Wouldn't they be the same thing by definition?
Regards
A.P.
----- Original Message -----
From: Steve Jacobs
To: vpFREE@yahoogroups.com ; ckbrune
Sent: Saturday, July 26, 2003 10:14 PM
Subject: Re: [vpFREE] Re: "minimize ROR" and/or "maximize EV"
I'm working to prove that min-cost and min-ROR are mathematically
equivalent in the general case, but I haven't got there yet.
[Non-text portions of this message have been removed]
Steve-
Thanks for your valuable posts.....it gives me something to ponder.
I also wanted to verify that the Min Cost analysis requires the
scaling factor to be added to/subtracted from all payoffs, rather
than multiplied by/divided by. Doing the latter would result in a
playing strategy that is identical to the Max EV strategy. I believe
this is what AJ was questioning.
-Chris
--- In vpFREE@yahoogroups.com, Steve Jacobs <jacobs@x> wrote:
On Saturday 26 July 2003 11:04 pm, AJ wrote:
> Are you saying that scaling ALL the payoffs by the same scaling
> factor will change the max-EV strategy?Yes, it will, provided that you do NOT scale losses equally.
This implies that cashback can cause the optimal strategy to change.
Steve-
Thanks for your valuable posts.....it gives me something to ponder.
I also wanted to verify that the Min Cost analysis requires the
scaling factor to be added to/subtracted from all payoffs, rather
than multiplied by/divided by.
No, it is multiplied, but I screwed up the description by omitting
some important details.
You scale only the winnings. I mentioned this, but I should have
elaborated. In VP, payoffs are "N for 1" instead of "N to 1" and
this needs adjusting. For example, with a 50/1 payoff on a 4/kind,
you treat the first unit as your original bet that is returned to you,
and the 49 other units as winnings, and scale only the 49 units.
A one unit payoff is treated as a "push," so no scaling is applied.
No scaling is applie to units lost.
Doing the latter would result in a
playing strategy that is identical to the Max EV strategy. I believe
this is what AJ was questioning.
No, it wouldn't give an identical strategy. I realize it might seem like
it should, but it doesn't work that way. I _think_ it would still give
a min-cost strategy after iterating to get a breakeven game, but I'm
not sure. Now, if you scaled the losses at the same rate, it would be
just like playing with larger/smaller coins, and then nothing would
change. Perhaps that was what AJ was thinking. The main driving
force that changes the strategy is the fact that losses are weighted
differently than wins.
There is another way of thinking about min-cost that might be
helpful (or not). Suppose you won some contest and the prize
was that you got to play through 100 "no cash value" units on
a special VP machine. The machine takes your silver coins, but
payoffs return one unit in silver coin(s) and the rest in gold coins.
You never play the gold coins, because the machine is -EV, and
the gold coins are too valuable. How valuable? You won't know
until the end. You are allowed to play until all the silver coins are
gone, then you take whatever you have in gold coins and they
spin a wheel to determine whether the gold coins are worth $1 or
$5 or $25 or $100 each.
The min-cost strategy is equivalent to exchanging silver coins for
the maximum number of gold coins, on average, without regard to
the relative worth of silver and gold. It seeks to get the most units
in gold won/purchased in exchange for silver lost/spent. This is
just like buying gas and shopping for the best price. Gas prices
fluctuate from week to week, but at any given time the best deal
is to exchange the fewest dollars for the most gas (assuming all
gas is equal).
In contrast, the max-EV strategy seeks to maximize the average
total number of coins after each round of play, without regard
to whether the coins are silver or gold. Max-EV is "blind" to
coin color, so it can't tell you when you've finished converting
silver into gold.
On Sunday 27 July 2003 08:06 am, ckbrune wrote:
Steve Jacobs wrote:
You scale only the winnings. I mentioned this, but I should have
elaborated. In VP, payoffs are "N for 1" instead of "N to 1" and
this needs adjusting. For example, with a 50/1 payoff on a 4/kind,
you treat the first unit as your original bet that is returned to
you and the 49 other units as winnings, and scale only the 49 units.
You know, Steve, I think I follow this intuitively (which at this
point should worry some).
There was a "high meter progressive" min cost discussion over on acvpp
recently that involved applying a strategy determined by a paytable at
which the RF value produced a B/E ER.
