vpFREE2 Forums

Lenny Frome Article

Elliot Frome wrote (snip):

I think the difference in our
points of view come from whether we consider playing a hand of VP
followed by Double-Up as a single gambling event or two distinct
events.

I agree with that. And since you can't Double Up until you have had a
win (or push) on the VP game, they are not two distinct events.
Linked events can not be treated as if they were independent. Ask a
statistician.

If I throw 5 coins into a VP machine and hit a High Pair returning
the 5 coins, and then decide to play Double-Up, in my mind I have
played 10 coins. If I win the Double-Up and stop, the machine
returns 10 coins. I consider this a return of 15 coins (5 for the
High Pair, an additional 10 for the Double-Up).

In your mind you have played 10 coins, but in reality you have
deposited only five coins. You risked five coins, were given an
option to get them back or to risk them again. That's only five coins
in, not ten.

Let's extend your example to higher paying hands. Suppose you insert
five coins, and you get a flush. The payoff is 30 coins. You choose
to Double Up. By your definition, you have now wagered a total of 35
coins. You have a 50% probability of receiving 60 coins, else you
receive nothing, so that's an average of 30 coins out. By your
definition, that's an average of 30 coins received for a total
35-coin wager, or only 85.7% payback. Does that convince you that
there's something wrong with this math?

If, on the other hand, you consider this to be a single event, you
will consider it to be a return of 10 coins with a coin-in of 5.

Exactly right. Twice the payoff with half the probability of receiving it.

I suggest you write a simulation program. To simplify things, you
could replace the Jacks or Better game with one that simply returns
your bet 99.54% of the time, and then Double Up with 50% probability
every time you do receive that payoff. Then compare coin in (not
action) with coin out. The payback will still be 99.54%.

Dan

···

--
Dan Paymar, author of "Video Poker - Optimum Play"
Editor and publisher of "Video Poker Times" newsletter
Visit my web site at www.OptimumPlay.com

"Chance favors the prepared mind"
- Louis Pasteur

[Non-text portions of this message have been removed]

I'm a novice at this cutting and pasting someone else's post and
then replying, so forgive me if I mess it up a bit.

Let's extend your example to higher paying hands. Suppose you

insert

five coins, and you get a flush. The payoff is 30 coins. You

choose

to Double Up. By your definition, you have now wagered a total of

35

coins. You have a 50% probability of receiving 60 coins, else you
receive nothing, so that's an average of 30 coins out. By your
definition, that's an average of 30 coins received for a total
35-coin wager, or only 85.7% payback. Does that convince you that
there's something wrong with this math?

There's a lot wrong with this math, but you haven't kept true to my
methodology. You wagered 5 coins and won 30. 30/5 = payback of
600%. you now wagered THAT 30 with a 50/50 shot of 0 or 60 for an
EV of 30.

You then calculated an 85.7% payback by saying 30/35, but that's not
how I would calculate it. I would use 60/35 (30 that you won at VP
and an EV of 30 for Double-Up). This calculates to 171.4% return.

I never said to use an average, but a WEIGHTED average. So, a 600%
return on a 5 coin wager and a 100% return on a 30 coin wager. So,
if you weight the 100% 6 times greater than the 600% (there are many
ways to do this, but let's go with writing 100 down 6 times and 600
once and the total is 1200% divided by 7 (6+1) and you get the same
171.4%!)

Nothing wrong with that math!

Let's look at this a different way. You put 1 coin in VP and you
hit a Flush paying 5 (you're stuck on an 8/5 machine!). You now bet
5 and get a Jacks or Better and you get your 5 back. What is the
return on your 'session'?

In my eyes, I wagered 6 coins and won back 10 (5 each time). 10/6
is 166.67%. I may have 5 times the amount of money I started with,
but that is not EV or payback.

A computer simulation will not solve our 'dispute'. We are
disagreeing on the calculation, not the result. In your example,
unless we both agree that coin in was 35 and returned was 60, no
computer simulation will do us any good.

