Having looked at many of the posts regarding this puzzle, first posed by
Nicholas Bernouli (not Daniel!), I'm a little confused as to how an
infinite expected value is arrived at.

The problem, simplified, is to find the willing price for playing the
game of flipping a coin until the first tails lands up. Each contiguous
flip (starting from the begining) landing heads pays the player $2^n.

The calculation of Expected value is the SUM $(2^n)/(1/2^n) = $1 + $1 ...
until the game finishes.

Many explanations of the game then simply state that the expected value
is infinite because the sum of 1's goes on forever (or could go on
forever?).

The problem I am having with this explanation is that the game has an
undefined state. If the coin flips to heads every time forever, no payout
is ever made. A payout is only made when the first tails appears. Thus
every defined state of the game requires a finite series of flips. Thus
it will never be the case that an infinite payout will occur.

Can someone explain how a finite series of flips results in an infinite
payout? Or how an infinite series of flips becomes a defined state?

The only explanation I can come up with (though it is unsatisfactory to
me) is that we take the coin flipping process as a partial recursive
function and enumerate an infinite number of games by the length of
contiguous heads as each game ends (throwing away repeat lengths). Thus
we will eventually hit every natural number in the series as we go,
(unless the function went on forever--but then the process wouldn't be
random would it?). Since the natural numbers are infinite so is the
expected value? I can't get around the fact that any particular game I
play, that game has to end at some point, or else there is no payout.
Thus my purchase price must be finite not infinite.

Thanks.

Without quibbling about formalities of infinite series, you pick any
value and it is possible to see how to get a larger sum.

I don't see how this answers the question.

There is a much bigger problem with the St. Petersburg paradox! There
is no way to collect winnings that are greater than some value. You
can't collect $897 billion from anyone. You probably can't collect a
few thousand from your buddy in the bar. The realistic expected
return is much smaller that the simple minded series sum.

Having looked at many of the posts regarding this puzzle, first posed by
Nicholas Bernouli (not Daniel!), I'm a little confused as to how an
infinite expected value is arrived at.

The problem, simplified, is to find the willing price for playing the
game of flipping a coin until the first tails lands up. Each contiguous
flip (starting from the begining) landing heads pays the player $2^n.

The calculation of Expected value is the SUM $(2^n)/(1/2^n) = $1 + $1 ...
until the game finishes.

Many explanations of the game then simply state that the expected value
is infinite because the sum of 1's goes on forever (or could go on
forever?).

The problem I am having with this explanation is that the game has an
undefined state. If the coin flips to heads every time forever, no payout
is ever made. A payout is only made when the first tails appears. Thus
every defined state of the game requires a finite series of flips. Thus
it will never be the case that an infinite payout will occur.

Can someone explain how a finite series of flips results in an infinite
payout? Or how an infinite series of flips becomes a defined state?

Jason Simons schrieb:
Having looked at many of the posts regarding this puzzle, first posed
by Nicholas Bernouli (not Daniel!), I'm a little confused as to how
an infinite expected value is arrived at.

The problem, simplified, is to find the willing price for playing the
game of flipping a coin until the first tails lands up. Each
contiguous flip (starting from the begining) landing heads pays the
player $2^n.

The calculation of Expected value is the SUM $(2^n)/(1/2^n) = $1 + $1
... until the game finishes.

Many explanations of the game then simply state that the expected
value is infinite because the sum of 1's goes on forever (or could go
on forever?).

The problem I am having with this explanation is that the game has an
undefined state. If the coin flips to heads every time forever, no
payout is ever made. A payout is only made when the first tails
appears. Thus every defined state of the game requires a finite
series of flips. Thus it will never be the case that an infinite
payout will occur.

Can someone explain how a finite series of flips results in an
infinite payout? Or how an infinite series of flips becomes a defined
state?

The problem is that you cannot put an upper bound on the expected
payout. Every game of the St Petersburg game has a defined payout. The
payout itself has no upper bound (at least not in theory), although it
is definitely finite. However, there's no maximum.

Every (finite) sequence of St. Petersburg games has a defined and
finite average value - just add all the payout values and divide by
the number of games. The problem is that as you keep playing this game
and some of the higher payouts show, the average value rises.

If we play simple game of coin flip, that is, you get a dollar when
heads comes up, and you get nothing when tails comes up, the average
value of a few games is close to half a dollar - and it will tend to
get closer to that the longer you keep playing. If you play the
Petersburg game, that is not the case.

Imagine we're plaing the Petersburg game with a conscientious random
number generator. It produces sequence lengths that corresponding to
the position of the lowest binary digit of the number of the game
we're playing to make sure that the probabilities come out as exactly
as they possibly can.

