Probability Tables on Wizardofodds site

I calculate that the Probaility Tables such as those regarding the first 4 cards dealt in Blackjack or the Tables refering to the Perfect Pairs probability do not factor in a cut. For example on the 8 deck BJ, the probability of the 1st 4 cards being the same are 0.000003 (rounded up from 0.00000295952), if however a deck is cut off the back, I calculate that this becomes 0.00000252964. This alters the odds from 337837 to 1 to 395256 to 1. Would someone please confirm this?
Thanks
beachboy

Is this the table you're talking about?

I agree that it appears that no cut card effect is accounted for. Since the level of penetration is unknown, it would be hard to accurately account for it.

Yes it is, thank you

Perhaps I'm missing something -- if you are only dealing the first four cards off the shoe, how could the cut card become an issue at all? If there are un-dealt cards of interest to you (they might possibly have formed a suited pair with a card you have seen, but didn't), you still have no idea whether those cards are ahead of the cut card or behind it. If they exist in the shoe, they just aren't in the first four cards, and you don't know where they are. They may not exist in the shoe. If there are only two decks, you can't have a third identical card (or a second identical card in a single deck).

Now if the cards are dealt as a suited pair (e.g., two 6s of clubs) within the first four cards, then that does give some indication of how many decks there are in the shoe -- you are more likely to get the pair if there are a larger number of 6s of clubs that have the possibility of being within the first four cards. That's what that table is all about, but I don't think the cut card comes into play at all in that table. Can you tell me how it might?

Most of my analysis of blackjack and blackjack side bets are based on a freshly shuffled shoe. The cut card effect indeed does change the odds slightly, except in the case of single-deck, the effect can be large. Personally, I prefer to use direct math whenever I can, which is based on a new shoe assumption.

The odds are determined by how many cards are available to complete the objective. In this case an 8 deck game would contain 8 of everything A's to K's. However if a Deck was cut off the back and we assume that there is an even distribution within that deck, then there are only 7 of each card within the accessible cards. I believe the calculation would be based on:
Instead of 415 [(52*8 no of decks)-1 (card to be matched)] cards to come there are 363 [(52*7)-1] and the calculation is then conducted as:
(6/363)*(5/362)*(4/361) not (7/415)*(6/414)*(5/413)
I think?

Thanks for the reply, I have always found your site to be extremely informative.

beachboy

Quote: beachboy

The odds are determined by how many cards are available to complete the objective. In this case an 8 deck game would contain 8 of everything A's to K's. However if a Deck was cut off the back and we assume that there is an even distribution within that deck, then there are only 7 of each card within the accessible cards. I believe the calculation would be based on:
Instead of 415 [(52*8 no of decks)-1 (card to be matched)] cards to come there are 363 [(52*7)-1] and the calculation is then conducted as:
(6/363)*(5/362)*(4/361) not (7/415)*(6/414)*(5/413)
I think?

Quite possibly, I am still missing something. Here is my difficulty with your explanation: You start with, "... if a Deck was cut off the back and we assume that there is an even distribution within that deck...." If we start by assuming that all of the types of cards are evenly distributed, then I don't think we should expect to ever encounter a matched pair. Identical cards will tend to be separated by a full deck. Instead, we typically assume that the cards are randomly distributed, and I have not yet seen how that leads to the same conclusion you reached. Note: I admit I have not thought through this thoroughly.

Dang, I had a heck of a time typing that last sentence with the thought/through/thoroughly phrasing! ;-)

Hi Doc, what I mean by even distribution is not the order in which they come out but the number of each individual card that would appear before and after the cutting card and have used one deck behind as a guide. If it were 2 decks then the number of cards available to make the match would shrink by 1 and the number of cards in total by 52. i.e (5/311)*(4/310)*(3/309)

hope that helps

beachboy