That is a bit of a simplified approach of what you describe by
"scaling" (correct me if I'm wrong). While the "high meter" approach
seeks to minimize loss while waiting for a RF, what you detail should
serve to minimize loss anytime while waiting for all hands. That's at
the expense of return, but serves as a more conservative strategy
designed to protect bankroll.
Let me know if I've run way off the beaten path here, Steve.
If not, then I have a couple of follow up questions to better
understand this approach.
From a practical standpoint, wouldn't this scaling best be applied to
those hands that present the greater risk of a shortfall in return,
say 4K's and above? If not, is it because either the preservation of
game ER would only be nominal, or because the resulting change of game
pressure on bankroll is appreciable?
Is the incentive to adopt a min loss strategy highest when the
bankroll margin for game play is thin?
Thanks, Steve,
- Harry
Perhaps, if the right "definitions" are used, and if you've got such
a pair of definitions I'd really like to hear about them. Perhaps I've
been over-complicating things all along.
To compute cost, you separate winning payoffs from losing payoffs,
and compute average loss and average win, then take the ratio
mean(loss)/mean(win) to find cost. That is my definition of cost.
To compute ROR, you solve a polynomial equation where the
probabilities are not used to weight the payoffs. The probabilities
weight terms for the form R^(payoff) where R is the ROR value
you're solving for.
Example: Suppose you have a game where you win 1 unit
with probability 0.30 and win 2 units with probability 0.15, and
lose 1 unit with probability 0.55.
EV = 0.55(-1) + 0.30*(1) + 0.15*(2) = 0.05 (5% edge)
Cost is (0.55) / [0.30*(1) + 0.15*(2)] = 11/12 = 0.91666
ROR comes from solving:
R = 0.55 + 0.30*(R^2) + 0.15*(R^3)
=> R = 0.93242
There are an infinite variety of games that will have this same
risk of ruin, but the games themselves will have very different
costs.
This ROR corresponds to a coin flip with p(win) = 0.51749
and p(lose) = 0.48251. For coin flips, cost = ROR.
This ROR also corresponds to a game with a fixed 2:1 payoff
where you win 2 units with probability p(win)=0.35691. Cost
for this game is p(lose)/[2*p(win)].
This can be generalized to a N:1 payoff. The corresponding
game has a loss/win ratio given by:
rho = [R / (1-R)]*(1 - R^N)
after computing this loss/win ratio, you get p(win), p(lose)
and cost from:
p(win) = 1 / (1 + rho)
p(lose) = rho / (1 + rho)
C = rho/N
Here's a cute trick that I don't think is widely known. You can
model games with the "risk equivalent" coin that corresponds to
ROR. Suppose you have a spinner with two sections that are
marked "play for 1:1" and "play for 2:1" and they are sized so
that "1:1" comes up 57.972% of the time and "2:1" comes up
the other 42.028% of the time. When the spinner lands on
"1:1" you take your biased coin and flip it once. You win
with probability 0.51749, and since you play this sub-game
with probability 0.57972 your overall probability of winning
one unit is 0.51749*0.57972=0.30. When the spinner lands
on "2:1" you take the coin and say "I'm going to treat this coin
as my total bankroll and play until I've either won 2 new coins
for a total bankroll of 3, or lose all". With this same biased coin
you have a 0.35691 probability of surviving until you reach 3
coins, and you play this sub-game 42.028% of the time for an
2:1 win rate of 0.35691*0.42028=0.15. In other words, this
spinner/coin-flip game has an overall probability distribution
that is identical to the "example game" above. Playing an
endless number of rounds of the example game is mathematically
equivalent to an endless number of flips of the risk-equivalent
coin. In terms of starting with a specific bankroll and shooting
for a specific target bankroll, the dynamics of these two games
are identical.
The same process can be applied to VP, for a given strategy.
The overall probability distribution can be used to compute
ROR, and the ROR value corresponds to a single biased coin
that can be used to model the more complex VP game.
Other "equivalent coin" models have been proposed in the
past, but they have been based on approximations derived
from mean and variance. This risk-equivalent coin model
can be used to exactly reconstruct the probability distribution
of the original game. Therefore, it includes not only ROR,
but can provide mean, variance, and all other statistical
moments by virtue of the probability distribution.
I'm planning to write up a more formal derivation of these
results, but I'm still trying to figure out just how "cost" fits
into this ROR picture. They are certainly closely related,
and I know that min-cost is equivalent to min-ROR for
simple games with an N:1 payoff, but I haven't been able
to prove the more general case.