It would appear that you are looking at the end result. I started
with 1 coin, I ended with 6 coins, thus a 600% return. This is a
correct calculation of the 'return of your money'.

let's say I walk into a casino with 4 quarters in my pocket. If I
put 1 coin into a VP machine and play 1000 hands. The first 999
miraculously come up High Pair and the last one comes up a loser?
What was my payback? 0 coins returned, 1 wagered? 0%? or I walk
away with 3 coins, started with 4 so it's 75%?

Let's go to the old days before 'credits' on the machine, and rather
than playing the same 1 'coin', I start with a bucket of 1000
quarters. I play them one at a time and for the first 999 hands I
get my High Pair and for the last one I get a loser. The bin at the
bottom now has 999 quarters. What's the return on my session?
would ANYONE argue it's 99.9%?

I realize the Double-Up option doesn't allow you this choice.
It 'holds' the money and you decide to play or not. Once you decide
to 'play' you start a whole new game with it's own EV and this EV
must be weighted in the overall payback. Because Double-Up has a
100% EV, it's easy to 'lose' it.

What if Double-Up had a 99% payback stand alone? what would its
impact to the overall experience be? It certainly couldn't have NO
impact to the combined game.

I'm glad a 10 year old article can generate such interest!

Elliot

Elliot Frome wrote (snip):
> I think the difference in our
>points of view come from whether we consider playing a hand of VP
>followed by Double-Up as a single gambling event or two distinct
>events.

I agree with that. And since you can't Double Up until you have

had a

win (or push) on the VP game, they are not two distinct events.
Linked events can not be treated as if they were independent. Ask

a

statistician.

>If I throw 5 coins into a VP machine and hit a High Pair returning
>the 5 coins, and then decide to play Double-Up, in my mind I have
>played 10 coins. If I win the Double-Up and stop, the machine
>returns 10 coins. I consider this a return of 15 coins (5 for the
>High Pair, an additional 10 for the Double-Up).

In your mind you have played 10 coins, but in reality you have
deposited only five coins. You risked five coins, were given an
option to get them back or to risk them again. That's only five

coins

in, not ten.

Let's extend your example to higher paying hands. Suppose you

insert

five coins, and you get a flush. The payoff is 30 coins. You

choose

to Double Up. By your definition, you have now wagered a total of

35

coins. You have a 50% probability of receiving 60 coins, else you
receive nothing, so that's an average of 30 coins out. By your
definition, that's an average of 30 coins received for a total
35-coin wager, or only 85.7% payback. Does that convince you that
there's something wrong with this math?

>If, on the other hand, you consider this to be a single event, you
>will consider it to be a return of 10 coins with a coin-in of 5.

Exactly right. Twice the payoff with half the probability of

receiving it.

I suggest you write a simulation program. To simplify things, you
could replace the Jacks or Better game with one that simply

returns

your bet 99.54% of the time, and then Double Up with 50%

probability

···

--- In vpFREE@yahoogroups.com, Dan Paymar <Dan@O...> wrote:

every time you do receive that payoff. Then compare coin in (not
action) with coin out. The payback will still be 99.54%.

Dan

--
Dan Paymar, author of "Video Poker - Optimum Play"
Editor and publisher of "Video Poker Times" newsletter
Visit my web site at www.OptimumPlay.com

"Chance favors the prepared mind"
- Louis Pasteur

[Non-text portions of this message have been removed]

Elliot Frome wrote (snip):
> I think the difference in our
>points of view come from whether we consider playing a hand of VP
>followed by Double-Up as a single gambling event or two distinct
>events.

I agree with that. And since you can't Double Up until you have had a
win (or push) on the VP game, they are not two distinct events.
Linked events can not be treated as if they were independent. Ask a
statistician.

