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 for any
value of n is 1/2.

Take it from there.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
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Jerry Avins said the following on 27/05/2006 22:45:
> 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 for any
> value of n is 1/2.
> 
> Take it from there.

A quick back of envelope summation hints to me that the answer is *a* 
*very* *large* *number* *indeed*?


-- 
Oli

Jerry Avins wrote:
> 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 for any
> value of n is 1/2.
> 
> Take it from there.

Jerry, you gave to many hints...

Anyway, I guess the (n-1) and (n) do not "match" really, do they?

I mean, summing up: (1/2)*1 + (1/4)*2 + (1/8)*4 + ... does not
stay too stable...

bye,

-- 

piergiorgio

Jerry Avins wrote:

> 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?

It's not bounded. However, in a certain sense, it is equal to -1/2.
Interested?

> What entry fee would make it a fair game?

Since the expected value is infinite, any price is too low to make it
fair.

> What would you pay?

That depends. How many times may I play? For a one off chance, this is
a complex psychological problem (sort of like the Prisoner's Dilemma).
If I can play repeatedly, the Law of Large Numbers should give a good
hint.

This reminds me of another probability puzzle (due to Mandelbrot, if I
remember correctly). Imagine an archer standing on the y axis and
facing an infinitely long wall along the x axis (Mandelbrot is a
mathematician:-). The archer is spun around blindfolded, which
effectively points him in a direction uniformly distributed between
-90=B0 and +90=B0 (while still facing the x axis from his position). He
then shoots his arrow. Two questions:
1=2E What is the expected x coordinate of the point of entry of the arrow
(remembering that the archer is standing on the y axis)?
2=2E What is the expected length of the flight path of the arrow?

(Hint: the two questions do not have the same answer).

Regards,
Andor

Jerry Avins wrote:

> 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 for any
> value of n is 1/2.
> 
> Take it from there.
> 
> Jerry

Hi Jerry.

Since the payout is 2^(n-1), a fair entry should be 2^(0-1), which is 1/2.
1/2 per toss! With an agreement to stop after the defined number of tosses.
A quick check reveals:

tosses fee   max_gain min_gain
1      0.5   1        0
2      1     2        0
3      1.5   4        0
4      2     8        0

However, if the number of tosses isn't limited, and a fixed entry fee has to
be paid in advance. Hmmm, I guess that there's no fair entry for this game.
It's a risk for both, and a chance. 
And I'd say there's no satisfying solution, since sum(2^(n-1)/2^n)) doesn't
converge.
Physical analogon: a moving sphere on a flat ground. How far will it roll?
Depends mainly on the friction.

Your problem lacks a sort of friction ...

Bernhard 

Andor wrote:

> Jerry Avins wrote:
> 
> 
>>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?
> 
> 
> It's not bounded. However, in a certain sense, it is equal to -1/2.
> Interested?
> 
> 
>>What entry fee would make it a fair game?
> 
> 
> Since the expected value is infinite, any price is too low to make it
> fair.
> 
> 
>>What would you pay?
> 
> 
> That depends. How many times may I play? For a one off chance, this is
> a complex psychological problem (sort of like the Prisoner's Dilemma).
> If I can play repeatedly, the Law of Large Numbers should give a good
> hint.

Try it and see. If a $64 payout would break the bank, how much would you
pay? $1,048,576? $1,099,511,627,776?

> This reminds me of another probability puzzle (due to Mandelbrot, if I
> remember correctly). Imagine an archer standing on the y axis and
> facing an infinitely long wall along the x axis (Mandelbrot is a
> mathematician:-). The archer is spun around blindfolded, which
> effectively points him in a direction uniformly distributed between
> -90� and +90� (while still facing the x axis from his position). He
> then shoots his arrow. Two questions:
> 1. What is the expected x coordinate of the point of entry of the arrow
> (remembering that the archer is standing on the y axis)?
> 2. What is the expected length of the flight path of the arrow?
> 
> (Hint: the two questions do not have the same answer).

Maybe next week.

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

Andor wrote:

> This reminds me of another probability puzzle (due to
> Mandelbrot, if I remember correctly). Imagine an archer standing
> on the y axis and facing an infinitely long wall along the x
> axis (Mandelbrot is a mathematician:-). The archer is spun
> around blindfolded, which effectively points him in a direction
> uniformly distributed between -90� and +90� (while still facing
> the x axis from his position). He then shoots his arrow.

