Alan: Perfect Pairs

[Alan]

In Australia, we have a side bet called Perfect Pairs. I've carefully read the article in Blackjack Attack on Royal Match and I was hoping someone might be able to extrapolate the house edge for me, as it is along similar lines, but maths is not my strength.

For the purposes of this side bet, a pair is a hand where the first 2 cards dealt, in the initial deal only, are either:

Mixed Pair - 2 cards of the same number or picture type but different colour (1 red, 1 black) - pays 5 to 1
Coloured Pair - 2 cards of the same number or picture type and colour but different suits (e.g. clubs and spades) - pays 10 to 1
Perfect Pair - 2 cards of the same number or picture type and suit - pays 30 to 1.

It's an 8 deck game, 60 to 66% penetration.

I normally dismiss side bets out of hand, but I see these "pairs" come up remarkably often, so it's something I'd like to have a quantifiable outcome for.

Would someone be able to explain to me what the house edge is in this side bet? Is there any way that this side bet is exploitable through some sort of counting system?

Thanks.

[Don Schlesinger]

> For the purposes of this side bet, a pair is
> a hand where the first 2 cards dealt, in the
> initial deal only, are either:

> Mixed Pair - 2 cards of the same number or
> picture type but different colour (1 red, 1
> black) - pays 5 to 1

The first card you receive, of 416, can be anything at all. To form a mixed pair, you now have to match the rank and be of a different color. As there are 32 of each rank, that makes 16 left of the "off" color. So, you have 16 chances out of 415 to make a mixed pair. And they're paying you FIVE to one????? Very generous of them!! For every 415 bets you make, you win 16 times and when you pick up your chips, you have 6 x 16 = 96 chips in front of you (we'll be needing 415, from all bets, to break even!).

On to the next part.

> Coloured Pair - 2 cards of the same number
> or picture type and colour but different
> suits (e.g. clubs and spades) - pays 10 to 1

Same beginning. Any card will do. This time, once the suit is established, we need to match the color, but NOT match the suit. So, only 8 cards will work this time. When I win the 8 times, I pick up 11 chips each time, for 88 more. That's a grand total of 96 + 88 = 184, on our way (ha!!) to 415, to break even.

Next part.
> Perfect Pair - 2 cards of the same number or
> picture type and suit - pays 30 to 1.

To match perfectly, after the "any" first card is drawn, we have to match it with one of the 7 identical cards that remain. Strange that this pays so much more than the above "colored pair," since the odds are almost identical. so we pick up 31 chips 7 times for another 217 in the till. Add that to the 184, above, and we get a grand total of 401, while we needed 415 to break even.

The shortfall is 14 out of 415, or 3.37%, which is the house edge for this side bet.

> I normally dismiss side bets out of hand,
> but I see these "pairs" come up
> remarkably often, so it's something I'd like
> to have a quantifiable outcome for.

Now you've got it! :-)

> Would someone be able to explain to me what
> the house edge is in this side bet?

Clear?

>Is there any way that this side bet is exploitable
>through some sort of counting system?

I doubt it. I won't say "no," flat out, but it doesn't seem like something you'd like to be spending your time on!

Don

[Alan]

Don,

Thanks for such a clear and prompt answer to my question.

I tried to work it out, along the lines of your Royal Match explanation, but I couldn't seem to get it right. My calculations appeared (and were) way off.

I appreciate your time on this, though you make it seem so simple now, the way you explained it.

Alan
> The first card you receive, of 416, can be
> anything at all. To form a mixed pair, you
> now have to match the rank and be of a
> different color. As there are 32 of each
> rank, that makes 16 left of the
> "off" color. So, you have 16
> chances out of 415 to make a mixed pair. And
> they're paying you FIVE to one????? Very
> generous of them!! For every 415 bets you
> make, you win 16 times and when you pick up
> your chips, you have 6 x 16 = 96 chips in
> front of you (we'll be needing 415, from all
> bets, to break even!).

> On to the next part.

> Same beginning. Any card will do. This time,
> once the suit is established, we need to
> match the color, but NOT match the suit. So,
> only 8 cards will work this time. When I win
> the 8 times, I pick up 11 chips each time,
> for 88 more. That's a grand total of 96 + 88
> = 184, on our way (ha!!) to 415, to break
> even.