Let's not confuse the concept of "independent" with the concept of
"distinct." The fact that two events are not independent does not imply
that they must be treated as a single event. The two events called
"conception" and "childbirth" can clearly be considered to be distinct
events even though one is dependent on the other. Similarly, the
doube-up may be treated either as a single event or as two distinct
where the second event is dependent on the first.

>If I throw 5 coins into a VP machine and hit a High Pair returning
>the 5 coins, and then decide to play Double-Up, in my mind I have
>played 10 coins. If I win the Double-Up and stop, the machine
>returns 10 coins. I consider this a return of 15 coins (5 for the
>High Pair, an additional 10 for the Double-Up).

In your mind you have played 10 coins, but in reality you have
deposited only five coins. You risked five coins, were given an
option to get them back or to risk them again. That's only five coins
in, not ten.

Deposited doesn't matter. You could deposit them then cash out
without playing, and lather/rinse/repeat until any desired level of
"deposited coins" has been reached. Not mathematically significant
in any meaningful sense.

Let's extend your example to higher paying hands. Suppose you insert
five coins, and you get a flush. The payoff is 30 coins. You choose
to Double Up. By your definition, you have now wagered a total of 35
coins. You have a 50% probability of receiving 60 coins, else you
receive nothing, so that's an average of 30 coins out. By your
definition, that's an average of 30 coins received for a total
35-coin wager, or only 85.7% payback. Does that convince you that
there's something wrong with this math?

You didn't count the 30 coins out from winning the flush. That was
the outcome of an event. Then he chose to risk 30 coins on a double
up proposition with a 100% ER. The true overall ER will depend on
the probability of winning the flush. The overall probability distribution
would include probability of -1 unit, + 30 units, and -30 units, and
the overall ER (if computed with proper accounting for all events)
would be different than the ER for just the flush alone. In this case,
where only favorable outcomes are used, the effect of the double up
would be to dilute the ER.

>If, on the other hand, you consider this to be a single event, you
>will consider it to be a return of 10 coins with a coin-in of 5.

Exactly right. Twice the payoff with half the probability of receiving it.

That is one correct way of viewing it, and nobody is claiming that
your way of viewing it is wrong. It just isn't the _only_ way.

I suggest you write a simulation program. To simplify things, you
could replace the Jacks or Better game with one that simply returns
your bet 99.54% of the time, and then Double Up with 50% probability
every time you do receive that payoff. Then compare coin in (not
action) with coin out. The payback will still be 99.54%.

Not if you _correctly_ account for _all_ dependent events separately
from the independent events.

···

On Monday 29 March 2004 11:33 am, Dan Paymar wrote:

I thank both poster for their insights regarding the double up option
on VP machine, both authors are very knowledgable VP writer. I am a
very new VP player, but after think about the postings I concluded
(for what it worths) that double up does not increase the EV. The VP
game and the double up are 2 different game and not one hand. They
are independent of each other. However this is not the reason that
double up does not increase the EV. Becasue if double up pay out
less 100% in the long run, it will negativly impact whatever VP game
you play ;Vice versa, if double up is a positive game (over 100%),
then it will improve whatever VP game EV you are playing. However,
because double up is a even (exactly 100%) game, that means in a long
run, the amount of money you double up (it doesn't matter you double
up everything even RF or just High card pair) the EV will come out
the same. So what double up does is just increase the standard
variation but not the pay out and there is no advantge to neither the
player or the casino.

I hope this help.

--- In vpFREE@yahoogroups.com, "Elliot Frome" <compuflyers@p...>
wrote:

I'm a novice at this cutting and pasting someone else's post and
then replying, so forgive me if I mess it up a bit.

> Let's extend your example to higher paying hands. Suppose you
insert
> five coins, and you get a flush. The payoff is 30 coins. You
choose
> to Double Up. By your definition, you have now wagered a total of
35
> coins. You have a 50% probability of receiving 60 coins, else you
> receive nothing, so that's an average of 30 coins out. By your
> definition, that's an average of 30 coins received for a total
> 35-coin wager, or only 85.7% payback. Does that convince you that
> there's something wrong with this math?