Call the archer's position y, the entry point x, the angle a and the 
path length p. We have a = atan x/y = asec p/y.

> 1. What is the expected x coordinate of the point of entry of the
> arrow (remembering that the archer is standing on the y axis)?

So P(x) = P(a) da/dx = 1/(2 pi) y/(x^2 + y^2)
and E(x) = int -inf..inf x P(x) dx,
which has principal value zero since the integrand is odd.

> 2. What is the expected length of the flight path of the arrow? 

Here, P(p) = 1/(2 pi) y/p 1/sqrt(p^2 - y^2)
and E(p) = 2 int y..inf p P(p) dp = inf
because of the pole at p = y.


Martin

-- 
Quidquid latine scriptum sit, altum viditur.

Martin Eisenberg wrote:

>> 2. What is the expected length of the flight path of the arrow?
> 
> Here, P(p) = 1/(2 pi) y/p 1/sqrt(p^2 - y^2)
> and E(p) = 2 int y..inf p P(p) dp = inf
> because of the pole at p = y.

Actually, the pole in the integrand
is integrable but its tail is not.


Martin

-- 
Quidquid latine scriptum sit, altum viditur.

Martin Eisenberg wrote:
> Andor wrote:
>
> > This reminds me of another probability puzzle (due to
> > Mandelbrot, if I remember correctly). Imagine an archer standing
> > on the y axis and facing an infinitely long wall along the x
> > axis (Mandelbrot is a mathematician:-). The archer is spun
> > around blindfolded, which effectively points him in a direction
> > uniformly distributed between -90=B0 and +90=B0 (while still facing
> > the x axis from his position). He then shoots his arrow.
>
> Call the archer's position y, the entry point x, the angle a and the
> path length p. We have a =3D atan x/y =3D asec p/y.
>
> > 1. What is the expected x coordinate of the point of entry of the
> > arrow (remembering that the archer is standing on the y axis)?
>
> So P(x) =3D P(a) da/dx =3D 1/(2 pi) y/(x^2 + y^2)
> and E(x) =3D int -inf..inf x P(x) dx,
> which has principal value zero since the integrand is odd.

Martin,

I like the way you avoid the problem :-). What does this tell us about
the expected value of x?

>>> 2. What is the expected length of the flight path of the arrow?
>> Here, P(p) =3D 1/(2 pi) y/p 1/sqrt(p^2 - y^2)
>> and E(p) =3D 2 int y..inf p P(p) dp =3D inf
>> because of the pole at p =3D y.
>
> Actually, the pole in the integrand
> is integrable but its tail is not.

This is the reason why it reminded me of Jerry's (Daniel's) puzzle: the
expected value of the random variable L=3D"length of flight path" is
larger (with probability 1) than any value in the range of L. This is
very counter-intuitive of what we usually understand by the "mean".

Regards,
Andor.

Andor wrote:
> Martin Eisenberg wrote:
>> Andor wrote:

>> > 1. What is the expected x coordinate of the point of entry of
>> > the arrow (remembering that the archer is standing on the y
>> > axis)? 
>>
>> So P(x) = P(a) da/dx = 1/(2 pi) y/(x^2 + y^2)
>> and E(x) = int -inf..inf x P(x) dx,
>> which has principal value zero since the integrand is odd.
> 
> Martin,
> 
> I like the way you avoid the problem :-). What does this tell us
> about the expected value of x?

Ah well, it was worth a try ;)

>>>> 2. What is the expected length of the flight path of the
>>>> arrow? 
>>> Here, P(p) = 1/(2 pi) y/p 1/sqrt(p^2 - y^2)
>>> and E(p) = 2 int y..inf p P(p) dp = inf

The factor two slipped in the wrong line.

> This is the reason why it reminded me of Jerry's (Daniel's)
> puzzle: the expected value of the random variable L="length of
> flight path" is larger (with probability 1) than any value in
> the range of L. This is very counter-intuitive of what we
> usually understand by the "mean". 

Indeed. Is yours an old Basel family that you're on such good terms 
with Daniel?


Martin

-- 
What is it: is man only a blunder of God,
or God only a blunder of man?
--Friedrich Nietzsche