vpFREE2 Forums

Real cost of an "error"

#1

When dealt two pairs Aces-up in 10/7 Double Bonus Poker, many players
have a strong inclination to hold just the pair of aces, even knowing
that holding the two pairs has a higher EV. At one time, someone on this
forum asked what it cost to hold just the aces. An analyst who is no
longer with us replied 0.01%, but I'm sure that was just an easy way to
say the cost was very small.

I now have software to accurately evaluate a strategy chart, and I
decided to find out. Surprise! There was no change in game's "payback"
(expected return) to five decimal places!!

Conclusion? On a 5-coin $1 machine, the "cost" of this error is about
1.24 cents each time it occurs, but the long-term cost is in the very
low noise level.

My suggestions. If you're a pro, hold the two pair. If you're in Las
Vegas for only an occasional weekend, hold just the pair of aces. For
anyone else, it's basically a toss-up.

Dan Paymar

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

#2

It costs 0.247% of your bet to hold just the aces (1.24 cents on a $1 game
as you say). Aces up are 19008 out of the 2,598,960 possible hands. So the
EV is lowered by 0.00247 * (19008 / 2,598,960) or 0.0018%. So it'll cost you
about $18 for every million dollars you put through the machine. At 1000
hands per hour you'd be throwing away 9 cents/hour. Some folks may have that
kind of money to waste but I don't.

Cogno

···

-----Original Message-----
From: vpFREE@yahoogroups.com [mailto:vpF…@…com] On Behalf Of
Dan Paymar
Sent: Thursday, February 3, 2011 10:18 AM
To: vpFREE@yahoogroups.com
Subject: [vpFREE] Real cost of an "error"

When dealt two pairs Aces-up in 10/7 Double Bonus Poker, many players
have a strong inclination to hold just the pair of aces, even knowing
that holding the two pairs has a higher EV. At one time, someone on this
forum asked what it cost to hold just the aces. An analyst who is no
longer with us replied 0.01%, but I'm sure that was just an easy way to
say the cost was very small.

I now have software to accurately evaluate a strategy chart, and I
decided to find out. Surprise! There was no change in game's "payback"
(expected return) to five decimal places!!

Conclusion? On a 5-coin $1 machine, the "cost" of this error is about
1.24 cents each time it occurs, but the long-term cost is in the very
low noise level.

My suggestions. If you're a pro, hold the two pair. If you're in Las
Vegas for only an occasional weekend, hold just the pair of aces. For
anyone else, it's basically a toss-up.

Dan Paymar

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

------------------------------------

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

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#3

I'm curious, Mr. Scienti?

Mr. Paymar has added this strategy evaluation feature to his OpVP software, that will be in the new version.

I assume you don't have a copy, since it's pre-release. How did you do it. By hand?

If so, kudos!

~FK

#4

I did figure the number of Aces up hands by hand. For the difference in the
value of holding AA v. AAxx, I just used VPW. Pretty much every VP program
has that feature.

Cogno

···

-----Original Message-----
From: vpFREE@yahoogroups.com [mailto:vpF…@…com] On Behalf Of
Frank
Sent: Thursday, February 3, 2011 1:06 PM
To: vpFREE@yahoogroups.com
Subject: [vpFREE] Re: Real cost of an "error"

I'm curious, Mr. Scienti?

Mr. Paymar has added this strategy evaluation feature to his OpVP software,
that will be in the new version.

I assume you don't have a copy, since it's pre-release. How did you do it.
By hand?

If so, kudos!

~FK

------------------------------------

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

Yahoo! Groups Links

#5

Cogno Scienti wrote:

It costs 0.247% of your bet to hold just the aces (1.24 cents on a
$1 game as you say). Aces up are 19008 out of the 2,598,960
possible hands. So the EV is lowered by 0.00247 * (19008 /
2,598,960) or 0.0018%. So it'll cost you about $18 for every
million dollars you put through the machine. At 1000 hands per hour
you'd be throwing away 9 cents/hour. Some folks may have that
kind of money to waste but I don't.

Along the same lines, in my book chasing higher variance at the expense of ER is seldom a smart wager.

Whether pro or rec player, bankroll should be conserved.

- H.

#6

--- In vpFREE@yahoogroups.com, Cogno Scienti <cognoscienti@...> wrote:

Aces up are 19008 out of the 2,598,960 possible hands.

This is the type of math I had to learn when I got into the gambling game. I'm a high school dropout that never even took algebra. But necessity is the mother of invention. I had to teach this stuff to myself past the age of 39 If I can do it anyone can. Trust me, being able to do this kind of math will set you free in the gambling world. Because once you learn it you will no longer be playing follow the leader. You will be the leader.

How did Cogno come up with 19008 as the number of five-card combinations representing Aces Up in a 52 card deck? Others may be able to do it faster but here's the uneducated way. I may be slow but I can get the job done.

There are 4 cards in each rank. How many combinations in that rank make a pair? To be dealt a pair you would have to catch any one of 4 cards then any one of 3 cards. 4 X 3 = 12 permutations. When you catch a specific pair, say 2 of diamonds and 2 of clubs it can come in two different orders. You could catch the diamond first and the club second or the club first and the diamond second. So we divied 12 by 2 which = 6. That's the number of pairs in a rank.

When catching Aces up we know that one of the pairs we're dealt will be Aces and the other pair could be anything from Kings down to deuces. So the equation looks like this:

6X72X44 = 19008

The 6 represents the combinations representing the pair of Aces. The 72 represents the number of pairs in the other ranks....12 ranks X 6 pairs. The 44 represents the number of sidecards one would catch with the hand....this number is easily derived; when you catch two pair the fifth card has to be a card of a different rank than the two pair. 52 minus 8 = 44.

Aces up is the most prevalent two pair hand. The equation for Kings Up is 6X66X44 = 17424. Queens up 6X60X44.... and so on down the line.

#7

Mickey wrote:

<<Aces up is the most prevalent two pair hand.>>

That's true! And it's also true that all two-pair hands are equally likely.
:slight_smile:

Cogno