There's a lot wrong with this math, but you haven't kept true to my
methodology. You wagered 5 coins and won 30. 30/5 = payback of
600%. you now wagered THAT 30 with a 50/50 shot of 0 or 60 for an
EV of 30.

You then calculated an 85.7% payback by saying 30/35, but that's

not

how I would calculate it. I would use 60/35 (30 that you won at VP
and an EV of 30 for Double-Up). This calculates to 171.4% return.

I never said to use an average, but a WEIGHTED average. So, a 600%
return on a 5 coin wager and a 100% return on a 30 coin wager. So,
if you weight the 100% 6 times greater than the 600% (there are

many

ways to do this, but let's go with writing 100 down 6 times and 600
once and the total is 1200% divided by 7 (6+1) and you get the same
171.4%!)

Nothing wrong with that math!

Let's look at this a different way. You put 1 coin in VP and you
hit a Flush paying 5 (you're stuck on an 8/5 machine!). You now

bet

5 and get a Jacks or Better and you get your 5 back. What is the
return on your 'session'?

In my eyes, I wagered 6 coins and won back 10 (5 each time). 10/6
is 166.67%. I may have 5 times the amount of money I started with,
but that is not EV or payback.

A computer simulation will not solve our 'dispute'. We are
disagreeing on the calculation, not the result. In your example,
unless we both agree that coin in was 35 and returned was 60, no
computer simulation will do us any good.

It would appear that you are looking at the end result. I started
with 1 coin, I ended with 6 coins, thus a 600% return. This is a
correct calculation of the 'return of your money'.

let's say I walk into a casino with 4 quarters in my pocket. If I
put 1 coin into a VP machine and play 1000 hands. The first 999
miraculously come up High Pair and the last one comes up a loser?
What was my payback? 0 coins returned, 1 wagered? 0%? or I walk
away with 3 coins, started with 4 so it's 75%?

Let's go to the old days before 'credits' on the machine, and

rather

than playing the same 1 'coin', I start with a bucket of 1000
quarters. I play them one at a time and for the first 999 hands I
get my High Pair and for the last one I get a loser. The bin at

the

bottom now has 999 quarters. What's the return on my session?
would ANYONE argue it's 99.9%?

I realize the Double-Up option doesn't allow you this choice.
It 'holds' the money and you decide to play or not. Once you

decide

to 'play' you start a whole new game with it's own EV and this EV
must be weighted in the overall payback. Because Double-Up has a
100% EV, it's easy to 'lose' it.

What if Double-Up had a 99% payback stand alone? what would its
impact to the overall experience be? It certainly couldn't have NO
impact to the combined game.

I'm glad a 10 year old article can generate such interest!

Elliot
> Elliot Frome wrote (snip):
> > I think the difference in our
> >points of view come from whether we consider playing a hand of VP
> >followed by Double-Up as a single gambling event or two distinct
> >events.
>
> I agree with that. And since you can't Double Up until you have
had a
> win (or push) on the VP game, they are not two distinct events.
> Linked events can not be treated as if they were independent. Ask
a
> statistician.
>
> >If I throw 5 coins into a VP machine and hit a High Pair

returning

> >the 5 coins, and then decide to play Double-Up, in my mind I have
> >played 10 coins. If I win the Double-Up and stop, the machine
> >returns 10 coins. I consider this a return of 15 coins (5 for

the

> >High Pair, an additional 10 for the Double-Up).
>
> In your mind you have played 10 coins, but in reality you have
> deposited only five coins. You risked five coins, were given an
> option to get them back or to risk them again. That's only five
coins
> in, not ten.
>
> Let's extend your example to higher paying hands. Suppose you
insert
> five coins, and you get a flush. The payoff is 30 coins. You
choose
> to Double Up. By your definition, you have now wagered a total of
35
> coins. You have a 50% probability of receiving 60 coins, else you
> receive nothing, so that's an average of 30 coins out. By your
> definition, that's an average of 30 coins received for a total
> 35-coin wager, or only 85.7% payback. Does that convince you that
> there's something wrong with this math?
>
> >If, on the other hand, you consider this to be a single event,

