vpFREE2 Forums

ER/ROR (Expected Return/Risk of Ruin) puzzle

#41

--- In vpFREE@yahoogroups.com, "Harry Porter" <harry.porter@v...>
Steve Jacobs wrote:
"Well, sort of, but not really. The players who don't bust have
bankrolls that grow without bound. On a percentage basis, as they
play indefinitely their actual outcome will still tend toward 1%.
The loss of those who bust out will dwindle to an insignificant
fraction of the total action."

Harry still hangs on to his strong views of:
"Fully agreed, if I understand this to say that the surviving players
in playing indefinitely will approach an ER just over 1%, but not be
equal to 1%."

The surviving players in playing indefinitely will approach an ER of
1%. The "not just over 1%" part is simply incorrect. The actual
results of the surviving group can be above, exact at, OR even below
the theoretical ER of 1%.

The games have no memory! Harry, please stop taking the position the
games have memory. The gambling gods DO NOT bless these players with
a DIFFERENT theoretical ER simply because a few unlucky souls busted
out while playing the game. This also applies for group results as a
whole.

Harry, this is my last post of this topic. I kindly afford you the
last words in our discussion of this tread.

Regards.

#42

I've used ER the way you *and* Steve Jacobs had used the term; I
tend to use "ER" in relative terms as opposed to absolute terms. So,
please don't use a double standard!

Am I being admonished here? Hate to see how hepped up you can get
when on the subject of politics :wink:

If the dispute here is over semantics, I shouldn't have butted in.
But there are occasions where partial information is known about a
situation, but not the full result. Case in point, a period of "101%
ER" play in which it's know that losses never exceeded a given
threshold. Expectations change when the assumptions for those
expectations change.

We don't know what the result was (or in a hypothetical, what the
result would be), but we're able to determine an expected result that
incorporates what we know about the loss limitation. I believe this
revised expectation under a set of assumptions is an "expected return"
in its own right.

But, having explained myself, let's just leave it that I'm writing out
of my own private dictionary.

- H.

#43

I don't think there's any doubt that vpFREE will give us the full Monty....

Chandler

···

At 12:28 PM 7/24/2004, you wrote:

I hope this doesn't start a long Monty thread....

#44

Groan!!!!
   Enough already! LMAD has been off the air for over 30 year's. None of us will EVER have to make this choice.

Ned C.
The Wild Joker

···

Chandler <omnibibulous1@comcast.net> wrote:
At 12:28 PM 7/24/2004, you wrote:

I hope this doesn't start a long Monty thread....

I don't think there's any doubt that vpFREE will give us the full Monty....

Chandler

vpFREE Links: http://members.cox.net/vpfree/Links.htm

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#45

Don't you get the Gameshow Network?

We can discuss all those reruns for years to come!

carlos

PS No, let's not....
  
Groan!!!!
Enough already! LMAD has been off the air for over 30 year's. None of us will EVER have to make this choice.

Ned C.
The Wild Joker

Chandler wrote:

···

The Wild Joker <jokerswild1203@yahoo.com> wrote:
At 12:28 PM 7/24/2004, you wrote:

I hope this doesn't start a long Monty thread....

I don't think there's any doubt that vpFREE will give us the full Monty....

Chandler

vpFREE Links: http://members.cox.net/vpfree/Links.htm

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#46

The ER of the game is independent of what happens to a particular
subset of players. The _condition_ ER given that the player comes
out ahead is still a distinct concept from the _total_ ER of the game.

···

On Saturday 24 July 2004 08:25 pm, Harry Porter wrote:

> Thank you for the answer! It never ceases to amaze me how people
> confuse expectations with actual results. If a game returns 101%
> with perfect play, the ER (Harry's lingo) is 1%. Whether Harry lost
> or gain $10,000 from this game so far, the ER remains 1%.
>
> I never could understand how people believe the ER will be greater
> than 1%!!! Did this game **mutate** somehow and someone forgot to
> tell the rest of us. If the game does mutate, can I get a ER of 3%,
> instead of 1%?

I'll stand by my statement (very carefully).

You defined your problem over a finite amount of play. The discussion
was regarding the expected return during that actual play, and not
the prospective ER of the game going forward.

If you have a group of players who've played for a given period with
an ER of 101%, and then you subtract out a subset of all players who
went broke, the balance of the group will have an ER in excess of 101%.

#47

--- In vpFREE@yahoogroups.com, "Harry Porter" <harry.porter@v...>
wrote:

Am I being admonished here? Hate to see how hepped up you can get
when on the subject of politics :wink:

No kidding, LOL. But hey, at least he's enthusiastic.

But there are occasions where partial information is known about a
situation, but not the full result. Case in point, a period

of "101%

ER" play in which it's know that losses never exceeded a given
threshold. Expectations change when the assumptions for those
expectations change.