Sorry for the huge tangent.
On Sunday 27 July 2003 12:20 am, Albert Pearson wrote:
Wouldn't they be the same thing by definition?
Steve Jacobs wrote:
> You scale only the winnings. I mentioned this, but I should have
> elaborated. In VP, payoffs are "N for 1" instead of "N to 1" and
> this needs adjusting. For example, with a 50/1 payoff on a 4/kind,
> you treat the first unit as your original bet that is returned to
> you and the 49 other units as winnings, and scale only the 49 units.You know, Steve, I think I follow this intuitively (which at this
point should worry some).There was a "high meter progressive" min cost discussion over on acvpp
recently that involved applying a strategy determined by a paytable at
which the RF value produced a B/E ER.
Right, that is how you minimize the overall cost of playing until you hit
the royal. If you have no competition, that min-cost strategy will maximize
your average final bankroll after playing however long it takes to hit the
royal. By contrast, playing a max-EV strategy will win the most $$/hour
while playing for the royal. You win more quickly, but on average you
leave the casino with fewer dollars in your pocket. I believe I originated
this concept several years ago.
That is a bit of a simplified approach of what you describe by
"scaling" (correct me if I'm wrong). While the "high meter" approach
seeks to minimize loss while waiting for a RF, what you detail should
serve to minimize loss anytime while waiting for all hands. That's at
the expense of return, but serves as a more conservative strategy
designed to protect bankroll.
Excellent summary. You seem to have this pegged.
If not, then I have a couple of follow up questions to better
understand this approach.From a practical standpoint, wouldn't this scaling best be applied to
those hands that present the greater risk of a shortfall in return,
say 4K's and above?
Maybe. If your goal was to play until you hit a "big payoff" then the
min-cost way of doing that would be to scale only the "big payoff"
dollars until you have a B/E game. Then, if you plan to stop playing
once you hit a big payoff, then this strategy would maximize the
average number of dollars in your pocket after you quit.
If not, is it because either the preservation of
game ER would only be nominal, or because the resulting change of game
pressure on bankroll is appreciable?
I don't understand what you mean by "preservation of game ER" or
"game pressure." Generally, min-cost doesn't care about ER except in
terms of final bankroll after reaching some predefined goal. One way
I think about this is that max-EV is "urgent" and cares about time -- it
wants results NOW. Min-cost is more laid-back, and doesn't care how
long it takes to reach the destination, only about how much is spent
along the way. Max-EV is like driving at maximum speed, no matter
how much gas is wasted. Min-cost is trying to maximize gas mileage,
no matter how long it takes to reach the goal. They are opposing
concepts, a yin and yang.
Is the incentive to adopt a min loss strategy highest when the
bankroll margin for game play is thin?
Not sure what "bankroll margin" means.
Perhaps just the opposite. If the max-EV strategy happens to give
a breakeven game, then the min-cost and max-EV strategies become
identical. As game EV strays further from breakeven, whether in the
positive or negative directions, the min-cost strategy will become more
and more different from the max-EV strategy.
But it you're completely free to either use a min-cost or max-EV strategy
or anything else, then the right choice depends on what you decide your
true objective should be. If you want to maximize your income rate in
$$/hour, then max-EV is probably what you want. If you don't care how
quickly you win the money, but want to maximize average final outcome,
then min-cost is probably what you want. If you want maximum bankroll
growth while shooting for an infinite bankroll, then log-optimal/Kelly play
is probably what you're after.
To me "incentive" is about your objective, and not about the characteristics
of the game. You can choose the best game from those offered, in terms
of reaching your objective. If you are only offered a single game, you can
find the strategy that is optimal for your objective. The optimal strategy
might change with bankroll, or it might not. For example, the min-cost
strategy for hitting a royal is a fixed strategy. If your goal is "hit the
royal or bust" then bankroll isn't a factor, and the strategy is the same
whether you are down to your last unit or you just hit a straight flush.
A variable strategy would be best for a goal like "play to increase your
bankroll from $10,000 to $30,000" As you get close to the $30K target,
you'd alter the strategy to forego big payoffs. The min-cost way to hit
a $30K target would probably treat any payoff that overshoots the goal
as if it were only large enough to reach the goal. If you have $29K and
the royal pays $2K, you'd treat the royal as a $1K payoff and compute
a min-cost strategy based on that. When your at $29,999 you might
treat all payoffs as equal, and play to minimize the cost of "buying" that
last dollar.