you

···

--- In vpFREE@yahoogroups.com, Dan Paymar <Dan@O...> wrote:
> >will consider it to be a return of 10 coins with a coin-in of 5.
>
> Exactly right. Twice the payoff with half the probability of
receiving it.
>
> I suggest you write a simulation program. To simplify things, you
> could replace the Jacks or Better game with one that simply
returns
> your bet 99.54% of the time, and then Double Up with 50%
probability
> every time you do receive that payoff. Then compare coin in (not
> action) with coin out. The payback will still be 99.54%.
>
> Dan
>
> --
> Dan Paymar, author of "Video Poker - Optimum Play"
> Editor and publisher of "Video Poker Times" newsletter
> Visit my web site at www.OptimumPlay.com
>
> "Chance favors the prepared mind"
> - Louis Pasteur
>
> [Non-text portions of this message have been removed]

I respectfully disagree with a portion of what you state.

While I obviously agree with the part of your post that considers
Double-Up to be a seperate game from VP, I disagree with the part
that says because Double-Up is a 100% game that it makes no
difference to EV.

I think we may be confusing a few terms. I agree that in the end
you will have (on average) the same number of coins in your bucket
whether you play Double-Up or not, but this means that your net
win/loss does not change. I'm not sure of the official term for
this... Net win/loss....Return on Investment....etc...

However, by choosing to play Double-Up, you are playing more games
(assuming you keep the number of VP hands constant as we have so far
in our examples), thus you are risking more while winning/losing the
same, which creates a new percentage.

Perhaps a new example will illustrate this better:

You and your buddy have 2 hours before you need to head to the
airport. A hand of VP takes 10 seconds (600/hr). A hand of Double-
Up takes 10 secons (600/hr). (Note: the math will work with any
rate/hr, even if the two games don't have the same rate. I'm just
trying to keep the algebra as simple as possible for this example)

So, you sit down at a 9/6 JOB machine and start playing. Your buddy
does the same. You decide to forgo the double-up option
completely. Your buddy decides to play it after every High Pair
(again, the math will work if he chose to play EVERY hand double-up,
but I'm using a realistic option here).

After 2 hours, you would have played 1200 hands, putting in 6000
coins and miraculously the machine played right to theoretical
numbers and you have 5976 coins. You've lost 24 coins.

Your buddy on the other hand will play 980 VP hands and 220 Double-
Up hands (using an assumption of approx 22.5% of all hands end in a
High pair). He will pump 4900 coins into VP and 1100 coins into
Double-Up. Again, miraculously both games paid out on their
theoretical averages. From VP he will have 4880 returned. From
Double-Up he will have 1100 coins returned. A total of 5980 coins.

You both wagered 6000 coins. You both played 1200 total games of
chance (and skill). Why did your buddy wind up with 4 more coins?

Your EV was 99.6. His was 99.66%

This isn't due to rounding or anything like this. It's a weighted
average 980 occurrences of 99.6 and 220 occurrences of 100.0 (this
works out to be 99.73 a smidge higher because in reality yoru buddy
would have won 4880.4 coins from VP, but you can't get back .4 of a
coin so I dropped it)

If I were to use 6/5 JOB for this example, the difference between
you and your buddy would become a lot more obvious than 4 coins.
You'll wind up with 5730 coins and your buddy with 5779. You with a
payback of .955 (5730/6000) and your buddy with 96.31666.

Elliot

I thank both poster for their insights regarding the double up

option

on VP machine, both authors are very knowledgable VP writer. I am

a

very new VP player, but after think about the postings I concluded
(for what it worths) that double up does not increase the EV. The

VP

game and the double up are 2 different game and not one hand.