This makes perfect sense to me, so I'd like to associate myself with
this - when I use ER in the future, this is the same way I will use
it. To take a more common case, I may occasionally calculate and
refer to an ER with a different-from-max-EV strategy, such as a
variance-minimizing strategy. I reserve the right to quote a
different ER for the same game when I assume strategies other than
max-EV strategy, because, as Harry says, the assumptions underlying
the expectations have changed.

In other words, ER is not a characteristic of a game, but of a game
AND a set of assumptions regarding play in that game.

#48

R is the risk of ruin for a player who starts with a single unit.

···

On Saturday 24 July 2004 07:27 pm, fordscks wrote:

--- In vpFREE@yahoogroups.com, Steve Jacobs <jacobs@x> wrote:
> Not so. EV and RoR are independent concepts, so house edge is
> not in any sense "due to" insufficient bankroll. If you take a

million

> players and start them with one unit each, they do not as a group
> experience a different house edge than a group of 10 players who
> each start with 100,000 units. But, the players who start with

only one

> unit have a much higher risk of ruin (by a factor of (1/R)

^100,000).

I'm sorry, but I didn't understand the "R" part in (1/R)^100,000.
What does the "R" mean? Thank you.

#49

PS I also come from a background where the rule is to question
things, but I use theory to tell me where or what the answer should
be.

But then how will you ever detect when your theories are wrong? No
scientist who ever achieved a breakthrough would say what you just
said.

blaw57:

If you don't like common sense, because you have problems with
common sense, then may I suggest you try logic. Let's go back to
your puzzle with the 101% game, assuming perfect play. Logic tells
us that the *expected* change in the bankroll at any point in time
is from this game: 101% * (the amount of dollars wagered). Logic
will say this is true for 1 person or for 100 people or for 10,000
people. Logic will also tell us, it really doesn't matter if you
chose 1 hour or 1 day or 1 month. If you continue this line
of "reasoning" using logic, it would have saved you a lot of time
and frustration.

What blaw57 did was to choose an arbitrary point in time and with
some arbitrary number of players. I couldn't understand how some
arbitrary number of players at some arbitrary time period would
transmute the relevant equation of "101% * (the amount of dollars
wagered)."

Cheers.

And here is where you miss in practice what you missed in theory in
what I quoted up top. I was questioning what ER was saying, and to do
so I needed empirical evidence. Since ER describes a real phenomenon,
namely finacial performance in a VP game over the long run, all I had
to do was set up a long run simulation.

And, in the end, that simulation proved that in this instance the
theory was correct, as it will in 99.99% of the cases I challenge.
But when I find that 0.01% of cases where theory is wrong, then I
will be making money off people like you who never question the
theory!

Hey, how old are you anyway? You're coming across as a 13-year old
who can't handle his temper. But I think I detect some smarts in what
you write, in between all the smart-aleck. I really wish you'd try
the dispassionate discourse style, because I think you may have
something to add around here. I'd hate to hear one day that you died
of apoplexy after reading one of my posts :slight_smile:

···

--- In vpFREE@yahoogroups.com, "fordscks" <jason_c_vp@y...> wrote:

#50

Nothing is ever perfect straightforward. At least one of the times that
this appeared in the "Ask Marilyn" column, the wording was ambiguous
enough that different answers were possible depending on your
interpretation of the wording.

···

On Saturday 24 July 2004 05:24 pm, Bill Velek wrote:

Steve Jacobs wrote:

snipped my bungled details on the Monty Hall dilemma.

> You've left out one important detail. After the initial choice, Monty
> opens
> one of the unchosen doors and shows that it doesn't contain the prize.
> Then the player is offered the choice to switch.

Yeah, I caught that 10 minutes later when I saw it posted, and then I
posted a correction. Duhhhh. I was half asleep. Sorry about that.
What's interesting is that there were some pretty impressive
mathematicians who attempted to argue the wrong side, based upon their
intuition. I can't remember the names of the principle parties, but
demands were being made for a columnist to resign over the whole
affair. Just a bunch of hot-heads -- nothing like the calm folks here
on vpFree. :wink:

#51

The correct answer here depends on your assumptions (if any) about
the nature of god. If you believe him when he says "I'll help you out"
then you may have picked wrong to begin with. If you picked correctly
to begin with, then you have the dilemma that god is trying to trick you
into being sent to hell. Now, he wouldn't do that, would he? :wink:

Then again, are you really sure which gates you are standing before?

In order to know that this is like the Monty Hall problem, you wouuld
have to know that all "contestants" are told a bad choice and offered
an opportunity to switch.

···

On Saturday 24 July 2004 12:56 pm, nightoftheiguana2000 wrote:

i would think it's 50/50 too, so it doesn't matter if you switch or
not, but:
http://tinyurl.com/48kuz
on a related question, you die and go to heaven and god says, ok you
pick either hinduism, christianity or islam, only one of them is
correct, pick the wrong one and you go to hell, you pick one and god
says hey you've been good i'll help you out, this other one here is
wrong, way wrong, now of the remaining two one is wrong and one is
right, care to change your choice? what do you do? is it 50-50 or 1/3-
2/3?