I hope something in there helped asnwer your questions. Some of this
gets into stuff that I don't think anyone has solved yet.
On Sunday 27 July 2003 12:05 pm, Harry Porter wrote:
Steve: I wouldn't attempt to argue with you on the math of the problem, I'm
sure you are correct.
Rats! I was hoping I'd missed some simple connection. Please don't
assume I'm correct -- I'm often wrong and even more often I don't quite
understand the questions that are posed.
When in doubt, question everything and everybody.
If you think in terms of goals of the average player, when they say risk of
ruin, they probably mean minimum cost.
i.e. Joe Player plans a trip to the casino and he has a fixed bankroll. The
only thing that he wants to know is how to maximize his playing time. He
doesn't care if it is called ROR or Minimum Cost strategy. He just wants
to get some action, have a chance at getting lucky and have a reasonable
estimate of what to expect.So, what do we call that ?
Sounds to me like "maximize playing time." For -EV games, the optimal
strategy for that would probably be "bet the minimum and use min-cost
playing strategy".
In contrast, the min-ROR approach for -EV games really only makes
sense if there is a finite goal like "double your bankroll" since -EV
implies that they player will eventually lose everything if they don't
limit their playing time. So, min-ROR for a "double your bankroll"
goal in a -EV game would use min-ROR playing strategy (same
as min-cost if my intuition is correct) and would employ large bets
in order to minimize total amount wagered while reaching for the
goal.
I suggest that there should be 4 strategies for every game.
At least. I'd claim there are an infinite variety of strategies, each
corresponding to a different objective.
1. The perfect strategy that maximizes E.V., for the Pros and those that
get pleasure out of playing as best as they can. i.e. advanced strategy
from VPSM.
I'm trying hard to dispell the myth that max-EV is THE optimal strategy. In
my view, it is only one of many different optimal strategies. Part of my
motivation for talking about min-ROR and min-cost strategies is to
illustrate that different objectives call for different approaches, and to
give concrete examples of optimal strategies that have an objective
different from max-EV.
In my view, the question "what is the best strategy" is an incomplete
question. When seeking optimal strategies, the question should always
have the form "what is the best way to maximize (or minimize) X" where
X represents a quantifiable objective. The best way to maximize EV is
different than the best way to minimize ROR. The shortest land path from
New York to LA is different than the path that minimizes travel time, and
both are different than the path which maximizes gas mileage along the
way. None are "superior" to the others, they are simply based on
different objectives that call for different paths.
2. The simplified strategy, which gives a high E.V. but sacrifices
perfection for the sake of simplicity, for the use of the majority. i.e.
basic strategy from VPSM.
Certainly a practical strategy. Another would be a simplified strategy
which is nearly min-cost.
3. The extremely simplified strategy which still has a good E.V. but is
extremely easy to learn, this would be good for beginners or for those that
don't want to spend hours in practicing , but still want a run for their
money.
The "quick and dirty" approach. Sure, why not.
4. The conservative strategy (minimum cost) , that gives up some E.V. for
the sake of increasing short term length of play, for the player with
limited funds.
Min-cost isn't just for players with limited funds, it is a completely
different philosophy than max-EV. Either objective gives one
consistent strategy that applies whether the player has only one
unit left, or has a large bankroll.
It is true that one must "give up" EV in order to minimize cost, but it
is equally true that one must "give up" average final bankroll (or
degree of certainty for reaching a target) in order to maximize EV.
In short, it is a trade-off. You can't optimize everything at the same
time, so you have to choose which objective best fits your personal
style. Max-EV isn't automatically the correct answer to every
question seeking the best way to play.
Unfortunately, gambling literature has been so focused on max-EV
for so long that the average educated gambler tends to think that
max-EV is the only rational objective. Some trade in their "max-EV
religion" for a "Kelly/log-optimal" religion, but still retain the flawed
notion that there is "one true way" to optimize play.
By suggesting the idea of using 4 different strategies for different
types of players, you've taken a rare step toward a broader view.
The tricky part is letting go of the idea that max-EV is somehow
superior to other objectives. That is a hard step to make because
most of the "experts" haven't taken that step themselves. My view
is definitely outside the main stream, but I have confidence that
the experts will eventually see the light 
Thanks for the feedback and discussion.
On Sunday 27 July 2003 10:15 pm, Albert Pearson wrote:
Steve,
Could you give us an example of an actual min-cost strategy? How difficult
is it to derive?