They

are independent of each other. However this is not the reason

that

double up does not increase the EV. Becasue if double up pay out
less 100% in the long run, it will negativly impact whatever VP

game

you play ;Vice versa, if double up is a positive game (over 100%),
then it will improve whatever VP game EV you are playing. However,
because double up is a even (exactly 100%) game, that means in a

long

run, the amount of money you double up (it doesn't matter you

double

up everything even RF or just High card pair) the EV will come out
the same. So what double up does is just increase the standard
variation but not the pay out and there is no advantge to neither

the

player or the casino.

I hope this help.

--- In vpFREE@yahoogroups.com, "Elliot Frome" <compuflyers@p...>
wrote:
> I'm a novice at this cutting and pasting someone else's post and
> then replying, so forgive me if I mess it up a bit.
>
> > Let's extend your example to higher paying hands. Suppose you
> insert
> > five coins, and you get a flush. The payoff is 30 coins. You
> choose
> > to Double Up. By your definition, you have now wagered a total

of

> 35
> > coins. You have a 50% probability of receiving 60 coins, else

you

> > receive nothing, so that's an average of 30 coins out. By your
> > definition, that's an average of 30 coins received for a total
> > 35-coin wager, or only 85.7% payback. Does that convince you

that

> > there's something wrong with this math?
>
>
> There's a lot wrong with this math, but you haven't kept true to

my

> methodology. You wagered 5 coins and won 30. 30/5 = payback of
> 600%. you now wagered THAT 30 with a 50/50 shot of 0 or 60 for

an

> EV of 30.
>
> You then calculated an 85.7% payback by saying 30/35, but that's
not
> how I would calculate it. I would use 60/35 (30 that you won at

VP

> and an EV of 30 for Double-Up). This calculates to 171.4%

return.

>
> I never said to use an average, but a WEIGHTED average. So, a

600%

> return on a 5 coin wager and a 100% return on a 30 coin wager.

So,

> if you weight the 100% 6 times greater than the 600% (there are
many
> ways to do this, but let's go with writing 100 down 6 times and

600

> once and the total is 1200% divided by 7 (6+1) and you get the

same

> 171.4%!)
>
> Nothing wrong with that math!
>
> Let's look at this a different way. You put 1 coin in VP and

you

> hit a Flush paying 5 (you're stuck on an 8/5 machine!). You now
bet
> 5 and get a Jacks or Better and you get your 5 back. What is

the

> return on your 'session'?
>
> In my eyes, I wagered 6 coins and won back 10 (5 each time).

10/6

> is 166.67%. I may have 5 times the amount of money I started

with,

> but that is not EV or payback.
>
> A computer simulation will not solve our 'dispute'. We are
> disagreeing on the calculation, not the result. In your

example,

> unless we both agree that coin in was 35 and returned was 60, no
> computer simulation will do us any good.
>
> It would appear that you are looking at the end result. I

started

> with 1 coin, I ended with 6 coins, thus a 600% return. This is

a

> correct calculation of the 'return of your money'.
>
> let's say I walk into a casino with 4 quarters in my pocket. If

I

> put 1 coin into a VP machine and play 1000 hands. The first 999
> miraculously come up High Pair and the last one comes up a

loser?

> What was my payback? 0 coins returned, 1 wagered? 0%? or I

walk

> away with 3 coins, started with 4 so it's 75%?
>
> Let's go to the old days before 'credits' on the machine, and
rather
> than playing the same 1 'coin', I start with a bucket of 1000
> quarters. I play them one at a time and for the first 999 hands

I

> get my High Pair and for the last one I get a loser. The bin at
the
> bottom now has 999 quarters. What's the return on my session?
> would ANYONE argue it's 99.9%?
>
> I realize the Double-Up option doesn't allow you this choice.
> It 'holds' the money and you decide to play or not. Once you
decide
> to 'play' you start a whole new game with it's own EV and this

EV

> must be weighted in the overall payback. Because Double-Up has

a

> 100% EV, it's easy to 'lose' it.
>
> What if Double-Up had a 99% payback stand alone? what would its
> impact to the overall experience be? It certainly couldn't have

NO

> impact to the combined game.
>
> I'm glad a 10 year old article can generate such interest!
>
> Elliot
> > Elliot Frome wrote (snip):
> > > I think the difference in our
> > >points of view come from whether we consider playing a hand

of VP

> > >followed by Double-Up as a single gambling event or two

distinct

> > >events.
> >
> > I agree with that. And since you can't Double Up until you

have

> had a
> > win (or push) on the VP game, they are not two distinct

events.