Game / Binary / Sequence length / Payout / sum / average
1 1 1 1 1 1.00
2 10 2 2 3 1.50
3 11 1 1 4 1.33
4 100 3 4 8 2.00
5 101 1 1 9 1.80
6 110 2 2 11 1.83
7 111 1 1 12 1.71
8 1000 4 8 20 2.50
9 1001 1 1 21 2.33
10 1010 2 2 23 2.30
11 1011 1 1 24 2.18
12 1100 3 4 28 2.33
13 1101 1 1 29 2.23
14 1110 2 2 31 2.21
15 1111 1 1 32 2.13
16 10000 5 16 48 3.00
17 10001 1 1 49 2.88

See how the average keeps on rising?

If you had gotten in at game 14, your average would have been even
higher than that.

Now, the expected value asks for the value the average would tend to
if we played an infinite number of games. There is no such number. Up
to game 7 you'd have thought it tends to 1.50; then you'd have thought
it tends to 2 (and kept on thinking that until game 31). You can't
even find an upper bound on the number the average will tend to. Put
"no bound" in Latin and you get "infinite". You can average ANY amount
in this game if you just keep playing long enough (and noone's messed
with your coin).

Someone who gets in on the game just before you make it big obviously
averages more than you; those that drop out before that get less.
That's why we average over an infinte number of games.

I guess the answer to your question is that the expected value doesn't
consider a finite series. It asks how the average would be if you
played infinitely long. In teh simple flip game, playing infinitely
long gets you a finite average: half a dollar. Playing the St
Petersburg game doesn't.

Does that help?
Michael

The problem, simplified, is to find the willing price for playing the
game of flipping a coin until the first tails lands up. Each
contiguous flip (starting from the begining) landing heads pays the
player $2^n.

The calculation of Expected value is the SUM $(2^n)/(1/2^n) = $1 + $1
... until the game finishes.

The problem I am having with this explanation is that the game has an
undefined state. If the coin flips to heads every time forever, no
payout is ever made. A payout is only made when the first tails
appears. Thus every defined state of the game requires a finite
series of flips. Thus it will never be the case that an infinite
payout will occur.

This is a good explanation! But now I wonder about the problem of a
defective game. Isn't it possible that the coin could flip heads forever?
Given that it is random, we can't say that it is in a loop, and we
wouldn't be able to tell if the next flip would end the game. So now we
still have this undefined state problem--what do we do if the game goes
on forever?

The problem, simplified, is to find the willing price for
playing the game of flipping a coin until the first tails
lands up.

Each contiguous flip (starting from the begining) landing
heads pays the player $2^n.

Many explanations of the game then simply state that the
expected value is infinite because the sum of 1's goes on
forever (or could go on forever?).

The problem I have with these explanations is that the very
notion of "expected value" can be broken when a unbounded
values are involved. There are some cases where an infinite
series converges and is thus the sum is well defined.
Generally, though, an infinite series does NOT converge and
thus the sum is NOT defined.

In this case, the relevant series is 1+1+1+1+..., and this
series simply does not converge. Calling the outcome of
this sum "infinity" doesn't really help because infinity
is not a number!

In real world situations, there are various practical
limitations which prevent this sort of bizarre non-converging
series from arrising. For example, there is a hard limit
on how much money exists to make the payoff. There's also
a practical limit on how many flips can be made before the
player dies of old age. There are also practical limits
on how much money is so much that twice as much money is
no longer worth twice as much to the player. Any of these
limitations make a real life game converge to a
expected value.

Isaac Kuo wrote:
The problem I have with these explanations is that the very
notion of "expected value" can be broken when a unbounded
values are involved. There are some cases where an infinite
series converges and is thus the sum is well defined.
Generally, though, an infinite series does NOT converge and
thus the sum is NOT defined.

It is defined, its just not finite. This is a
theoretical objection--but not sustainable.

In this case, the relevant series is 1+1+1+1+..., and this
series simply does not converge. Calling the outcome of
this sum "infinity" doesn't really help because infinity
is not a number!

True, its more a destination than a number, but what is
important is that it is not divergent (i.e. the values
don't vacillate farther and farther apart), or non-convergent
(i.e. the values don't simply vacillate back and forth).

That is why we say it converges to absolute infinity.

In real world situations, there are various practical
limitations which prevent this sort of bizarre non-converging
series from arrising. For example, there is a hard limit
on how much money exists to make the payoff. There's also
a practical limit on how many flips can be made before the
player dies of old age. There are also practical limits
on how much money is so much that twice as much money is
no longer worth twice as much to the player. Any of these
limitations make a real life game converge to a
expected value.

This is a practical consideration. In math practical
considerations don't matter. Math is theoretical.

