Effect of number of decks

[Whatsmytotal]

Forgive me and please redirect if this post is found elsewhere on the site.

A friend of mine, a math major, has a hard time believing that one of the main reasons that the house edge is lowered as the number of decks decrease, (assuming the same rules) is because of the higher frequency of BJ's. He insists that the calculation is relevant only on the first hand of a shoe (1 or 6 decks).

Incidentally, how much does the BJ frequency with a large number of decks (a million or a billion), and how is it calculated?

 

[zengrifter]

A friend of mine, a math major, has a hard time believing that one of the main reasons that the house edge is lowered as the number of decks decrease, (assuming the same rules) is because of the higher frequency of BJ's. He insists that the calculation is relevant only on the first hand of a shoe (1 or 6 decks).

He is wrong, but I cannot give you a math proof. Anyone? zg

 

[BJLFS]

The more number of decks the greater the HA. See Stanford Wong's PBJ.

 

[moo321]

It's not just the blackjacks. It's the fact that your blackjacks tie less, the effect of removal when you double down; i.e. if you double down 5,6 vs. a 5 in a single deck game, then 3/20 low cards are removed, but if you do it in a shoe, then only 3/120 of them are removed.

Nothing really massively important on its own, but they add up to several tenths of a percent.

 

[Jack_Black]

He insists that the calculation is relevant only on the first hand of a shoe (1 or 6 decks).

your "friend" is talking about off the top HE. meaning, what is the HE before any cards are removed from the shoe. If you were to take a SD vs 6D game with the exact same rules, you will find the the HA is lower on the SD game. But as the game is dealt, the advantage is constantly changing with each card that is coming out. Each card has an effect of removal value, but that value is also dependent on the number of cards that are still left in the shoe. Say for example, a SD game. there are only 52 cards. 16 tens, 4 aces, and 32 cards that hurt you or are neutral. say you take out the 5s, the worst card you could have, and the edge swings heavily in your favor. The good card bad card ratio is now 28:20. But take out 4 5value cards in a 6D game, and the ratio is only 188:140.

 

[NightStalker]

Off the top HE Vs Decks

Forgive me and please redirect if this post is found elsewhere on the site.

A friend of mine, a math major, has a hard time believing that one of the main reasons that the house edge is lowered as the number of decks decrease, (assuming the same rules) is because of the higher frequency of BJ's. He insists that the calculation is relevant only on the first hand of a shoe (1 or 6 decks).

Incidentally, how much does the BJ frequency with a large number of decks (a million or a billion), and how is it calculated?

Probability of getting blackjack =
Single Deck =2*4/52*16*1/(1*52-1)
Six Deck =2*4/52*16*6/(6*52-1)

n-deck = 2*16n/(52n-1)
As n approaches infinity, value approaches 32/52
For n=1, value = 32/51

A little difference, but should be enough to prove him wrong.. Other edge comes from double down.. If your double down 11Vs6 gets a small card, then dealer is more probably to bust in single deck than multi deck games.. This is what we call EoR of cards played during the hand before the round is complete..

 

[tobe1]

Frequency of blackjacks

I am the "friend" referenced in the top post. The question I have is does the frequency of blackjacks decrease as the number of decks increase?

I understand that if you deal one random hand of bj from 1 deck, the probability of blackjack is 4.83%; the 8 deck probability is 4.75%.

I find it impossible that if you deal through all of the cards (not just one hand) that there will consistently be a greater number of blackjacks in 8 single decks, than in 1 8-deck shoe. You can take this to the extreme and say 100,000 single decks vs. a 100,000 deck shoe.

 

[Jack_Black]

I understand that if you deal one random hand of bj from 1 deck, the probability of blackjack is 4.83%; the 8 deck probability is 4.75%.

I find it impossible that if you deal through all of the cards (not just one hand) that there will consistently be a greater number of blackjacks in 8 single decks, than in 1 8-deck shoe. You can take this to the extreme and say 100,000 single decks vs. a 100,000 deck shoe.

Am I reading this question right? It is impossible to have a greater number of snappers in a SD vs 8D?! What do you mean 8SD vs 8D shoe? As in 8 separated SD vs an intermixed 8D?

Please refer to nightstalker's formula. Also, think on a basic probability level. Let's pretend there is a casino game in which you roll one six sided die and win if you roll 1-4, but lose on 5-6. as you can see, anyone could play this game and win tons of money. then the casinos came up with a 60 sided die. You win if you roll 1-40, but lose on 41-60. The ratio of winning/losing sides are the same, but the frequency of wins(winrate) per trial is less. Now would you rather play the 6 sided dice game 100 times or would you rather play the 60 sided dice game 10 times? what would be your winrate if you played both games 100 times?

 

[QFIT]

I am the "friend" referenced in the top post. The question I have is does the frequency of blackjacks decrease as the number of decks increase?

I understand that if you deal one random hand of bj from 1 deck, the probability of blackjack is 4.83%; the 8 deck probability is 4.75%.

I find it impossible that if you deal through all of the cards (not just one hand) that there will consistently be a greater number of blackjacks in 8 single decks, than in 1 8-deck shoe. You can take this to the extreme and say 100,000 single decks vs. a 100,000 deck shoe.

Same thing happens in the extreme example. It's just that the difference is smaller and smaller as you increase number of decks.

 

[Canceler]

I understand that if you deal one random hand of bj from 1 deck, the probability of blackjack is 4.83%; the 8 deck probability is 4.75%.

I find it impossible that if you deal through all of the cards (not just one hand) that there will consistently be a greater number of blackjacks in 8 single decks, than in 1 8-deck shoe. You can take this to the extreme and say 100,000 single decks vs. a 100,000 deck shoe.

The bolded part would be right if every A matched up with a T to produce a blackjack. But in the sentence above that, you acknowledge that it's less likely to happen with more decks.

 

[tobe1]

What I acknowledge.

I agree with the formula for the first hand dealt. It is fairly obvious that in the first hand dealt, the probability of a blackjack is better with a smaller deck. What is not as obvious is what occurs in the hands in the future.

There are four cases for the first hand.

1. Blackjack
2. Ten, No ace
3. Ace, No ten
4. No ten, No Ace.

If number 1, 2 or 3 occur in both a 1 deck shoe and an 8 deck shoe then the probability of a future blackjack decreases at a quicker rate in the smaller shoe.

The problem comes in calculation the conditional probabilities after hand 1 for each and every subsequent hand.

Frequency: the number of occurences per given unit of time (in this case number of trials).

 

[Jack_Black]

I agree with the formula for the first hand dealt. It is fairly obvious that in the first hand dealt, the probability of a blackjack is better with a smaller deck. What is not as obvious is what occurs in the hands in the future.

There are four cases for the first hand.

1. Blackjack
2. Ten, No ace
3. Ace, No ten
4. No ten, No Ace.

If number 1, 2 or 3 occur in both a 1 deck shoe and an 8 deck shoe then the probability of a future blackjack decreases at a quicker rate in the smaller shoe.

The problem comes in calculation the conditional probabilities after hand 1 for each and every subsequent hand.

Frequency: the number of occurences per given unit of time (in this case number of trials).

You're still forgetting that after 4 aces are gone from the SD, the deck is reshuffled vs 4 aces leaving an 8D shoe, the game keeps going. So the frequency of getting a BJ is still higher, since you will be reshuffling the aces back into the single deck game(which they usually reshuffle after 26 cards have been dealt) Compare that to 8D in which they don't reshuffle after 26 cards, not even 52 cards, but after a monstrous 364 cards being dealt.

You're also forgetting the increasing number of non tens and aces as you add decks, which makes it harder to land a snapper.

 

[NightStalker]

Correct, agree with your thinking

I agree with the formula for the first hand dealt. It is fairly obvious that in the first hand dealt, the probability of a blackjack is better with a smaller deck. What is not as obvious is what occurs in the hands in the future.
There are four cases for the first hand.

1. Blackjack
2. Ten, No ace
3. Ace, No ten
4. No ten, No Ace.

If number 1, 2 or 3 occur in both a 1 deck shoe and an 8 deck shoe then the probability of a future blackjack decreases at a quicker rate in the smaller shoe.

The problem comes in calculation the conditional probabilities after hand 1 for each and every subsequent hand.

Frequency: the number of occurences per given unit of time (in this case number of trials).

But some hands will lower the probability in the next round while others will increase.. But in the long run, it should average to the initial probability- that is and will always be the mean of all probabilities at every point of the shoe..

The only point you need to focus on is:: Effect of Removal of cards in the hand..For a blackjack, you need an ace/face...

Consider probability of getting blackjack is p assuming n cards remaining in the deck with T faces and A aces..

Now your first card is an ACE.
Now probability of getting blackjack will be increased to p+x..
where x= T/(n-1)

With any number of decks, T/n ratio is constant but
T/(n-1) ratio is not constant..
Hence more blackjack for smaller n..
But effect is very low...

 

[tobe1]

You're still forgetting that after 4 aces are gone from the SD, the deck is reshuffled vs 4 aces leaving an 8D shoe, the game keeps going. So the frequency of getting a BJ is still higher, since you will be reshuffling the aces back into the single deck game .

I am not forgetting that. For my arguement, the deck would be ran to completion. Upon recieving the last card, the deck would be reshuffled the first card off the deck would continue the hand if it was not finished. I am not worried about profitability or stopping card counters, just the frequency of blackjacks.

 

[blackjacktilt]

Ace Tracking

You people just ruined my thoughts on Ace Tracking...:grin:

 

[Jack_Black]

I am not forgetting that. For my arguement, the deck would be ran to completion. Upon recieving the last card, the deck would be reshuffled the first card off the deck would continue the hand if it was not finished. I am not worried about profitability or stopping card counters, just the frequency of blackjacks.

So is there still confusion on the issue? Do you not agree with this equation:

Probability of getting blackjack =
Single Deck =2*4/52*16*1/(1*52-1)
Six Deck =2*4/52*16*6/(6*52-1)

You said earlier in another post about how you believe the probability of getting Naturals early on in an SD game will be higher than in an 8D game. So you should agree that it is better to play a SD game 8 times, than it is to play an 8D game once.

 

[tobe1]

Naturals early

I Do agree that it is easier to get naturals earlier in a single deck game. For counting purposes I agree, play the single deck game. There are only 32 blackjacks possible in and 8 deck game. I can't see how they would be less likely to occur in an 8deck game with the same proportion of tens.

 

[Jack_Black]

how do you disagree with the equation I gave? Also, how do you disagree with yourself, if you acknowledge how easier it is to get a natural in a SD vs 8D???????

yes the proportion of tens are the same, but the increase in non tens/tens is what makes it harder to pick a blackjack out of the deck.

remember the dice game example? everything increased proportionally but now the winrate/frequency has decreased because the number of possible outcomes increased. IOW, playing each game 1000 times will show the decrease in frequency of winners(bjs or otherwise) of the 60 sided dice game/8D shoe because of the increase in possible outcomes in the "bigger" games vs. "smaller" games.

 

[Whatsmytotal]

The problem seems to be that, it's agreed that the chances for BJ on the first hand in a SD is higher than tha chances for a BJ for the first hand in 8D, but it's not clear that it stays higher on any given hand. How can this misunderstanding be cleared up?

 

[London Colin]

The problem seems to be that, it's agreed that the chances for BJ on the first hand in a SD is higher than tha chances for a BJ for the first hand in 8D, but it's not clear that it stays higher on any given hand. How can this misunderstanding be cleared up?

It's the same (on average) on every hand, not just the first hand, just as the overall house edge is the same on every hand (if you always follow basic strategy).

Griffin gives a simple proof (that even I can understand) of why the 2nd hand must have the same overall expectation as the 1st hand -

This is easily proven by imagining all possible permutations of the deck and recognizing that, for any first and second hand that can occur, there is an equiprobable reordering of the deck which merely interchanges the two.