> Next part.

> To match perfectly, after the
> "any" first card is drawn, we have
> to match it with one of the 7 identical
> cards that remain. Strange that this pays so
> much more than the above "colored
> pair," since the odds are almost
> identical. so we pick up 31 chips 7 times
> for another 217 in the till. Add that to the
> 184, above, and we get a grand total of 401,
> while we needed 415 to break even.

> The shortfall is 14 out of 415, or 3.37%,
> which is the house edge for this side bet.

> Now you've got it! :-)

> Clear?

> I doubt it. I won't say "no," flat
> out, but it doesn't seem like something
> you'd like to be spending your time on!

> Don

[Susan]

There are many possible subsets of cards that would make this bet attractive. However, tracking them seems impossible.

If you remove exactly one entire rank of cards (e.g. like in Spanish 21) your edge would be 4.7%. Removing two ranks would get you edge of 14.25%.

Let?s look at some slightly more possible, though still unrealistic subsets.
Removing (tracking) of sixteen cards: four cards of four ranks only (one of each suite) gets you EV of 0.50% if rank of the first card is different than the ranks of the removed cards and negative EV of 13.03% if it is one of the four ranks, for total EV of minus 3.29%.
Removing exactly thirty two cards, eight cards of four ranks, two of each suite, would provide edge of 4.7% (first card is of undiminished ranks) and edge of negative 23.5% (!) if rank of the first card is of the four diminished ranks, for total EV of minus 2.35%.

PS
I guess I had 20 minutes to kill ;-),
Susan

[jack in the black]

> In Australia, we have a side bet called
> Perfect Pairs. I've carefully read the
> article in Blackjack Attack on Royal Match
> and I was hoping someone might be able to
> extrapolate the house edge for me, as it is
> along similar lines, but maths is not my
> strength.

> For the purposes of this side bet, a pair is
> a hand where the first 2 cards dealt, in the
> initial deal only, are either:

> Mixed Pair - 2 cards of the same number or
> picture type but different colour (1 red, 1
> black) - pays 5 to 1
> Coloured Pair - 2 cards of the same number
> or picture type and colour but different
> suits (e.g. clubs and spades) - pays 10 to 1
> Perfect Pair - 2 cards of the same number or
> picture type and suit - pays 30 to 1.

> It's an 8 deck game, 60 to 66% penetration.

> I normally dismiss side bets out of hand,
> but I see these "pairs" come up
> remarkably often, so it's something I'd like
> to have a quantifiable outcome for.

> Would someone be able to explain to me what
> the house edge is in this side bet? Is there
> any way that this side bet is exploitable
> through some sort of counting system?

> Thanks.

I did lots of research into this a while back, whilst advantages of 5% or so do occur, they are almost impossible to predict - a counting system would have to more or less keep track of every rank of card. Using a hi-lo the odds were in the players favour at something like true +8, but nothing worthwhile - especially with the penetration at Star City.

[paranoid android]

Is the reason the hi-lo count works at +8 because if we've remove an inordinate number of small cards, it increases the likelihood that the high card values will match? If so, a count of -8 or less should provide the same edge, no?

> I did lots of research into this a while
> back, whilst advantages of 5% or so do
> occur, they are almost impossible to predict
> - a counting system would have to more or
> less keep track of every rank of card. Using
> a hi-lo the odds were in the players favour
> at something like true +8, but nothing
> worthwhile - especially with the penetration
> at Star City.

[alienated]

Nice post. You should kill 20 minutes more often. ;-)

The following is a generic remark, not particularly related to Alan's game/s which may or may not be amenable to this approach. Side bets like the perfect pairs can be attacked through sequencing and/or tracking methods.

With tracking, you can note any clumps (!) of like cards that are discarded together during the current shoe, track them through the next, and hope that two of them fall in the one hand. The viability of this approach obviously depends on the characteristics of the shuffle. The key issues are the number of riffles, determining dispersion, and the type (if any) of stripping. (Stripping does not necessarily increase dispersion.)

