vpFREE2 Forums

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

#21

Bill Velek wrote:

snip

... the question is: in Let's Make a Deal, after a person chooses one
door out of three, and Monty Hall doesn't tell them yet whether they've
won or not, but instead asks them if they would like to change their
minds, are they mathematically better off to change their minds or stay
with their original choice? _Intuitively_, since they don't know what
is behind any door yet, most folks think that there is still a one in
three chance, but mathematically speaking the odds are better if they
always change their minds.

Dang it, I was up until real late last night and I'm still half asleep. Sorry about this. Monty Hall always opens one of the doors that is not a winner, leaving the contestants choice and one other door still unopened. The question is still: should the contestant change his mind. Most people think the odds are still the same: 50/50, but mathematically speaking the contestant has better odds if s/he changes his/her mind. I think I'm awake now. :wink:

Cheers.

Bill

#22

it's a short term effect, as you say only for positive games

RoR is absolutely not a short term effect. Computing RoR is
exactly equivalent to computing the probability that your
finite bankroll will allow you to play *forever*. Clearly
forever and "short term" are mutually exclusive concepts.

a classic example would be a small casino that has always offered
full pay deuces in quarters and nothing playable in dollars, they
decide to put in dollar deuces, they do quite well at first because
they bust out many of the quarter players who are insufficiently
bankrolled - for example a quarter player might quit when he looses
$4,000 and go back to quarters - if players quit when they are on a
losing streak the casino wins even if the game is positive and the
players are playing perfect strategy - but, this is a short term
effect, soon the only players left are the ones who have sufficient
bankroll to ride out the negative streaks and thus the casino begins
to lose money near the expected rate

Nonsense. If all players used a perfect strategy, then the casino
can expect to lose an amount equal to the player's edge multiplied
by the number of bets placed. It makes no difference what "type"
of player feeds bills into the machine. What you describe defies
mathematical reality. There is no mathematical mechanism which
creates a tendency for the casino to tend to win shortly after
such a switch in denomination and lose later.

You are essentially predicting a non-random effect. You are saying
the casino will tend to win for the first N plays, but eventually
lose for blocks of N plays that occur later. In reality, every block
of N plays will tend to have the same distribution of outcomes,
without regard to where the block occurs relative to ANY starting
point you care to choose (including "right after a switch in denomination")

another way to think of it:
suppose the rule is play dollar deuces until you lose $4,000. for
that the risk of ruin in 60%. even if the other 40% of surviving
players are up an average of $4,000, the casino is still ahead

Nope. The 40% of surviving players, as a group, go on to have their
bankrolls grow without bound. On average, the non-ruined players
are up plus infinity.

(it's a short term effect, but risk of ruin is a short term effect,
for positive games)

I'm sorry, but this is absolutely false, and repeating it won't make
it true. Risk of ruin is not a short term effect. It is simply equal
to the probability that you _don't_ get to play forever. The
bigger your stake, the better your chance of playing forever.

Suppose you start with a large number of expert players and give
each a starting bankroll large enough that they have a 60% risk of
ruin. Then you say "go play" and watch as more and more players
bust out. You'll have to wait a while for the first player to bust. If
he/she loses at an average rate of 5% then it will take about 76
hours of play (at 600 plays/hour). On average, players will take
much longer than this to bust out. But, the longer players survive,
the more likely they are to be ahead of their starting point, which
reduces their probability of going bust. The bust out rate will
tend to decrease. It is take a long time to reach the 40% bustout
mark, and even longer to reach the 50% bustout. As the bustout
rate approaches 60%, the number of bustouts will decline from
"hardly ever" to "never".