#52

This would only be a new event if they moved the prize after showing
an empty door. Think of it this way: if you decide in advance to
never switch, then your original pick will win 1/3 of the time and since
you don't switch your overall outcome is to win 1/3 of the time.

The fact that there are now two choices does not imply that they have
equal probability. To never switch is to choose a 1/3 probability of
winning. To alwasy switch is to choose a probability of winning that
is [1 - p(winning if you never switch)] = 2/3.

···

On Saturday 24 July 2004 11:46 am, blaw57 wrote:

--- In vpFREE@yahoogroups.com, Steve Jacobs <jacobs@x> wrote:
> People tend to think that since the choice have been reduced from
> three to two, they must be 50/50 so it doesn't matter whether they
> switch or not. But, if you always stick to your original choice,

you

> will win with probability 1/3 and nothing can change that. So, if
> you switch doors, you increase your chances from 1/3 to 2/3.

I disagree. After Monty opens one door and eliminates it, the player
has a completely new event with 50-50 odds. Choosing the door that
coincidentally happens to be the same one that he chose before, at
lower odds, now has the same odds (50%) as choosing the other door.

#53

Steve Jacobs wrote:

The ER of the game is independent of what happens to a particular
subset of players. The _condition_ ER given that the player comes
out ahead is still a distinct concept from the _total_ ER of the
game.

Steve, I think we're making the identical point. "E.R." is a value
that's expressed in a given context.

For example, if I tell you that a player is less than perfectly
skilled and can be expected to suffer strategy errors that will reduce
his return by .2% against perfect play ER, would anyone insist that
the ER applicable to this players results is the "perfect play" ER.

Another example, we discuss expected return under a max ER strategy,
but also reference expected return of alternate strategies (such as
min loss).

There's no question, whenever the term "E.R." is used without further
qualification, it's a reference to perfect play of a game using a max
ER strategy.

However, the term E.R. can be used in a qualified manner (e.g. the ER
of all players who play through 10000 hands of a game without
suffering a loss position greater than 2500 credits) and still be a
reasonable and valuable application of the term. As you say, ER is
being referenced in a different conceptual context, but the use
preserves the basic meaning of "expected return" in every sense.

- Harry

#54

blaw57 wrote:

In other words, ER is not a characteristic of a game, but of a game
AND a set of assumptions regarding play in that game.

Gee, if I'd bothered to read this succinct explanation before replying
to Steve's post I'd have saved a lot of bandwidth (and words :wink:

(and, because nuance can be sometimes unclear in this environment, I'm
not being at all facetious ...)

- H.

#55

Steve Jacobs wrote:
> Well, sort of, but not really. The players who don't bust have
> bankrolls that grow without bound. On a percentage basis, as they
> play indefinitely their actual outcome will still tend toward 1%. The
> loss of those who bust out will dwindle to an insignificant fraction
> of the total action.

Fully agreed, if I understand this to say that the surviving players
in playing indefinitely will approach an ER just over 1%, but not be
equal to 1%.

Nope. What would "just over 1%" mean? 1.1%? 1.01%? 1.00000001%?
How would you find the right amount of "overage" for players whose
bankrolls become infinitely large? Their _expected_ outcome will approach
1% of their total action.

Once we're told something retroactively about the performance of this
group of players (that none of them suffered losses sufficient to ruin
them) which distinguishes them from the population as a whole, then
their ER as a group shifts.

ER never shifts. It is a prediction about future performance, and is not
influenced by past results. If you restrict the problem to a subset of the
outcomes, then the resulting _conditional_ ER may be different than the
_total_ ER but this is an apples/oranges comparison. Still, the conditional
ER given that the player never goes broke is 1%.

···

On Saturday 24 July 2004 08:35 pm, Harry Porter wrote:

#56

Steve Jacobs wrote:

Nope. What would "just over 1%" mean? 1.1%? 1.01%? 1.00000001%?
How would you find the right amount of "overage" for players whose
bankrolls become infinitely large? Their _expected_ outcome will
approach 1% of their total action.

That's right. As their play increases, with prospective play having
an ER of 1%, ER of all play considered would converge on 1% asymptoticly.

ER never shifts. It is a prediction about future performance, and
is not influenced by past results. If you restrict the problem to a
subset of the outcomes, then the resulting _conditional_ ER may be
different than the _total_ ER but this is an apples/oranges
comparison.

The discussion wasn't a matter of comparing conditional ER with total
ER. Both values are are valid for consideration in their own right
and reference to either one would reflect the context of the question
posed.

Beyond this assertion, I think we both agree that this is a discussion
of semantics.

- Harry