One sequencing method is to key the first of a discarded pair and tailor the number of boxes to the number of riffles used in the shuffle. (On the off chance you're interested, I posted at length on certain aspects of this method quite some time ago at CCCafe using my other handle, Ted Forrester. In any search the following key words could be used: keying, naturals, pairs.)

In short, there are a variety of methods based on shuffle analysis that can be used to predict when it is most probable that numerous like cards will be dealt near each other. Devices would of course be the most effective tool for this endeavour, though the least recommended for obvious reasons. But even without a device much can be achieved with a little imagination and a close study of the specific shuffle. Other bodies, even of unskilled players, can be useful to lock up a table.

[Susan]

> Is the reason the hi-lo count works at +8
> because if we've remove an inordinate number
> of small cards, it increases the likelihood
> that the high card values will match? If so,
> a count of -8 or less should provide the
> same edge, no?

You are right that for a balanced, single level count average EV of this side bet is the same whether the count is negative or positive.

What needs to be noticed here is fact that with lesser number of decks house edge gets bigger and bigger.
For example at the top of 8, 4, 2 decks, and single deck, the house edge is respectively 3.34%, 10.63%, 23.8% and 54.9%. The reason for it is that with smaller number of decks your ?first? card is actually removed from the rest of the deck therefore chances of receiving card of the same rank, colour(black or red) and suite becomes proportionally smaller. In the case of one deck the chance of receiving a matching rank and suite is exactly zero.

Different TC will have different effect on EV depending on number of decks left. More decks left and higher TC make smaller (but not small) house edge.

Simple counting is absolutely unproductive here.

With four decks left, even with true counts as high as + (or -) 10 your edge on average would be minus 7.17%, and minus 33% if the first card is from the five depleted ranks, and plus 11.7% if the first card is from the undepleted ranks, and minus 10.6% for neutral cards.

You may want to read post above in the same thread by alienated and his other posts mentioned there, this game could be attacked through sequencing and/or tracking methods.

Even in precisely and well tracked segments positive EV would be not common in this game.

Let?s take one complete deck with additional 3 cards of the same rank and suite. For example, full deck with extra three Aces of clubs, this would give you EV of minus 57.4% if the first card is not an Ace, minus 24% if the first card is a red Ace, only measly 3.7% if the first card is an Ace of spades and great but rare 215% if the first card is an Ace of clubs.

However, imagine a subset of 13 cards that will conclude your hand: two Aces of clubs, two Aces of spades and nine other non-Aces cards each of a different rank. This subset provides you with EV of zero if the first card is non-Ace, 883% if the first card is an Ace, for the total EV of 272%. Nice edge, isn?t it?

Good luck,
Susan Yama

[jack in the black]

[past crazy]

> You are right that for a balanced, single
> level count average EV of this side bet is
> the same whether the count is negative or
> positive.

> What needs to be noticed here is fact that
> with lesser number of decks house edge gets
> bigger and bigger.
> For example at the top of 8, 4, 2 decks, and
> single deck, the house edge is respectively
> 3.34%, 10.63%, 23.8% and 54.9%. The reason
> for it is that with smaller number of decks
> your ?first? card is actually removed from
> the rest of the deck therefore chances of
> receiving card of the same rank,
> colour(black or red) and suite becomes
> proportionally smaller. In the case of one
> deck the chance of receiving a matching rank
> and suite is exactly zero.

> Different TC will have different effect on
> EV depending on number of decks left. More
> decks left and higher TC make smaller (but
> not small) house edge.

> Simple counting is absolutely unproductive
> here.

> With four decks left, even with true counts
> as high as + (or -) 10 your edge on average
> would be minus 7.17%, and minus 33% if the
> first card is from the five depleted ranks,
> and plus 11.7% if the first card is from the
> undepleted ranks, and minus 10.6% for
> neutral cards.

> You may want to read post above in the same
> thread by alienated and his other posts
> mentioned there, this game could be attacked
> through sequencing and/or tracking methods.

> Even in precisely and well tracked segments
> positive EV would be not common in this
> game.

> Let?s take one complete deck with additional
> 3 cards of the same rank and suite. For
> example, full deck with extra three Aces of
> clubs, this would give you EV of minus 57.4%
> if the first card is not an Ace, minus 24%
> if the first card is a red Ace, only measly
> 3.7% if the first card is an Ace of spades
> and great but rare 215% if the first card is
> an Ace of clubs.