Dick
Steve: I have to clarify my example.
The subject is a player with a fixed amount of money say $200 . The player is going to go to a casino for the day and wants to get let's say a minimum of 8 hours play for his money. The player wants to play max coin quarters, so that he may be able to get lucky and walk out with a nice profit.
Our player doesn't mind if he loses his money, but he does want to get a good days worth of action.
In your example of min cost, he would last but couldn't in any likelihood make a reasonable score.
Does he just go for the game with the lowest variance ?
Does he go with a higher variance and change strategies ?
What do we call this type of play ?
Regards
A.P.
----- Original Message -----
From: Steve Jacobs
To: vpFREE@yahoogroups.com
Sent: Monday, July 28, 2003 1:20 AM
Subject: Re: [vpFREE] Re: "minimize ROR" and/or "maximize EV"
> If you think in terms of goals of the average player, when they say risk of
> ruin, they probably mean minimum cost.
> i.e. Joe Player plans a trip to the casino and he has a fixed bankroll. The
> only thing that he wants to know is how to maximize his playing time. He
> doesn't care if it is called ROR or Minimum Cost strategy. He just wants
> to get some action, have a chance at getting lucky and have a reasonable
> estimate of what to expect.
>
> So, what do we call that ?
Sounds to me like "maximize playing time." For -EV games, the optimal
strategy for that would probably be "bet the minimum and use min-cost
playing strategy".
In contrast, the min-ROR approach for -EV games really only makes
sense if there is a finite goal like "double your bankroll" since -EV
implies that they player will eventually lose everything if they don't
limit their playing time. So, min-ROR for a "double your bankroll"
goal in a -EV game would use min-ROR playing strategy (same
as min-cost if my intuition is correct) and would employ large bets
in order to minimize total amount wagered while reaching for the
goal.
[Non-text portions of this message have been removed]
I need to make some modifications to my VP program in order to support
"general" min-cost strategies. I have no time tonight and I'll be offline for
the next week (going to Vegas for BARGE).
The phrase "min-cost" can be used to describe a lot of different strategies.
The first form of min-cost strategy that I computed was for minimizing the
cost of playing for a royal flush. I'm sure the min-cost-of-royal strategy
could be worked out with VPSM, since you just have to adjust the payout
for the royal flush until the max-EV strategy gives a breakeven game.
I posted messages in the last couple of days that outlined the procedure
for computing the more general min-cost strategy. It is an iterative process,
where the positive payoffs are scaled uniformly while the loss payoff is
left alone. You can use trial-and-error to narrow down the scaling factor
to find the number that gives a breakeven game when computing the
max-EV strategy. However, this requires a program that allows non-integer
payoffs, and you need to scale only the fraction of the payoff that represent
game (i.e. scale only 49 units of a 50-for-1 payoff).
So, the process isn't difficult, just a little bit tedious -- for me it is
easier to just modify my program, but my program doesn't support
wild cards.
I don't think this will cause a huge shift in the strategy for most
VP games. I would expect a few plays to trade places in a
prioritized list.
On Monday 28 July 2003 01:01 pm, Dick Kalagher wrote:
Steve,
Could you give us an example of an actual min-cost strategy? How difficult
is it to derive?
Steve: I have to clarify my example.
The subject is a player with a fixed amount of money say $200 . The player
is going to go to a casino for the day and wants to get let's say a minimum
of 8 hours play for his money. The player wants to play max coin quarters,
so that he may be able to get lucky and walk out with a nice profit. Our
player doesn't mind if he loses his money, but he does want to get a good
days worth of action.
I'm afraid that description is too vague for me to formulate a mathematical
model of this player's objective.
In your example of min cost, he would last but
couldn't in any likelihood make a reasonable score.
That coin-flip example was extreme. The shift in strategy for a VP game
would be much more subtle, and would likely just cause a few plays
to trade positions in the priority list.
Does he just go for the game with the lowest variance ?
Probably not, min-ROR doesn't translate directly into variance, except
by a (potentially poor) approximation.
Does he go with a higher variance and change strategies ?
Same answer as above. I'm sorry, I know that isn't very helpful, but
variance is "helpful" in some situations but not in others.
What do we call this type of play ?
What you really need, to get a feel for the difference, is a real min-cost
strategy that can be compared to the corresponding max-EV strategy
in order to contrast the difference in results. I'm going offline for a week,
so I can't do that right away, but I'll try to work out a specific example
after I return. Don't expect an earth-shaking difference, the strategy
isn't likely to look all that different than the max-EV strategy.
