)*[(1/2)+(1/2)^2+(1/2)^3)+....]+
(1/2)^2*[(1/2)+(1/2)^2+(1/2)^3)+....]+
(1/2)^3*[(1/2)+(1/2)^2+(1/2)^3)+....]+
(1/2)^4*[(1/2)+(1/2)^2+(1/2)^3)+....]+
....

Now [(1/2)+(1/2)^2+(1/2)^3)+....]=1, so the expected return is 1



I agree the game is fair.

The second game has something to do with St Petersburg
IIRC

David

Last fiddled with by davieddy on 2007-10-24 at 23:52 Reason: I could have omitted brackets round (1/2) with the advantage of brevity
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Old 2007-10-25, 00:12   #4
Zeta-Flux
 
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The first game is "fair" but ultimately loses for the casino if players are smart. (Yes, the expected return is what you pay, but eventually with 100% probability, there is some point at which the casino will owe you more than you paid. So you stop at that point. Of course, this assume you have an infinite amount of money, and that you keep playing when you are VERY down. Although, with a fair coin, it is unlikely that you will have to get down too much before you are up at least once.)
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Old 2007-10-25, 00:41   #5
davieddy
 
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ZetaFlux I think you need to read the original problem
more carefully. That said your observations have some relevance
to the second problem.
In the first case, quitting before a tail turns up would be a silly
move (unless the bouncers were getting trigger happy)

Last fiddled with by davieddy on 2007-10-25 at 00:42
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Old 2007-10-25, 02:15   #6
Kevin
 
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I think the second version would be more fun if he got paid (2^N)/N dollars if it took N tosses to get a tail...
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Old 2007-10-25, 03:20   #7
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Quote:
Originally Posted by davieddy View Post
The second game has something to do with St Petersburg
IIRC

David
That the expected payoff of the second game is infinite, is known as the St. Petersburg Paradox.
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Old 2007-10-25, 09:08   #8
davieddy
 
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Quote:
Originally Posted by Kevin View Post
I think the second version would be more fun if he got paid (2^N)/N dollars if it took N tosses to get a tail...
Does this make for a finite expected return? I think not.
David
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Old 2007-10-25, 09:33   #9
davieddy
 
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Quote:
Originally Posted by Zeta-Flux View Post
The first game is "fair" but ultimately loses for the casino if players are smart. (Yes, the expected return is what you pay, but eventually with 100% probability, there is some point at which the casino will owe you more than you paid. So you stop at that point. Of course, this assume you have an infinite amount of money, and that you keep playing when you are VERY down. Although, with a fair coin, it is unlikely that you will have to get down too much before you are up at least once.)
I see what you are getting at now.
May I recommend "Gamblers Anonymous"?

David
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Old 2007-10-25, 23:09   #10
Zeta-Flux
 
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I do not gamble. So don't worry about me going out and spending all my money to win it back. ;)
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Old 2007-10-26, 11:17   #11
davieddy
 
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Quote:
Originally Posted by Zeta-Flux View Post
I do not gamble. So don't worry about me going out and spending all my money to win it back. ;)
But suggesting "smart " gamblers can beat the house isn't
helping the addicts one iota
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