> > Linked events can not be treated as if they were independent.

Ask

> a
> > statistician.
> >
> > >If I throw 5 coins into a VP machine and hit a High Pair
returning
> > >the 5 coins, and then decide to play Double-Up, in my mind I

have

> > >played 10 coins. If I win the Double-Up and stop, the machine
> > >returns 10 coins. I consider this a return of 15 coins (5

for

the
> > >High Pair, an additional 10 for the Double-Up).
> >
> > In your mind you have played 10 coins, but in reality you have
> > deposited only five coins. You risked five coins, were given

an

> > option to get them back or to risk them again. That's only

five

> coins
> > in, not ten.
> >
> > Let's extend your example to higher paying hands. Suppose you
> insert
> > five coins, and you get a flush. The payoff is 30 coins. You
> choose
> > to Double Up. By your definition, you have now wagered a total

of

> 35
> > coins. You have a 50% probability of receiving 60 coins, else

you

> > receive nothing, so that's an average of 30 coins out. By your
> > definition, that's an average of 30 coins received for a total
> > 35-coin wager, or only 85.7% payback. Does that convince you

that

> > there's something wrong with this math?
> >
> > >If, on the other hand, you consider this to be a single

event,

you
> > >will consider it to be a return of 10 coins with a coin-in of

5.

> >
> > Exactly right. Twice the payoff with half the probability of
> receiving it.
> >
> > I suggest you write a simulation program. To simplify things,

you

> > could replace the Jacks or Better game with one that simply
> returns
> > your bet 99.54% of the time, and then Double Up with 50%
> probability
> > every time you do receive that payoff. Then compare coin in

(not

···

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

> --- In vpFREE@yahoogroups.com, Dan Paymar <Dan@O...> wrote:
> > action) with coin out. The payback will still be 99.54%.
> >
> > Dan
> >
> > --
> > Dan Paymar, author of "Video Poker - Optimum Play"
> > Editor and publisher of "Video Poker Times" newsletter
> > Visit my web site at www.OptimumPlay.com
> >
> > "Chance favors the prepared mind"
> > - Louis Pasteur
> >
> > [Non-text portions of this message have been removed]

--- In vpFREE@yahoogroups.com, "Elliot Frome" <compuflyers@p...>
wrote:

While I obviously agree with the part of your post that considers
Double-Up to be a seperate game from VP, I disagree with the part

this is first year prob&stats
independent v. dependent events
independent probabilities are averaged
dependent probabilities are multiplied

igt double up as it exists today is clearly a dependent event because
you can only double up wins and only in the amount of the win, to be
independent you would have to be allowed to double up after any hand,
win or lose, and in any amount of coins, i.e. bet 4 coins on poker,
oh well lost, now i'll bet 3 coins on double up ...

there might be different "opinions" but there is only one
mathematical truth

treating dependent probabilities as independent would be a rookie
mistake in a first year prob&stats class

if trump was the teacher, he would have to say "you're fired"

And the real world is not always reflected in a Math Class.

Because you have the opportunity to opt out of playing Double-Up they
are not completely dependent.

You and Buddy each play one hand of VP starting with 1 coin. You
each hit a flush for 5 coins. 1 of you chooses to play a 2nd game of
VP, the other choosed to Double-Up.

Are you telling me the expected value is equal for both of you?
Sorry, they're not.

