Weekend Puzzle

This is an old puzzle, due to Daniel Bernoulli. It has a name I'll
withhold for a while. One plays this game: a fair coin is tossed until
it comes up tails, the payout being 2^(n-1), where n is the number of
tosses. What is the expected value of winnings assuming many games are
played? What entry fee would make it a fair game? What would you pay?

Observe that the probability of n tosses is 1/2^n, and that the payout
in that case is 2^(n-1). In other words, the expected payout is always 1/2.

Take it from there.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

the expected payout isn't 1/2!

the probability of n tosses isnt 1/2^n, but that of getting a head(or
tail) is.
the probab depends on the number of tosses(n) and is 1/2^(n-1)
if a player wants to play the game, he is assured of winning 1 buck, as
2^(1-1)=1
then, if a head appears, the minimum payout becomes 2, then 4 and so on

thus, the expected payout is
         =  1 + 1/2 * 2 + 1/2^2 * 2^2 + ...
         = inf
the mean value that can be won is itself infinity

however there is a 50% chance that a tail comes up in the first toss
itself and hence, if the entry fee is greater than 1(=a, say), there is
still a nonzero chance of making a profit
the mean profit = (a-1)*0.5 + (a-2)*(1/2)^2 + (a-4)*(1/2)^3 + ......

again a should be infinity, but practically, if we limit the number of
tosses to a fixed value e.g. 50, we can hope to limit a.

i would pay '1' to be assured of no loss

"Sukrut" <sukrut1729ramanujan@gmail.com> wrote in message 
news:1148891883.581321.118180@u72g2000cwu.googlegroups.com...
> the expected payout isn't 1/2!
>
> the probability of n tosses isnt 1/2^n, but that of getting a head(or
> tail) is.
> the probab depends on the number of tosses(n) and is 1/2^(n-1)
> if a player wants to play the game, he is assured of winning 1 buck, as
> 2^(1-1)=1
> then, if a head appears, the minimum payout becomes 2, then 4 and so on
>
> thus, the expected payout is
>         =  1 + 1/2 * 2 + 1/2^2 * 2^2 + ...
>         = inf
> the mean value that can be won is itself infinity
>
> however there is a 50% chance that a tail comes up in the first toss
> itself and hence, if the entry fee is greater than 1(=a, say), there is
> still a nonzero chance of making a profit
> the mean profit = (a-1)*0.5 + (a-2)*(1/2)^2 + (a-4)*(1/2)^3 + ......
>
> again a should be infinity, but practically, if we limit the number of
> tosses to a fixed value e.g. 50, we can hope to limit a.
>
> i would pay '1' to be assured of no loss
>
You are assured of getting something back for your bet but there is a 
vanishingly small probability of getting a really staggeringly high amount. 
Math looks really abusrdly complicated so I'll guess that if I stake 
anything less than 2*e^2 coins to get 2^(n-1) coins of equal value back I'd 
play; although I suspect that even if they were pennies I could go broke ( 
or one or the other of us might die) before I broke even or saw a profit.
Best of Luck - Mike

Sukrut wrote:

> the expected payout isn't 1/2!
> 
> the probability of n tosses isnt 1/2^n, but that of getting a head(or
> tail) is.
> the probab depends on the number of tosses(n) and is 1/2^(n-1)
> if a player wants to play the game, he is assured of winning 1 buck, as
> 2^(1-1)=1
> then, if a head appears, the minimum payout becomes 2, then 4 and so on
> 
> thus, the expected payout is
>          =  1 + 1/2 * 2 + 1/2^2 * 2^2 + ...
>          = inf
> the mean value that can be won is itself infinity

You get the right sum for the wrong reason. There a lot of ways to use
an ellipsis ("...") to reach infinity.

> however there is a 50% chance that a tail comes up in the first toss
> itself and hence, if the entry fee is greater than 1(=a, say), there is
> still a nonzero chance of making a profit
> the mean profit = (a-1)*0.5 + (a-2)*(1/2)^2 + (a-4)*(1/2)^3 + ......
> 
> again a should be infinity, but practically, if we limit the number of
> tosses to a fixed value e.g. 50, we can hope to limit a.
> 
> i would pay '1' to be assured of no loss
> 


-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

I think we can solve this by treating n is a random variable.

In this case n is a negative binomial random variable, with 1 sucess in
n trials.
The mean of the negative binomial in this case is 2.

So that:
What is the expected value of winnings assuming many games are
played?  2

What entry fee would make it a fair game?  2*(2^(2-1))=4

What would you pay? 
1 (I am cheap)

I think we can solve this by treating n is a random variable.

In this case n is a negative binomial random variable, with 1 sucess in
n trials.
The mean of the negative binomial in this case is 2.

So that:
What is the expected value of winnings assuming many games are
played?  2

What entry fee would make it a fair game?  2*(2^(2-1))=4

What would you pay? 
1 (I am cheap)

Ikaro wrote:

> I think we can solve this by treating n is a random variable.
> 
> In this case n is a negative binomial random variable, with 1 sucess in
> n trials.
> The mean of the negative binomial in this case is 2.
> 
> So that:
> What is the expected value of winnings assuming many games are
> played?  2
> 
> What entry fee would make it a fair game?  2*(2^(2-1))=4
> 
> What would you pay? 
> 1 (I am cheap)

The expected return on any bet is the sum of (the return for outcome n
times the probability of outcome n) over all n. It is therefore infinite
in the case of unlimited payout. If the payout is limited, the expected
value of a game is finite. Even for large limits, the expected value is
surprisingly small. http://en.wikipedia.org/wiki/St._Petersburg_paradox
has this table:

     Backer           Bankroll      Expected value of lottery
Friendly game               $64          $3.50
Millionaire          $1,050,000         $10.50
Billionaire      $1,075,000,000         $15.50
Bill Gates      $51,000,000,000 (2005)  $18.00
U.S. GDP        $11.7 trillion (2004)   $22.00
World GDP       $40.9 trillion (2004)   $23.00
Googolnaire     $10^100                $166.50

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

So it seems like a fair fee to charge would be an infinite amount as
well (since the expected value of winning is infinity).

Like Sukrut said, if the fee is less than or equal to one the player is
guaranteed a profit..

So it seems like a fair fee to charge would be an infinite amount as
well (since the expected value of winning is infinity).


And for the last question, like Sukrut suggested, if the fee is less
than one the player is 
guaranteed a profit...

So it seems like a fair fee to charge would be an infinite amount as
well (since the expected value of the return is infinity).

And for the last question, like Sukrut suggested, if the fee is less
than one the player is 
guaranteed a profit...