Humean inductive skepticism leaves the possibility that
some physical process, we have yet to see, could require
the calculation of such an average in a game.

Jason Simons <Simons@Burnaby.com> wrote in message
True, its more a destination than a number, but what is
important is that it is not divergent (i.e. the values
don't vacillate farther and farther apart), or non-convergent
(i.e. the values don't simply vacillate back and forth).

In real world situations, there are various practical
limitations which prevent this sort of bizarre non-converging
series from arrising. For example, there is a hard limit
on how much money exists to make the payoff. There's also
a practical limit on how many flips can be made before the
player dies of old age. There are also practical limits
on how much money is so much that twice as much money is
no longer worth twice as much to the player. Any of these
limitations make a real life game converge to a
expected value.

This is a practical consideration. In math practical
considerations don't matter. Math is theoretical.

In mathematics, there are many many things which are bizarre
and non-intuitive. In the real world, we know that
probabilities and expected values HAVE to exist.

In the
theoretical mathematical world, we know that it's entirely
possible for probabilities or expected values to NOT exist.

Humean inductive skepticism leaves the possibility that
some physical process, we have yet to see, could require
the calculation of such an average in a game.

It is mathematically provable that such an average cannot
be calculated. If we find some physical process which
seems to model the game, and we find some "average" via it,
then that merely proves that the physical process doesn't
actually model the game.

Thank you for clearing that up. I did err in not including this series in
the definition of divergence. I guess I have been thinking about this
long enough that I forgot what distinction was needed. While all the
series diverge, there is an important difference. The vacillating series
have different answers all the way along the series. The series that
diverges off to infinity does so the same way all the way out.

In any case, you clearly think that your made up definitions
are somehow useful. As a mathematician, I disagree.

I think the distinctions however do matter. If the game were to diverge
in the "vacillating" sense there would be no way to find a correct answer
anywhere in the game.

I have another slightly diverging view.

mechdan@yahoo.com (Isaac Kuo) wrote in
news:acc26c07.0411040925.79bb4941@posting.google.com:

It is mathematically provable that such an average cannot
be calculated.

I suppose you mean a finite average value. Many
mathematicians find nothing pathalogical about an
infinite average (cf. Cohen, Diaconis--both at Stanford).

If you can prove that such a value is pathological
you've got a nice paper here; one that would be able
to settle this little paradox once and for all.

There are two camps. Those who say this is not a paradox
simply because of the practical limitations of money
supply or the practical limitations imposed in our
decision processes, and those who say its not paradoxical
to have an infinite expected value (regardless of what
practical limitations there might be).

ObPuzzle:
a) Name a sequence s(i) such that for any n, the subsequence starting with
s(n) contains all integers!

b) Can this question be extended (and answered) for fractions or reals?

Isaac Kuo schrieb:
Michael Mendelsohn wrote:
Out of curiosity, can a dirgent series do anything but diverge
to +/- infinity or oscillate?



I've been thinking some more, and I think I can classify seequences most
easily by determining whether they have exact upper and lower bounds. I
take "exact" to mean that for increasing n, the lowest upper (or
greatest lower) bound for the sequence converges to a constant. If you
have a monotonous sequence

that diverges to oo, you have a lower bound
for all n, but the higher the n you start a subsequence with, the higher
the lower bound can be. If you interleave such a sequence with a
sequence of constant values, the lower bound can't be higher than that
constant value, for any n, and thus is "exact". (Excuse my terminology;
I hope I made myself clear).

+UB +LB oscillates in "the intuitive sense"
+UB -LB oscillates "to -oo"?
-UB +LB oscillates "to +oo"?
-UB -LB oscillates all over the place

I doubt these distinctions are particularly useful, but they serve (me)
to broaden intuition.

ObPuzzle:
a) Name a sequence s(i) such that for any n, the subsequence starting
with s(n) contains all integers!

b) Can this question be extended (and answered) for fractions or reals?





And, for an early ObPuzzle, is there a simple game for which the
expected value oscillates?

Sure, just flip a fair coin until it comes up heads,
after n flips. At that point, the game is over and
the jackpot is (-4)^n dollars. Note that the player
pays the bank if n is odd.

The expected value is the series summing up (-2)^n,
which violently oscillates up and down in an unbounded
fashion.

Replace the original (-4)^n with (-2)^n if you don't
want an unbounded oscillation.

The oscillating St Petersburg game? ;)

However, divergent series can actually be "summed"
rigorously by using extensions to the usual summation rules
(e.g., so-called Abel and Cesàro sums).
For example, the divergent series 1-1+1-1+1-...
has both Abel and Cesàro sums of 1/2.
http://mathworld.wolfram.com/DivergentSeries.html

Does 1/2 have any meaning that can be transferred back to the game?

Cheers
Michael