Does that sound like "short term"? I don't think so.

using http://www.lotspiech.com/GamblersRuin.html i can run a sim
example, quarter deuces, $50 stake, $2300 retire, 2000 hands, results:
prob win product
81% -$50 -$40.50
10% $67.50 $6.75
5.4% $302.50 $16.34
1.3% $537.50 $6.99
0.59% $772.50 $4.56
0.8% $1007.50 $8.06
0.39% $1242.50 $4.85
0.1% $1477.50 $1.48
0.02% $1712.50 $0.34
0.02% $1947.50 $0.39
0.01% $2182.50 $0.22
---
net=$9.48 (expected win = 2000 x $1.25 x .0076 = $19)

granted this is not an idea example, ideally you would run out
several royal cycles but such a sim would take a while, but hopefully
i've demonstrated that having a low cutoff decreases the average
return

A single simulation of 2000 hands is statistically meaningless. Run
the same simulation 10 times and you are likely to get a wide range
of results. In addition, the actual simulator doesn't give a single dollar
value as you've listed above -- is shows ranges such as "-$50 to +$165"
and "+$165 to +$380". Without knowing how the outcomes are
distributed within those ranges, you can't compute a meaningful
average outcome. The dollar values you list don't even fall at the
middle of the ranges, so I'm not sure how you chose them. Anyway,
even if you consistently use the same approach, you're likely to see
vastly different net values from other simulations.

It is always dangerous to try to draw conclusions from such small
sample sizes. I'm afraid you've drawn incorrect conclusions here.

···

On Saturday 24 July 2004 02:19 am, nightoftheiguana2000 wrote:

#23

>the extra money goes to the casino
>gambler's risk of ruin favors the house
>if a lot of people are busting out due to insufficient bankrolls, the
>house edge is increased

Have to admit I've only been reading bits and pieces of this thread, but
the above statement sounds counter intuitive. Why would a casino benefit
from multiple people with insufficient bankrolls?

Why indeed.

Would not multiple
people with insufficient bankrolls at some point be equal to one person
with sufficient bankroll?

Yes.

IOW, wouldn't the casino loss or win be
determined by the total number of games played (assuming proper strategy)
without regard to who and which person's money played the hand?

Yes.

Please use little words because I'm slow;-)

Ah, but you have good instincts!

···

On Saturday 24 July 2004 08:55 am, Chandler wrote:

At 05:47 PM 7/23/2004, you wrote:

#24

Heh, heh. Intuition throwing mathematically intelligent people?? True
enough that that happens; it sounds like the Monty Hall dilemma which
had math professors quibbling. For those of you who are unfamiliar with
that, the question is: in Let's Make a Deal, after a person chooses one
door out of three, and Monty Hall doesn't tell them yet whether they've
won or not, but instead asks them if they would like to change their
minds, are they mathematically better off to change their minds or stay
with their original choice?

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.

_Intuitively_, since they don't know what
is behind any door yet, most folks think that there is still a one in
three chance, but mathematically speaking the odds are better if they
always change their minds.

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

···

On Saturday 24 July 2004 09:15 am, Bill Velek wrote:

#25

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.

···

On Friday 23 July 2004 09:41 am, Harry Porter wrote:

blaw57 wrote:
> ... It is a positive game returning 101%, and a group of 10,000
> rocket scientists who always play perfect set about playing it with
> $10,000 bankrolls, which are deemed adequate for this game using a
> 10% ROR assumption. Well, in the course of play 1000 players will go
> bust and 9000 will make 1% of $1 million or $10,000.

Under these general assumptions, it's the group of 10,000 players as a
whole who have a 1% return expection. Of the subset of players who
don't bust, they're ER will be greater than 1%.

#26

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.

···

--- In vpFREE@yahoogroups.com, Steve Jacobs <jacobs@x> wrote:

#27

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?

>
> 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

···

--- In vpFREE@yahoogroups.com, "blaw57" <blaw57@y...> wrote:

--- In vpFREE@yahoogroups.com, Steve Jacobs <jacobs@x> wrote:
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.

#28

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.

It is correct to switch. Let me give you an example to explain why
non-mathmatically. Suppose you initially choose door #1 and Monte
shows you the loser at door #2. That leaves the winner behind door
#3 or #1. However if it were behind #3, Monte would be forced to
show you #2; whereas if it were behind #1, Monte (assuming he
randomized) would choose #2 half the time and #3 half the time. Thus
it is twice as likely the winner is behind #3 than #1.

