Cutting the house advantage to 0.7868% - guaranteed without even Basic Strategy

[KOLAN]

I'm sorry, I don't think I understand. Please elaborate (and tell me what part of my calculations it pertains to).


I would like to note a general undertone of the responses here. People seem to be disregarding the fact that the only statistical advantage the house has over the player, is the situation when dealer AND player bust. The replies I'm seeing only look at player busts, and that is not what I'm talking about.

Yes, the player busts sometimes and the dealer wins with a hand <= 21. Also, sometimes the dealer busts and the player wins with a hand <= 21. The point is that the probability of either happening is EXACTLY THE SAME! That's why in the calculations I've been leaving out those situations - they're irrelevant. It's the situation when both bust, that skews the 50/50 chance of winning.

Licentia:


If my calculations are correct, from there it would put you at a chance to win of about 51.3%! a 2.1% spread against the house. It's unbelievable

thanks man it is good i understend whot you mine .

 

[iCountNTrack]

Then an important question remains: what is the source of the house advantage in black jack?

You are too focused on bust rate, what really matters is the expectation value of your hand for a given playing strategy (and given house rules), playing more hands will never change that. When you mimic the dealer your ev is -3.42% per hand, it doesnt matter how many hands you play each, each hand will have the same horrible ev.

 

[snipe44]

You are too focused on bust rate, what really matters is the expectation value of your hand for a given playing strategy (and given house rules), playing more hands will never change that. When you mimic the dealer your ev is -3.42% per hand, it doesnt matter how many hands you play each, each hand will have the same horrible ev.

We have to be careful though. There are lots of unintuitive situations in mathematics where the placement of decision affects probability. Yes, I can understand that playing continually like normal, over and over again will not cut the house advantage. But, according to the numbers that I've provided (and have yet to be actually refuted, or shown erroneous) playing 4 hands at the same time for every round will cut the house edge.

Take the famous monty hall problem, for example. http://en.wikipedia.org/wiki/Monty_Hall_problem

Here's a quick breakdown for the lazy people:

You're on a game show with 3 doors. Behind one of the doors is the prize, and the others have nothing. You pick a door (say, number 1). The game show host reveals the contents of one of the other doors (2 or 3), and it is empty. At this point, you are given the option of staying with what you picked, or switching to the other unrevealed door.

Many people, smart mathematical people even, will say that it doesn't matter. They'll say that your probability to find the prize is 1/3 no matter what - but they are wrong. You have a higher statistical probability to win if you switch at the end. Read the link if you don't believe me.

 

[KOLAN]

We have to be careful though. There are lots of unintuitive situations in mathematics where the placement of decision affects probability. Yes, I can understand that playing continually like normal, over and over again will not cut the house advantage. But, according to the numbers that I've provided (and have yet to be actually refuted, or shown erroneous) playing 4 hands at the same time for every round will cut the house edge.

Take the famous monty hall problem, for example. http://en.wikipedia.org/wiki/Monty_Hall_problem

Here's a quick breakdown for the lazy people:

You're on a game show with 3 doors. Behind one of the doors is the prize, and the others have nothing. You pick a door (say, number 1). The game show host reveals the contents of one of the other doors (2 or 3), and it is empty. At this point, you are given the option of staying with what you picked, or switching to the other unrevealed door.

Many people, smart mathematical people even, will say that it doesn't matter. They'll say that your probability to find the prize is 1/3 no matter what - but they are wrong. You have a higher statistical probability to win if you switch at the end. Read the link if you don't believe me.

it is look laik you bust transfer to tie it is :1st:bet need check

 

[snipe44]

haha thank... you... kolan? I think :laugh:

 

[QFIT]

We have to be careful though. There are lots of unintuitive situations in mathematics where the placement of decision affects probability.

They are only "unintuitive" to non-mathematicians. This thread is filled with complete nonsense. Ignore bust rate. This stat is used in thousands of stupid gambling systems created over centuries. It has no use. There is a reason that casinos have made billions for centuries. There is no easy method of beating them.

 

[snipe44]

QFIT, if mutual busting is not the source of house edge in blackjack, then what is?

If you can tell me, then I can go and investigate other hypotheses, however if mutual busting IS the source of house edge, then surely looking into how it may be changed and reduced has merit.

Please, answer my question so that if this is nonsense, it can stop.

 

[QFIT]

QFIT, if mutual busting is not the source of house edge in blackjack, then what is?

If you can tell me, then I can go and investigate other hypotheses, however if mutual busting IS the source of house edge, then surely looking into how it may be changed and reduced has merit.

Please, answer my question so that if this is nonsense, it can stop.

EV is calculated by simulation (or combinatorial analysis if fixed rounds are played). You must generate a strategy based on all rules. You are trying to pretend that all that matters is bust rate. Bust rate is a small part of hit/stand, double, split, surrender and insurance decisions. What matters is if you win the hand. Why would you think this is totally based on bust rate of the dealer? You are trying to win, not bust the dealer. You can win without the dealer busting. If you want to increase the dealer bust rate, this is easy. Never hit. The dealer bust rate will increase. But, you will lose your shirt.

 

[snipe44]

The reason why I've been ignoring everything else besides the bust rate is that it's irrelevant to the calculation at hand. I'm isolating the variables at hand, because I'm targeting a very specific aspect of the game!

It doesn't seem like people fully understand what I'm saying. I don't even know if they're actually reading my hypothesis or bothering to comprehend and understand my calculations. It doesn't seem like it either.

Yes, I know bust rate is only one piece of the puzzle for blackjack strategies, but this is NOT what I'm talking about. I'm talking about reducing the factor which causes the house to have an edge in the FIRST PLACE. This has nothing to do with the strategies which pull a player's chance at winning back to 50%. This can be done (I STILL believe, since people refuse to refute what I post, or even address it) WITH those other strategies.

Please QFIT, answer my question! What is the source of house edge in blackjack?!

 

[QFIT]

The reason why I've been ignoring everything else besides the bust rate is that it's irrelevant to the calculation at hand.

It doesn't seem like people fully understand what I'm saying. I don't even know if they're actually reading my hypothesis or bothering to comprehend and understand my calculations. It doesn't seem like it either.

Yes, I know bust rate is only one piece of the puzzle for blackjack strategies, but this is NOT what I'm talking about. I'm talking about reducing the factor which causes the house to have an edge in the FIRST PLACE. This has nothing to do with the strategies which pull a player's chance at winning back to 50%. This can be done (I STILL believe, since people refuse to refute what I post, or even address it) WITH those other strategies.

Please QFIT, answer my question! What is the source of house edge in blackjack?!

The house edge in Blackjack depends on a large number of factors. You must take into account ALL of the rules. Yes, the fact that the dealer gets to play last is a major part of the house edge. But, isolating that makes no sense. And, EV is NOT correlated to dealer bust rate. Think about it. Name one single hand where it matters. One player hand and dealer hand. Give one single example of a player and dealer hand where the number of players matters. There is no such hand. (without counting)

 

[snipe44]

The house edge in Blackjack depends on a large number of factors.

What factors? As far as I can tell there are NO other factors aside from the situation where both dealer and player bust.

You must take into account ALL of the rules.

What other rules? If you're talking about player options, then no. Those only serve to boost the player's chance of winning, and I am talking about one single way to boost performance - not an overall complete strategy. This is but a component (resulting in -5% spread)

Yes, the fact that the dealer gets to play last is a major part of the house edge.

So there IS merit in investigating it then, isn't there? It's not all bunk, since if a flaw or circumvention was found, there could be great benefits awarded.

But, isolating that makes no sense.

But isn't that what you have to do, if you want to see the exact impact that that one factor has on the game? It's inductive reasoning by addition.

And, EV is NOT correlated to dealer bust rate. Think about it. Name one single hand where it matters.

It's not? What if the dealer's strategy was to hit until 30 (so a guaranteed bust every time)? Would that not affect EV?

 

[QFIT]

I repeat "Give one single example of a player and dealer hand where the number of players matters." You did not. There is no such hand.

 

[FLASH1296]

Some peeps need to take a deep breath and relax.
While they are relaxed, they need to take the opportunity
to read QFit's fine on-line work that he ha made available.

I have a great deal of respect for him and for his work.

It is important to understand that the fact that the player must
make his play - even if it means busting, BEFORE the dealer flips
her hole card is what make the game possible in the first place.

That is the nexus of the House Advantage.

 

[snipe44]

I repeat "Give one single example of a player and dealer hand where the number of players matters." You did not. There is no such hand.