--Steve
On Monday 28 July 2003 08:26 pm, Albert Pearson wrote:
Albert....your friend would like to 1) Lose no more than $200.
2) Play for at least 8 hours. 3) Have a chance to get "lucky"
and win. The only way I see to satisfy all three criteria would
be a form of reverse martingale. Let him start max quarters till
he has lost $100. Then go to the dime machine. When he has lost
$50. playing max dimes, he takes his last $50. to the nickel
machines. (We will skip going to the penny machines as you now
get the idea.) He can easily spend 8 hours doing this. If he is
really "lucky", he may hit early and hard for that is what "luck"
is all about. With some luck he may get back at the nickel machines
what he lost at quarters and dimes. In any case, he had his "chance"
at a good win, and satisfied the other criteria. If he were to just
stay at max quarters for $200, unless he was super "lucky" that
day, he will lose it all in less than two hours.
Elliot
--- In vpFREE@yahoogroups.com, "Albert Pearson" <a-p@s...> wrote:
Steve: I have to clarify my example.
The subject is a player with a fixed amount of money say $200 .
The player is going to go to a casino for the day and wants to get
let's say a minimum of 8 hours play for his money. The player wants
to play max coin quarters, so that he may be able to get lucky and
walk out with a nice profit.
Our player doesn't mind if he loses his money, but he does want to
get a good days worth of action.
In your example of min cost, he would last but couldn't in any
likelihood make a reasonable score.
Does he just go for the game with the lowest variance ?
Does he go with a higher variance and change strategies ?
What do we call this type of play ?Regards
A.P.
Steve Jacobs wrote:
I'm trying hard to dispell the myth that max-EV is THE optimal strategy. In
my view, it is only one of many different optimal strategies.
I'm with you here. I'm surprised by how inconsistent even successful
gamblers are in how they treat the value of a game and how they treat
the strategy to the game. If a strategy for a particular hand has a
tiny increase in EV, they'll religiously do it even if it involves a
huge increase in fluctuation, but they'd never dream of playing, say,
a $100 machine with a .1% advantage that would dramatically increase
their hourly EV.
Steve Jacobs wrote:
The exact ROR for VP games can be computed in a more direct fashion,
using the "characteristic equation" which represents the game. I don't
have time right now to go into detail (and I'll be at BARGE next week,
and thus offline) but I feel this is a vastly superior approach for answering
questions that involve risk of ruin. In addition, this approach allows any
game to be modelled with a "risk-equivalent coin".
I'm interested in learning about this approach. How to incorporate EV
and the negative effect of fluctuation into an optimal strategy has
always been a struggle for me. I've also struggled with whether how
much less large payoffs should be valued in estimating overall EV is
the same as how much less they should be valued in determining optimal
strategy.
Steve Jacobs wrote:
Example: Suppose you have a game where you win 1 unit
with probability 0.30 and win 2 units with probability 0.15, and
lose 1 unit with probability 0.55.EV = 0.55(-1) + 0.30*(1) + 0.15*(2) = 0.05 (5% edge)
Cost is (0.55) / [0.30*(1) + 0.15*(2)] = 11/12 = 0.91666
I'd like to be sure I understand how you arrive at the "min-cost"
strategy. Would the "min-cost" strategy assume that the possible wins
were, rather than 1 and 2, 55/60 and 55*2/60?
Here's a cute trick that I don't think is widely known. You can
model games with the "risk equivalent" coin that corresponds to
ROR. Suppose you have a spinner with two sections that are
marked "play for 1:1" and "play for 2:1" and they are sized so
that "1:1" comes up 57.972% of the time and "2:1" comes up
the other 42.028% of the time. When the spinner lands on
"1:1" you take your biased coin and flip it once. You win
with probability 0.51749, and since you play this sub-game
with probability 0.57972 your overall probability of winning
one unit is 0.51749*0.57972=0.30. When the spinner lands
on "2:1" you take the coin and say "I'm going to treat this coin
as my total bankroll and play until I've either won 2 new coins
for a total bankroll of 3, or lose all". With this same biased coin
you have a 0.35691 probability of surviving until you reach 3
coins, and you play this sub-game 42.028% of the time for an
2:1 win rate of 0.35691*0.42028=0.15. In other words, this
spinner/coin-flip game has an overall probability distribution
that is identical to the "example game" above. Playing an
endless number of rounds of the example game is mathematically
equivalent to an endless number of flips of the risk-equivalent
coin. In terms of starting with a specific bankroll and shooting
for a specific target bankroll, the dynamics of these two games
are identical.The same process can be applied to VP, for a given strategy.