In the example I gave, the Payback of the person's session is altered
by their choice to play Double-Up. If the EV was totally unaffected,
this could not be, but it is.

If in my example, the 2nd person had played an additional 220 hands
of Video Poker (without doubling up), then his loss would've been the
same as his buddies. BUT, he would've played an additional 220 hands
of Double Up, putting at risk an additional 1100 coins. So he
wagered 7100 coins losing 24, while his buddy wagered 6000 coins and
lost 24.

We're not trying to calculate expected net loss or win. We're
calculating a payback which is based on the amount wagered. Once you
have the opportunity to opt out of the game, and choose to do so, the
amount wagered changes.

And as far as I'm concerned, someone needs to fire Donald.

Elliot

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

--- In vpFREE@yahoogroups.com, "Elliot Frome" <compuflyers@p...>
wrote:
> While I obviously agree with the part of your post that considers
> Double-Up to be a seperate game from VP, I disagree with the part

this is first year prob&stats
independent v. dependent events
independent probabilities are averaged
dependent probabilities are multiplied

igt double up as it exists today is clearly a dependent event

because

you can only double up wins and only in the amount of the win, to

be

independent you would have to be allowed to double up after any

hand,

···

win or lose, and in any amount of coins, i.e. bet 4 coins on poker,
oh well lost, now i'll bet 3 coins on double up ...

there might be different "opinions" but there is only one
mathematical truth

treating dependent probabilities as independent would be a rookie
mistake in a first year prob&stats class

if trump was the teacher, he would have to say "you're fired"

One additional thought on this topic.

I think we're simply not comparing apples and apples.

If I start with 5 coins, play a hand of VP, win some coins back and
play Double-Up, I agree that the EV based on the initial 5 coins is
the multiplication of the two events (99.6% * 100%).

However, this is simply NOT the number my father (or I) are
attempting to capture. My father defined Expected Value (for
gambling) as total coins returned over total coins wagered. Any
chance you have to take your coins back, in essence ENDS a wager.
If you choose to leave them up or re-bet them, a new wager takes
place, EVEN if there is some dependence (which is why you need to
take a weighted average based on the number of hands played on
each 'game', as opposed to a standard average).

For a game like Let It Ride, you put up 3 bets. Much of the time,
you pull back 2 of them. Thus, those 2 are only counted as coins
wagered IF they truly go at risk (you choose to 'let it ride').

This is the same reason why there are different numbers given as the
payback or cost/hour of games like Three Card Poker and Caribb.
Some analysts consider only the original bet, some consider the
entire amount put at risk, etc...

Different numbers have different values for different people under
different circumstances.

I guess the good news is that a new article should be out tomorrow
so we can start all over again....

Good luck everyone!

Elliot

--- In vpFREE@yahoogroups.com, "Elliot Frome" <compuflyers@p...>
wrote:

And the real world is not always reflected in a Math Class.

Because you have the opportunity to opt out of playing Double-Up

they

are not completely dependent.

You and Buddy each play one hand of VP starting with 1 coin. You
each hit a flush for 5 coins. 1 of you chooses to play a 2nd game

of

VP, the other choosed to Double-Up.

Are you telling me the expected value is equal for both of you?
Sorry, they're not.

In the example I gave, the Payback of the person's session is

altered

by their choice to play Double-Up. If the EV was totally

unaffected,

this could not be, but it is.

If in my example, the 2nd person had played an additional 220

hands

of Video Poker (without doubling up), then his loss would've been

the

same as his buddies. BUT, he would've played an additional 220

hands

of Double Up, putting at risk an additional 1100 coins. So he
wagered 7100 coins losing 24, while his buddy wagered 6000 coins

and

lost 24.

We're not trying to calculate expected net loss or win. We're
calculating a payback which is based on the amount wagered. Once

you

have the opportunity to opt out of the game, and choose to do so,

the

···

amount wagered changes.

And as far as I'm concerned, someone needs to fire Donald.

Elliot