The same logic applies to Bridge in a situation where you are
missing Q J and and opponent drops one offside after you play the A
or K. In such case the odds favour playing the opponent who dropped
the Q or J as having a singleton, since if he had both he would play
the Q first half the time and the J half the time, assuming proper
randomization. This is called the law of restricted choice.

David

···

--- In vpFREE@yahoogroups.com, "blaw57" <blaw57@y...> wrote:

#29

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
<nightoftheiguana2000@y...> wrote:

i would think it's 50/50 too, so it doesn't matter if you switch or
not, but:
http://tinyurl.com/48kuz

So it is.... very neat question, I like it. Since Monty always shows
you the remaining bad prize, you're really getting the choice between
the best of two versus the remaining one. Believe it or not, I've
never come across that one before.

#30

for what it's worth here's my solution to the mystery:
your initial odds are 2/3 of picking a dog, of the remaining two
choices, the odds are 100% that there is at least one dog and 1/3
that they both have dogs, so monty reveals a dog, the odds that the
remaining choice also is a dog is still 1/3, meaning the odds are 2/3
that it is gold, so your odds of picking gold are 1/3 if you stick
with your original choice, 2/3 if you switch

http://www.google.com/search?hl=en&lr=&ie=UTF-8&q=monty+hall+problem
http://www.google.com/search?hl=en&ie=UTF-8&q=Bayes'+Theorem

#31

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:

Cheers.

Bill

#32

"To go back to the original puzzle, the 9000 survivors could not
have normalized in a time period so short that the losers losses
were a significant percentage of coin-in. The survivors will
intially be ahead by the amount of the losers' losses, and the above
average return would be the beneficial effect of playing on a 90%
ruin-free bankroll instead of a 100% ruin-free one. In the many
millions of more hands required for normalization, the average
returns achieved will gradually bring down this premium to the point
of insignificance.

Thus it all comes back to common sense in the end, and no doubt
there will be many who wonder if there was any value in pursuing
that circular path at all. Well, all I can say is I come from a
background where the rule is to question everything, even common
sense, as there are occasionally flaws even in common sense. If a
common sense math propositon is really true, then it should be
possible to prove it mathematically as well as common-sensically."

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.

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.

···

--- In vpFREE@yahoogroups.com, "blaw57" <blaw57@y...> wrote:

#33

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.

···

--- In vpFREE@yahoogroups.com, Steve Jacobs <jacobs@x> wrote:

#34

Harry Porter wrote:
"Under these general assumptions, it's the group of 10,000 players
as a whole who have a 1% return expection. Of the subset of players
who don't bust, they're ER will be greater than 1%."

Steve replied
"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."

Steve:

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%?

···

--- In vpFREE@yahoogroups.com, Steve Jacobs <jacobs@x> wrote:

#35

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%.

But just to be clear --- we're now talking about the ER of the group
who played a 101% game AND didn't bust ... not the ER of the group as
a whole. That's how the ER "mutates" here; you've added information
to the problem.

- Harry

#36

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%.

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.

- H.

#37

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%.

But just to be clear --- we're now talking about the ER of the group
who played a 101% game AND didn't bust ... not the ER of the group as
a whole. That's how the ER "mutates" here; you've added information
to the problem.

Harry, I think that they are speaking about ER = the 'statistically' _Expected_ Return, and I think that you are speaking about the _ACTUAL_ Return that players, in fact, realize. But 'ER' is a _constant_ for a given game and strategy when consistently used without error. Luck enters into the equation, and some of the players should realize a return that is less than statistically expected, and they go bankrupt. The others, given infinite play, would be _expected_ to have an actual return that has a _tendency_ toward the statistically expected return, ... _BUT_ ... obviously there is no mathematical or physical laws that actually _guarantee_ that that will happen. It is _POSSIBLE_ that every single hand drawn by every single person on earth for the rest of our existence -- in fact for all of the rest of time, for infinity -- will be a Royal Flush ... sequential in hearts. :-/ And 10,000 years from now, despite all math concepts, a Heart Royal would come to be _expected_ -- not because of math but rather from empirical data -- although no one would ever actually 'play' cards anymore because it would be pointless, and card games and strategies as we know them today would be considered myths, and the unexplainable 'constant Royal' from a fully shuffled 52-card deck would simply be considered to be a divine mystery, for lack of any other possible explanation. :wink:

Anyway, I'd have to agree with the others that you are a bit wrong on this, Harry, although I'd chalk it up to semantics as much as anything.

