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 2007-10-24, 17:44 #1 davar55     May 2004 New York City 2×2,099 Posts 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?
2007-10-24, 19:36   #2
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17·251 Posts

Quote:
 Originally Posted by davar55 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.

2007-10-24, 22:57   #3
davieddy

"Lucan"
Dec 2006
England

6,451 Posts

Quote:
 Originally Posted by Mini-Geek 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

 2007-10-25, 00:12 #4 Zeta-Flux     May 2003 2×761 Posts 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.)
 2007-10-25, 00:41 #5 davieddy     "Lucan" Dec 2006 England 6,451 Posts 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
 2007-10-25, 02:15 #6 Kevin     Aug 2002 Ann Arbor, MI 1B116 Posts 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...
2007-10-25, 03:20   #7
Mr. P-1

Jun 2003

7×167 Posts

Quote:
 Originally Posted by davieddy 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.

2007-10-25, 09:08   #8
davieddy

"Lucan"
Dec 2006
England

6,451 Posts

Quote:
 Originally Posted by Kevin 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

2007-10-25, 09:33   #9
davieddy

"Lucan"
Dec 2006
England

6,451 Posts

Quote:
 Originally Posted by Zeta-Flux 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

 2007-10-25, 23:09 #10 Zeta-Flux     May 2003 2×761 Posts I do not gamble. So don't worry about me going out and spending all my money to win it back. ;)
2007-10-26, 11:17   #11
davieddy

"Lucan"
Dec 2006
England

645110 Posts

Quote:
 Originally Posted by Zeta-Flux 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

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