The overall probability distribution can be used to compute
ROR, and the ROR value corresponds to a single biased coin
that can be used to model the more complex VP game.
I don't see how this approach can be applied to video poker, since
there are no subsets of video poker that have varying numbers of
trials as was the case in your example.
Steve Jacobs wrote:
>Example: Suppose you have a game where you win 1 unit
>with probability 0.30 and win 2 units with probability 0.15, and
>lose 1 unit with probability 0.55.
>
>EV = 0.55(-1) + 0.30*(1) + 0.15*(2) = 0.05 (5% edge)
>
>Cost is (0.55) / [0.30*(1) + 0.15*(2)] = 11/12 = 0.91666I'd like to be sure I understand how you arrive at the "min-cost"
strategy. Would the "min-cost" strategy assume that the possible wins
were, rather than 1 and 2, 55/60 and 55*2/60?
In order to have a strategy, we'd need two or more different plays
to compare in order to choose which play minimizes the overall
cost. The example above merely shows how cost is computed
from a probability distribution. The cost is the mean value of
losses divided by the mean value of wins.
>Here's a cute trick that I don't think is widely known. You can
>model games with the "risk equivalent" coin that corresponds to
>ROR. Suppose you have a spinner with two sections that are
>marked "play for 1:1" and "play for 2:1" and they are sized so
>that "1:1" comes up 57.972% of the time and "2:1" comes up
>the other 42.028% of the time. When the spinner lands on
>"1:1" you take your biased coin and flip it once. You win
>with probability 0.51749, and since you play this sub-game
>with probability 0.57972 your overall probability of winning
>one unit is 0.51749*0.57972=0.30. When the spinner lands
>on "2:1" you take the coin and say "I'm going to treat this coin
>as my total bankroll and play until I've either won 2 new coins
>for a total bankroll of 3, or lose all". With this same biased coin
>you have a 0.35691 probability of surviving until you reach 3
>coins, and you play this sub-game 42.028% of the time for an
>2:1 win rate of 0.35691*0.42028=0.15. In other words, this
>spinner/coin-flip game has an overall probability distribution
>that is identical to the "example game" above. Playing an
>endless number of rounds of the example game is mathematically
>equivalent to an endless number of flips of the risk-equivalent
>coin. In terms of starting with a specific bankroll and shooting
>for a specific target bankroll, the dynamics of these two games
>are identical.
>
>The same process can be applied to VP, for a given strategy.
>The overall probability distribution can be used to compute
>ROR, and the ROR value corresponds to a single biased coin
>that can be used to model the more complex VP game.I don't see how this approach can be applied to video poker, since
there are no subsets of video poker that have varying numbers of
trials as was the case in your example.
The subgames are based purely on the probability distribution, so
any game with more than one level of win will have multiple
subgames.
For video poker, you treat the different possible outcomes as
"subgames" where you risk your original bet for a N:1 payoff.
For example, with 9/6 JOB the payoffs (converted from N-for-1
to N-to-1) are:
799 Royal
49 Str-Flush
24 4/kind
8 Full House
5 Flush
3 Straight
2 3/kind
1 Two Pair
(push) High Pair
I don't have the exact numbers for dividing up the "spinner",
but as an example if you spin and it lands on 4/kind, you would
model this subgame as a series of flips of a biased coin, where
the series ends when you either accumulate 24 extra coins,
or lose all. If the first flip loses, you count the outcome as
a loss of one coin. If you happen to win 24 consequetive
flips, the outcome is a win. If you win the first two flips then
lose three in a row, it is an overall loss. There are many
winning sequences and many losing sequences for each
of the subgames, but they are all based on repeated flips
of the same biased coin.
I'll try to put together a complete example of this when I have
more time. The 9/6 JOB game is so close to breakeven that
the ROR of the equivalent coin will be 0.9999... -- the first
several digits will be nines. Perhaps it would be best if I
show a nearly breakeven game like 9/6 JOB and compare
that to a game that isn't so close to breakeven.
On Wednesday 13 August 2003 03:18 am, Tom Robertson wrote: