Probability question

What is the probability of each value card being the last card dealt in the first, second, third hands etc. of a shoe given 1, 2, 3, etc. numbers of players all playing basic strategy, liberal LV strip rules and 100% penetration?

Thanks.

Hi nanumula, welcome to the forums.

So your question doesn't really matter about blackjack so much as given 6 decks of cards, what's the probability that each of the 3 players match their own card (or that they get the card given last to the player in front of them on the original deal)? Just trying to clarify your question, doesn't sound like a difficult probability problem as of now though.

Perhaps you can provide an example for clarity?

This could be a very complicated question to calculate depending on various factors.
For the simple case of the first hand of the shoe but the last hand of the first player you can look at it as follows
(1) Give the dealer their upcard (whether you give them the first or second card in the deck shouldn't matter)
(2) Give the player(s) their first two cards
(3) If the player would split cater for how many splits might have happened before playing the last hand. For example with 8-8 cater for starting the last hand being an 8 knowing some 8's have already been removed. (I am assuming mathematically what happens to the earlier player hands doesn't affect the last hand.)
(4) Look at all the ways the last hand could be played out.
There probably is some corelation between what the first player did and what might happen to the second player, but this starts getting very complicated.

I suspect the simple case can be solved fairly easily, but looking at how this might affect subsequent players or hands would be better solved via simulation. It's possible to run a simulation and just remember which cards were last for the player (personally my simulation runs only deal to one player but do allow infinite splits).

What is the purpose of this?

The easy answer is: approximately 1 in 13, except 10's, that one is approximately 4 in 13

First of all, thank you for the warm welcome.

Yes, I just mean the last card out in the hand with basic strategy play. For example, if the dealer pulls a 6 as their last card it would be 6. If the dealer has blackjack, with Ace as the hole card, it would be Ace. If the last player takes the last card, it would be whatever that card is.

I agree that it would be fairly easy to figure out based on simulation. I would think the last card out would be relatively equal, but my gut tells me there may be subtle and interesting differences. I am not sure the # of splits and players makes a difference either, but the simulation would be helpful nonetheless.

I was thinking this could be used to help guide betting, although I am not sure.

nanumula, sorry for my confusion, but I'm still not getting the specific question you're asking. In your OP it sounded like you were saying what are the odds that each of the players gets the same card the previous player got as their 2nd (i.e. "last" card on the deal) card... now it sounds as though you're referring to 1 instance of a card repeating at the end of a hand.

So am I correct to assume you're stating: Whatever the VERY LAST CARD of this round is, what are the odds that's the first card of the next round? That's pretty easy, given that all the other cards don't matter. Assuming they're shuffled randomly and come out randomly, we're only concerned with the last card repeating twice, essentially... which is really easy.

So say a round is dealt, and the dealer has an ace as the last card pulled from the shoe. The odds of the next card being an ace are as follows:

6 decks (312 cards), 24 aces, dealer had 1, 23 aces remain... to get "exact" you'd have to know the number of cards from each players hand and if they had any aces... BUT this will only vary the numbers SLIGHTLY. The average number of cards per hand in blackjack is 2.75 (for an average 4 players, dealer included at the table). Thus if you had 3 players and 1 dealer hand, with 2.75 cards (on average) that would equate out to 11 cards being dealt on the first round.

1) For only 1 ace (the dealers last card) that means... there are 301 cards remaining and 23 aces remaining. The odds of the next card being an ace are 23/301 = .0764, or about 7.64% chance.

2) Say you observed another ace, then that means there are 301 cards remaining and 22 aces remaining. The odds of the next card being an ace are 22/301 = 7.31%.

So I hope you can see something here.. regardless if there's 1 or 2 MORE or LESS cards out, or 1 or 2 more aces that come out... the odds are going to be ABOUT 7% on average. Now, realize this holds true for any specific card. Thus, whatever the last card to come out is, there's 'about' a 7% chance for it to repeat on the next card. Of course the deeper in the shoe and the more information you gather the more the numbers could change (say the deck is ace heavy with only 1 deck remaining for example)... BUT that's where card counting comes in to play. Tracking the odds of "repeat cards" is a fruitless effort in order to gain any kind of real advantage in the game. There are a bunch of archaic counting systems that track fewer cards, such as the Ace-5 count, but they are quite ineffective and require one to learn all the other disciplines of card counting to have any kind of real effect. If you try to keep track of individual cards you'll find that composite play is near impossible for humans, and computers are banned 'almost' everywhere... thus it all comes back to card counting, time and time again... and might I recommend the Hi/Low count.

I think he’s asking the probability that the last card pulled dealt in a round (he calls it hand) of BJ is a particular rank. Examples:

1) Dealer has BJ with Q showing. Ace hole card was last dealt so Ace is last card.

2) Dealer has T showing and 3rd base hits a 12 and pulls a 7. Dealer has 19 and stands. 7 was last card of round.

Seems to me without thinking about it too much that the “last card” will be weighted to high cards that are more likely to bust the last hitter (third base or dealer). Whether it’s H17 or S17 probably affects the probabilities also.

Ah, I did not gather that unJon, perhaps you're correct. IF that were the case, the card "should" average out to the median value of all of the cards, BUT I would indeed expect it to weigh a bit higher due to a few cards not qualifying the dealer for standing... so instead of an 8, maybe a 9 would be the 'average of averages' for the last card (without worrying about repeating/etc like was earlier).

Yes, UnJon is correct in what I am asking. What is the probability of each rank being the last card in the round?

^ I was assuming incorrectly that the orignal question was the last card a player would take.

If I understand it now, the question means if the dealer takes a second card (US rules) and also peeks for Blackjack what is the last card dealt (if there is late surrender then it is similar to them having a Blackjack since the player's hand is finished with) -
(1) The dealer will not be drawing further cards
(i) It creates a dealer Blackjack - the dealer's 2nd card
(ii) where all player hands are out of the game with Blackjacks or having surrendered - the dealer's 2nd card (assuming it isn't completed for sidebets)
(iii) all players are now out of the game and at least one player busted - the last card drawn by the player who busted last
(iv) where the dealer has a stiff (i.e. the dealer will stay on their original 2 cards)
(a) all players stand on their initial two cards - the dealer's 2nd card
(b) at least one player draws cards then - the last card used by the latest player to draw cards
(2) The dealer will be drawing a 3rd card (and a player is still in the game)
(v) where the dealer does not have a stiff hand and at least one player has a non-busted hand (not Blackjack) - the last (3rd or subsqeuent) card drawn by the dealer.

In some scenarios you just look at the initial cards as no further ones will be dealt. In others it's the last player to act or what the dealer draws to a non-stiff.

In some cases there is some correlation, for example opposite a dealer's 2 for all players to have busted their last cards would have been a ten and all gone via 12. The chances are affected by the knowledge that when the last player busted all the previous players had.

Yes, doing the simulation for just 1 player would be easiest.

I would divide this "last card analysis" into 4 scenarios:

1. The facedown card of the dealer is the last card because all players had initial hands of 17-21 and dealer has a 17 -20.

2. Dealer is last to draw a card because 1 or more players have not busted and the dealers original 2 card hand was 16 or less.

3. A player is the last to draw a card because player had a hand that required a Hit and:
a) dealer's first two cards are 17 -20
b) OR, dealers first two cards were <17 but all players HIT and busted.

4. Dealer gets a blackjack, peeks, and turns his cards over ending the hand. (Last card is Ace or Ten with equal probability.)

Under scenario 1, there is a bias: the last card must be a 7, 8, 9, T or A.

In scenarios 2 and 3 there is also a bias towards high cards because of the possibility of multicard draws. When the dealer or player is hitting a hand that 7-11, than a high card may possibly end the drawing process immediately, while a low card will require the dealer/player to hit again. Even if say, dealer or player has a hard 15 and hits, then a 2-T will end the hand ,whereas an Ace will require a re-draw - so in that scenario a 2-T has a higher probability of being the last card than an ace would.

I will note that the probability of scenarios 1 - 3 are strongly dependent upon the number of players at the table.

Edited: inserted scenario 4, Dealer blackjack.

I imagine someone will just simulate this problem and get the bottom-line answer, but I find that working through this kind of problem is educational.

Let's assume 1 player, infinite deck and H17, DOA2

For scenario 1: The facedown card of the dealer is the last card because all players (the only player) stood on initial hands of 17-21 and dealer has a H17-20.

This scenario will occur with a probability of 0.067295.

The last card drawn in this scenario is

Ace: 0.005042_____7.4922%
Ten: 0.037289____ 55.411%
9: 0.011834_____17.5858%
8: 0.009839_____14.6202%
7: 0.003291______4.8907%

This was tricky to calculate, because of the need to factor in that the following hands are not in the scenario: 99 vs 8,9 (Split) and A8 vs 9,T,A (HIT)

Is this for Stand 17 or Hit 17 game, b/c you said if dealer had hard 17 through 20, but the last card could also be 6 with Ace showing. Also, I agree that the scenarios are likely player # dependent for the first hand in a shoe, but may eventually equalize in infinite deck and nearly equilibrate in a shoe game. I would be interested in these 4 scenario results for 6D, S17, DAS, peek, late surrender game, but this looks like a good start.

First, the result was for Hit 17.

Nanumula, welcome to the forums. You posted an interesting question and some of us have spent a little time looking at aspects of it. But also realize that we are thinking about the question you posted out of:
1. Good will
2. Curiosity.

It is a bit unreasonable to suppose that forum members are going to do complex calculations/simulations as a function of number of decks and rule variations, etc. simply because a new member has posted and asked.

It would help if you would tell us why you are asking. Maybe tell us a little background info about yourself.

Quote: gordonm888

First, the result was for Hit 17.

Nanumula, welcome to the forums. You posted an interesting question and some of us have spent a little time looking at aspects of it. But also realize that we are thinking about the question you posted out of:
1. Good will
2. Curiosity.

It is a bit unreasonable to suppose that forum members are going to do complex calculations/simulations as a function of number of decks and rule variations, etc. simply because a new member has posted and asked.

It would help if you would tell us why you are asking. Maybe tell us a little background info about yourself.

If I were a betting man I would wager the OP has an idea for a BJ bonus bet.

Thank you for the warm welcome, gordonm888.

Okay, I thought you had said the dealer would stand on Hard 17 to 20, in which 6 could be the last card, or maybe it is hard 18-20. I usually only play Stand 17 b/c of the better odds, so I am not as familiar with Hit 17.

I agree discussing it with other ppl is helpful. I thought I may be able to use this information to help guide strategy to play in a mid-shoe game, as I often don't want to wait sometimes 1 hour for a shoe, or if i don't want to play no-mid-shoe.

I can try to run my own simulation, but I figure there are longer members with more experience who can offer some greater insight. I greatly appreciate all of the posts thus far and they already provided some of the answer to my opening question. If will try to search the forums as well, to see if there is anything else on this.

Best regards.

Quote: unJon

If I were a betting man I would wager the OP has an idea for a BJ bonus bet.

Lol, that could also be a good idea for the future. Although, I can't implement it since I don't work for the casino. It would be fun though to design new sidebets with huge house advantages and just rake in the dough from the casino side haha.

Quote: gordonm888

...interesting question...we are thinking about the question you posted out of:
1. Good will
2. Curiosity. ...

I agree it is an interesting puzzle in terms of working out how one might attack the calculation for a single player. However I think it gets too complicated with multiple players. So thanks for exercising the brain cells!

If I was continuing past that, I would write code to work out all the reasonable permutations (initially ignoring the effect of re-splits) and list the last card used assuming there was only one player (having more players is too much work at this stage).

If it was a phD then I should run simulations to confirm the calculations of one player and then add more players to see whether that affects anything (in practice it would mean the dealer plays out their cards more often).

Given the last part there might be a thought that one works out, given a number of players, how often the dealer does or doesn't play out their hand given any of their possible upcards (and not having BJ). For instance if the dealer has a 4, 5 or 6 faceup (except 6A s17) then the only thing to consider is all players get a BJ, otherwise whatever happens they don't bust and the dealer will act last.

Using UK rules (no peek, DDAS, s17, 6 decks) I got the following (you can see from the different counts for Kings, Queens etc how much of an estimate this gives).
Note I merely added code to set lastcarddrawn in the getacard routine and when the hand was over tallying the values of lastcarddrawn before moving onto the next hand; so I haven't checked the results in any detail.

Quote: nanumula

...new sidebet...[on the last card drawn]

You could design a sidebet based on the last card the dealer drew (that is easy to work out assuming you always play the dealer's hand to a finish, similar to the "22" bet in freebet). The problem if you includes the player's cards is it might affect their decisions, for instance the last player to act has a clear advantage as they could stand on their 2-card hand or deliberately take another card.

It appears there is a very high preponderance of 10 value cards, which confirms our initial suspicions. This explains to me why mid-shoes are much more often negative than positive in a neutral system.

Quote: nanumula

It appears there is a very high preponderance of 10 value cards, which confirms our initial suspicions. This explains to me why mid-shoes are much more often negative than positive in a neutral system.

I’m missing the logic. Can you explain please? The preponderance of tens is due I think to that card being most likely to make a hand or bust a hand. On the other hand it is much more likely that a 2 or A is followed by another hit.

Quote: unJon

I’m missing the logic. Can you explain please? The preponderance of tens is due I think to that card being most likely to make a hand or bust a hand. On the other hand it is much more likely that a 2 or A is followed by another hit.

I believe that is likely the reason. And also 10 ends the round in dealer blackjack.

Edit: while I was posting this, another member posted a simulation.

Here is an interesting analysis. This is for infinite decks, Hit Soft 17.

When the dealer has a H16 or lower (or soft17 or lower) and must draw cards, here is the probability of each of the ranks being the last card drawn:

Ten: 0.385849
9: 0.092798
8: 0.90542
7: 0.087497
6: 0.083659
5: 0.07761
4: 0.064854
3: 0.049568
2: 0.033664
A: 0.033959

However, this assumes that all dealer hands that are <17 will have an equal probability of needing to draw cards. That's not true, a dealer hand of 8-7 will always cause players to bust out more frequently than a dealer hand of 6-5, because the 8-7 hand will always have either a 7 or 8 face up. And when all palyers bust out, the dealer does not need to draw to their hand. If I were to take that into account, the distribution of "last card drawn" would be even more heavily skewed towards the high cards than what I have shown above.

The issue of how count varies in a shoe is complicated and is best studied by simulation. For instance, hands that require many cards are, by definition, using up small cards and thus raise the count.

^ Also remember I'm using UK rules (as that is how my simulation has been written) so dealing the dealer's second card after the player has acted. Thus many times where the dealer would stop on the second card but the player drews cards I would count the dealer's second card.
I've tried to simulate (by making the program put aside a card for the dealer before the player acts) and get the following:

Charlie, can you do the following? Using something like simple Hi-LO count, can you perform a simulation and compile a probability distribution of how often the cards in a "single player vs dealer" hand have a count of -8,-7. . .-4, -3, . . . -1, 0, 1 . . . +8, +9, etc.?

That is really what OP should be asking.

This seems obvious (my simulation doesn't change the player's strategy) but you get more 10s and As on a higher count.

No, I meant like this.

Whenever the hand is TT vs TT, that's a count of -4

T9 vs T8; count is -2
T67 vs 98; count is 0
44534 vs 73335; that a count of +9

How often is a hand a count of 0? +1? -1? etc.

I think that more hands are a negative count than a positive count, but that occasionally some hands are massively positive.

Here are some figures. 42.105% were negative and 38.280% were positive. Note my simulation allows infinite splits (as is common in UK) so I can imagine things like Aces being split numerous time.

Count	Hands	%
-15	1	0.000%
-12	4	0.000%
-11	6	0.000%
-10	20	0.000%
-9	58	0.000%
-8	341	0.001%
-7	1 323	0.003%
-6	8 968	0.023%
-5	32 570	0.085%
-4	956 681	2.501%
-3	3 153 518	8.245%
-2	5 277 621	13.799%
-1	6 672 251	17.446%
0	7 502 045	19.615%
1	6 015 167	15.728%
2	4 250 330	11.113%
3	2 397 981	6.270%
4	1 213 116	3.172%
5	507 761	1.328%
6	179 944	0.470%
7	55 862	0.146%
8	15 386	0.040%
9	3 780	0.010%
10	916	0.002%
11	267	0.001%
12	70	0.000%
13	34	0.000%
14	10	0.000%
15	3	0.000%
16	2	0.000%

Is it possible to correlate the true count results compared to the last card drawn from the shoe? Like, for example, if a 5 was the last card out of the shoe, would the average count be positive or negative and by how much is what I think I am looking for. I think it would be less negative than average, but I am not sure. I am not sure if this makes any sense.

Very interesting, Charlie. Thanks. Nice job.

So, there is a small effect! Hands that are negative in count occur more frequently than hands that are positive in count, especially when considering the 91% of the hands with a count -3 to +3. There may be a very slight tendency for the count to go down as you penetrate the shoe, until a positive count hand with +4 or more occurs (I.e., a hand with lots of small cards). It is a small effect, but statistically real.

Of course, this doesn't have the natural feedback mechanism: when the shoe becomes more rich in small cards, there is more of a tendency for small cards to come out. But its nice to isolate effects like this and analyze them.

Thanks Charlie. I guess what that means is that if everyone is playing basic strategy as most ppl do, the shoe will generally go negative, so it makes sense to go against basic strategy, so called "hunch bet," if you are not counting the cards.

Again, I am also curious if there are certain cards, or card combinations on the previous hand that on average would result in a positive count. For example, if player 1 has 5,5 doubles down receives a 5. Dealer flips over two 5s and draws two more 5s for a total of 20, what are the chances the overall count is positive as the shoe is penetrated. In other words, is there a ratio of small to large cards generally or specific ranks on the previous hand that would suggest the overall count is likely positive and what % penetration.