Perfect Cut %

Good Day,

I have a question regarding the perfect cut. Take a 8Deck, DAS, LS, S17, with 75% pen. game. Then lets say you only play perfect basic strategy. No spread and no play alterations. I think it is a -.39 game. Then lets say you are a very very lucky person and you always happen to get the cut card and always happen to cut at the exactly the best spot. What would be (dis)advantage of the game on average, and what would be the average count on the 6 Decks in play? I have seen various numbers and calculations but on given this exact scenario. I can probably workout the math but I am hoping someone has already done the sim or math required and could possibly direct me to a post, forum, or reference. (I already have the cookbook.)

Just looking to see the change that would happen if you were always really lucky.... ;)

Cheers,

Unless I misunderstand what you are asking, which is likely since I'm new, it seems as if you are assuming that there is one particular place in each deck that is the "perfect" place to cut a shoe. You are also then asking what % advantage it would give you as a player if you happened to cut at that point consistently.

I don't think the underlying premise holds up. If a shoe is truly randomly shuffled, any cut should be as good as another. Also, if there was a cut that was best for the player on a given shoe, that would change if another player dropped in or out of the game or if the player played one hand or two, or if a dealer change occurs and the new dealer burns a card, etc...

So, my ignorant guess would be that a) There is no "best spot" to place the cut card and b) even if there were, the other variables in the game would eliminate any tiny advantage it might give. I could be wrong however....

I also do not really know what you are talking about especially because you mention luck.
A shuffle tracker might be able to follow the dealers shuffle and then make sure he cuts a clump of aces and faces into play, otherwise I am at a loss here.

I was mentioning luck on purpose as it would obviously be impossible to know where to cut all of the time..... and if there were such a way I would think it would not be discussed here.....

In regard to my original question. All things being equal, yes there would be one spot to cut, for each shuffle occurrence, that would essentially cut out of play the most low cards. Given a specific game, of say -.39%, depending on a particular cut, we can recalculate the advantage or disadvantage depending.

My questions are meant to not be about ST or methods but rather pure math and simulation based. If you cut the two worst decks off a 8 deck shoe, over say a million shuffles, what is the average count of the remaining 6 decks, and therefore what would be the change in the overall average game percentage?

Thank you,

If you have found desirerable shuffled games and are skillfull enough to cut off negative card clumps, why then flat bet and play basic strategy?

Estimate your count and procede from there using card counting strategies. Get the money while you can.

Frenchman

My questions are meant to not be about ST or methods but rather pure math and simulation based. If you cut the two worst decks off a 8 deck shoe, over say a million shuffles, what is the average count of the remaining 6 decks, and therefore what would be the change in the overall average game percentage?

Without counting, you have no knowledge of what the "two worst decks" are. Therefore, the average HiLo count will be 0.

If I understand correctly, the poster is NOT asking about a ST skill. Rather, he's asking a question about the pure mathematical distribution of randomly shuffled 8-deck shoes, and it's a question I've never seen before.

He's asking, if you were omniscient, or had x-ray eyes, and could always know the composition of the shoe, such that you could always insert the cut-card at a spot that would cut away the worst 2-deck slug, what would be the mean, or average, value of that slug, and, therefore, what positive edge would the remaining 6-decks confer to the player lucky enough to get to play them.

It's a rather fascinating, if not, unfortunately, somewhat useless question, and I'll admit that I don't know the answer.

Don

See my post above.

As soon a someone asks a question whose answer has no practical value, but rather is just a mathematical curiosity, people don't want to bother to answer, or don't understand the original question.

I'm reasonably sure that my interpretation is the correct one, but we'll wait to hear back for confirmation.

And, again, if I'm correct, it's a question whose answer I've never seen before.

Don

Yes Don, you are correct. Thank you for explaining it further. If you could see all the cards in the shoe and be able to cut out the 2 deck slug with the most low cards, how would it change the percentage...(all other being equal, no advantage play, flat betting).

I think it is an interesting question, at least from a math perspective. The most difficult part would be estimating the average or the "worst two deck slugs." (I am not sure if one of the CV suite could do this?) I am also assuming a random shuffle. Looking at different types of shuffles and "clumping" of actual played shoes, etc. would get too complicated and theorectical most likely. I might and I stress might be able to work this out if we were paying attention to strict cut points, like it must be between full decks of the original shuffle, however with the sliding window of being able to cut even inbetween full decks, make it difficult to estimate. We could also obviously look at the best case and the worst case but the answer is obviously not the average of these. (That would be a random cut).

I am a math guy but as I started doing out a formula for this it starting growing quite long and complex (especially considering the 8 decks). If nothing exists and there is no sim out there to do it, then I would be willing to work with someone to accomplish it. A small VB proggy may do quite nicely.