> However, imagine a subset of 13 cards that
> will conclude your hand: two Aces of clubs,
> two Aces of spades and nine other non-Aces
> cards each of a different rank. This subset
> provides you with EV of zero if the first
> card is non-Ace, 883% if the first card is
> an Ace, for the total EV of 272%. Nice edge,
> isn?t it?

> Good luck,
> Susan Yama

if i understood you correctly you are saying that as you progress through an 8 deck shoe the house edge goes up?
that sounds crazy to me, maybe i misunderstood

[Susan]

> if i understood you correctly you are saying
> that as you progress through an 8 deck shoe
> the house edge goes up?
> that sounds crazy to me, maybe i
> misunderstood

Yes, we are talking about the matching pairs bet of course.

Look at it this way:
Your first card could be of any rank and color. If you use 1,000 decks there will be almost 1000 times two cards of the same rank and opposite color (black/red), 1000 times one cards of the same rank and same color but not the same suite, and 1000 times one cards same rank and suite ? to be received as a bonus combination.
In a single deck game your first card could be of any rank and color, but now to get a card to match it would be more difficult as you depleted the deck.
In single deck game your chance of getting a pair of same rank and color is decreased by 50% and chance of a pair of same rank and suite is decreased by 100%, compared to infinite deck.
For example if your first card is Ace of clubs you can not get another Ace of clubs to match it, as well, as your chance of getting a black Ace is reduced by 50% because only one Ace of spades is available.

The house edge could be written: [{n*2*6+(n-1)*11+(n-1)*31}-(n*52-1)]/(n*52-1)
Where n is the number of decks used.

Small correction. In the last part of my post, when dealing with the subset of 13 cards and the first card is non-Ace it is the chance for any win that is zero, the EV is minus 100%.

Happy New Year,
Susan Y

[past crazy]

using this sidebet with 1 deck remaining in an 8 deck shoe is not the same as using it against a single deck game

without any form of tracking, the house edge will remain the same all the way through the shoe

if there is 1 deck remaining of an 8 deck shoe the make up of that one deck will not be perfectly balanced with 1 card of every rank and suite. On average there will be enough cards of same rank in the last deck to keep the EV the same as it was off the top of the shoe.

> Yes, we are talking about the matching pairs
> bet of course.

> Look at it this way:
> Your first card could be of any rank and
> color. If you use 1,000 decks there will be
> almost 1000 times two cards of the same rank
> and opposite color (black/red), 1000 times
> one cards of the same rank and same color
> but not the same suite, and 1000 times one
> cards same rank and suite ? to be received
> as a bonus combination.
> In a single deck game your first card could
> be of any rank and color, but now to get a
> card to match it would be more difficult as
> you depleted the deck.
> In single deck game your chance of getting a
> pair of same rank and color is decreased by
> 50% and chance of a pair of same rank and
> suite is decreased by 100%, compared to
> infinite deck.
> For example if your first card is Ace of
> clubs you can not get another Ace of clubs
> to match it, as well, as your chance of
> getting a black Ace is reduced by 50%
> because only one Ace of spades is available.

> The house edge could be written:
> [{n*2*6+(n-1)*11+(n-1)*31}-(n*52-1)]/(n*52-1)
> Where n is the number of decks used.

> Small correction. In the last part of my
> post, when dealing with the subset of 13
> cards and the first card is non-Ace it is
> the chance for any win that is zero, the EV
> is minus 100%.

> Happy New Year,
> Susan Y

[Susan]

You are correct.
When we have no information on cards used, regardless how deep we are into the shoe, we have to calculate the EV as if it was the beginning of the original configuration.

So in eight decks game, without knowing any specifics of cards left, total payouts should be calculated as 6*(8*2)+11*(8*1)+31*(8*1-1)

The EVs I provided are good for situations at the top of complete decks. Sorry for the confusion.

In eight decks shoe (original Allen?s post) this side bet becomes positive for player at TC above 8.5. At TC =12 it is plus 3.3%.

Happy New Year,
Susan Y