Cheers.

Bill Velek

#38

Bill Velek wrote:

Anyway, I'd have to agree with the others that you are a bit wrong on
this, Harry, although I'd chalk it up to semantics as much as
anything.

Anything's possible :wink:

But I don't think it's semantics to say: Someone who plays 10000
hands of a game with an ER of 101%, obviously has an ER of 101% on
that play. However, if you add the fact that it's known that at no
time in their actual play were they down more than 1000 credits, the
expected return for that play is now greater than 101%.

- H.

#39

Harry Porter wrote:

Bill Velek wrote:
> Anyway, I'd have to agree with the others that you are a bit wrong on
> this, Harry, although I'd chalk it up to semantics as much as
> anything.

Anything's possible :wink:

But I don't think it's semantics to say: Someone who plays 10000
hands of a game with an ER of 101%, obviously has an ER of 101% on
that play. However, if you add the fact that it's known that at no
time in their actual play were they down more than 1000 credits, the
expected return for that play is now greater than 101%.

Not to flog a dead horse ... and I could also _possibly_ be wrong about this ... but these are the definitions as I see them (and I'm sure that these are not the official definitions listed on vpFree, but I'm too lazy to look them up right now):

ER = Expected Return -- the mathematically/statistically anticipated return for a particular game and pay table, given faultless play over the long term.

AR = Actual Return (I'm coining this term if it hasn't already been done ... although I'm sure it, or something like it, has): this is 'Coin-Out' divided by 'Coin-In', multiplied by a hundred to express it as a percentage. AR has nothing whatsoever to do with ER, i.e., regardless of what the ER actually is for a given game and strategy, any player at any time, no matter how skilled or inept, can have an AR that is astronomically above or astronomically below the ER. ER can predict the likelihood of an AR, but that's it.

EV = Expected Value -- the present value of a hand _prior_to_the_draw_, based on the combination of probable outcomes multiplied by the various values of those outcomes.

AV = Actual Value -- what you are paid by the machine upon the conclusion of the draw (and any double-down play); this, again, is related to EV only insofar as EV is based upon certain possibilities, and the final hand and its AV is simply one of those possible results. AV is also completely independent of ER.

Using those definitions, I have to say that a given paytable with long term perfect play has a constant ER. There are a wide range of ARs among all players which collectively TEND toward the ER, but those with low ARs as well as those with high ARs, nonetheless all share one, and only one, ER.

Harry, I don't really see this as being all that important, and if you or others want to persist in your position, you all have a right to your own opinions. No big deal, really.

Cheers.

Bill Velek

#40

--- In vpFREE@yahoogroups.com, "Harry Porter" <harry.porter@v...> >
Bill Velek wrote:
Anyway, I'd have to agree with the others that you are a bit wrong
on this, Harry, although I'd chalk it up to semantics as much as
anything.

Harry responded:
"But I don't think it's semantics to say: Someone who plays 10000
hands of a game with an ER of 101%, obviously has an ER of 101% on
that play. However, if you add the fact that it's known that at no
time in their actual play were they down more than 1000 credits, the
expected return for that play is now greater than 101%."

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!

Harry, you are confused when you say: "[i]f you add the fact that
it's known that at no time in their actual play were they down more
than 1000 credits, the expected return for that play is now greater
than 101%." The confusion is akin to a priori versus a posteriori.
The *expected* return is 1% [a priori]; the actual return is greater
than 1% [a posteriori].

Harry, may I suggest you look up the term "expected"
or "expectations." Clearly, these terms are causing you much
consternation. These machines have no memory!! Actual results have
no bearing on the expected results. The fact the actual results [a
posteriori] are 5 std dev from the mean does not change the
expectations.