Thanks,

Interesting question. I'm only guessing, but I imagine that someone knows what the typical distribution variance is in an 8 deck shoe, ie what is the average highest running count in any consecutive 104 card series in an 8 deck shoe? You could then use that running count (ie the average number of low cards that would be cut out)to give you a TC for the rest of the deck. Given the TC, you can then come up with the correct advantage to player or house.

The hard bit would be to come up with the highest running count in any consecutive 104 cards. Not too tough for someone who could write a simulation I suppose but impossible for me. :-)

You get it ... partially. You don't want to know the highest such running count that is ver encountered in normal play; rather, you wan't to know that highest running count for every random shuffle you do, and then you want to know the average of all such counts.

Obviously, it's not the same question.

I'll ask Norm if he can sim this.

Don

I wrote: "You get it ... partially. You don't want to know the highest such running count that is ver [ever] encountered in normal play; rather, you wan't [want] to know that highest running count for every random shuffle you do, and then you want to know the average of all such counts."

Didn't proofread. No excuse. Sorry.

Don

In almost all casinos, you are required to cut a minimum of one full deck, or at least one-half deck. They won't let you cut, say, 10 cards.

So, do you want to know the answer within the confines of these constraints, or do you want to know the "pure" answer (since the whole thing is fiction anyway), allowing even a cut of a single card?

Don

Don,

I was thinking about that before in an indirectly related matter. I would like it to be realistic and thus confined to the proper cut depth at either end. I was wondering what that was however and if it changes for different casinos. Is it a full deck or a half or quarter (from either end)? I know that just the other day the need/want arose for me to cut very close to the front. I did about 1/4 deck and the dealer went with it. (Not going to say where.) My feeling is that this particular time the dealer was either lazy, liked me, toke hustling etc, etc. Anyway the short answer is to use the common realistic (and allowed) distance, please.

Thank you for your help Don. I appreciate your assistance. Ultimately the question is if a fictional lucky guy got a lucky cut in a real casino and situation and played normal, what is the edge. I think it is fascinating to see how much the cut (or luck of the cut) really affects things, all else being equal.

Thanks,

Greatings from Sigmund Freud

I sent the following e-mail to Norm:

"Suppose, in an 8-deck game, that you had x-ray vision and could, every time, place the cut card in the exact location such that you would always cut away the worst two-deck slug. On average, what would be the RC of that slug?

"I imagine that this could be simulated. If you can do it, I'd like to take a guess before you furnish the answer; if you can't, I won't bother."

Norm replied: "Easy to sim. Depends on the minimum cards you can cut if you want to include the fact that the worst slug could be a combination of the top and bottom of the decks."

That prompted our exchange on this board, re cut-card position.

I replied: "The poster says to use a real-world cut, so I'd go with a half-deck minimum.

"I'm guessing that the average best two decks will have a Hi-Lo RC of about +15. Wanna take a guess?"

Then, Norm ran the sim and e-mailed me this:

Norm: "Good guess. Assuming a random shuffle, +15.5."

Don: "Sometimes, I scare myself!" :-)

Norm: "However, the answer is shuffle-dependent since you will be forcing the cards left in the shoe to be poor and they will end up in the same general areas post-shuffle. Using a simple one-pass R&R, I get +15.0. Of course you will have to alter your betting. If you do not, you will actually do worse since your TC will nearly always be negative and you will end up flat-betting. If you use a 1-15 spread and an initial running count of +15, I get a SCORE of 189. If you used an IRC of the actual cards cut out, the SCORE would be substantially higher."

My final comments: 1) It's always fun to research the answer to a question we've never seen before, even if this one has no practical value whatsoever and is simply a curiosity. And 2) If you'll pardon the immodesty, sometimes my blackjack intuition scares me! :-)

Thanks to Norm. Hope readers have found this interesting.

Don

I wrote a simple simulation where I calculate the maximum 108 card moving average running count for an 8 deck shoe. Note that the moving average wraps around the deck and assumes you choose not to cut the deck if that's favorable.

The calculation is set for a 1 billion shuffles, but these results seem to have stabilized after 8 million: mean = 16.68, sd=11.09.

Following are the results for various minimum cut sizes from either end of the 8 decks. The calculations are based on 5 million shuffles.

 

Cut  Mean   SD 

  0: 16.68 11.09 

 13: 16.22 10.96 

 26: 15.73 10.83 

 39: 15.22 10.69 

 52: 14.70 10.54 

 65: 14.18 10.38 

 78: 13.67 10.21 

 91: 13.18 10.03 

108: 12.59  9.80