I can give you an example, but it's not a good illustration because this isn't a card strategy. So, here's a poor example for you:

Top of the deck: 10, A, 7, K, 5, 10, Q, J, 7, K, Q

I omitted the suits. If it's just the player and the dealer, the dealer wins with blackjack. If its the player and 3 friends, they all win when the dealer busts (using dealer or basic strategy).

FLASH1296:

It is important to understand that the fact that the player must
make his play - even if it means busting, BEFORE the dealer flips
her hole card is what make the game possible in the first place.

That is the nexus of the House Advantage.

Exactly. And that's what my idea capitalizes on. I'm saying the house advantage is a result of the occurrence of mutual busts (players and dealer), and if you increase your team size from 1 to 4 (or more) then you decrease the likelihood of mutual bust, and therein decrease the house edge! But don't take that one, oversimplified sentence as is - look back through my posts and you'll see I'm trying to back this up with solid stats!

 

[Katweezel]

Dealer bust rate is an irrelevant number. Ignore it, just as most serious books ignore it.

This 28.20% figure you and the books say to ignore, because it is irrelevant? What an incredibly high figure to ignore and be irrelevant. Perhaps you are correct when it comes to single-box BJ. But if you are playing MHBJ, that is an entirely different kettle of fish. Then, the wonderfully-high figure becomes lucratively relevant, indeed.

 

[k_c]

Assume bust rate = 28% for both player and dealer:

PB=player bust
PNB=player non-bust
DB=dealer bust
DNB=dealer non bust

1 hand
(PB+DB or DNB) (.28)*(1)*(-1)
(PNB+DB) (.72)*(.28)*(+1)
(PNB+DNB) (.72)*(.72)*(0)
(1 hand expectation) -.28 + .2016 + 0 = -.0784

2 hands
(PB+PB+DB or DNB) (.28)*(.28)*(1)*(-2) = -.1568
(PB+PNB+DNB) (.28)*(.72)*2*(.72)*(-1) = -.290304
(PB+PNB+DB) (.28)*(.72)*(2)*(0) = 0.00
(PNB+PNB+DB) (.72)*(.72)*(.28)*(+2) = +.290304
(PNB+PNB+DNB) (.72)*(.72)*(.72)*(0) 0.00
(Expected 2 hands) = -.1568 - .290304 + .290304 = -.1568

Player loses .0784 units for each hand played.
Creative betting schemes do not alter the odds. That is the gambler's fallacy.

Above assumes player only gets even money for blackjack (so that he is on the same plane as dealer) and this will reduce his disadvantage by over 2% but the point is that the expected values are not altered by simply playing more hands.

I hope this answers your question. Basically it's a simple question.

 

[KenSmith]

Snipe44, I understand exactly what your argument is. Unfortunately, it's not right, so let's see why.

Yes, the entire house edge in blackjack is due to the "double bust" scenario, where both the player and the dealer bust. (The players loses anyway.)

OK, let's backtrack to snipe44's mythical game, we'll call it EvenJack.
The player and dealer both must play Stand on all 17s strategy, and the double bust is a tie. Snipe44 left out an important rule change that is necessary to even the game. Blackjacks must pay even money, not 3:2. It should be apparent to most that this is an even game, with a house edge of 0%.

Now, let's switch to the normal rules where if the player busts, they lose regardless, even if the dealer also busts. The house edge here is indeed the 0.28 * 0.28 = ~ 7.84% (If blackjack pays 3:2 for the player, the house edge is approximately 5.5%. I mention this because some books have published the house edge for a Mimic the Dealer strategy.)

It's better if we break it down this way though:

Player and Dealer bust: 0.28 * 0.28 = 7.84% of the time. (Player loses)
Player busts, and dealer does not: 0.28 * 0.72 = 20.16% of the time (Player loses)
Player does not bust, dealer busts: 0.72 * 0.28 = 20.16% of the time (player wins)
Neither player nor dealer bust: 0.72 * 0.72 = 51.84% of the time (player wins half, dealer wins half)

Player's expectation = -7.84% -20.16% + 20.16% + (51.84%/2) - (51.84%/2) = -7.84%

With two players, what happens?

When the dealer busts (28% of the time),

• Both players bust (0.28 * 0.28 = 7.84%) [-2 units]
• Both players do not bust (0.72 * 0.72 = 51.84%) [+2]
• Player 1 busts, player 2 not (0.28 * 0.72 = 20.16%) [Tie]
• Player 2 busts, player 1 not (0.72 * 0.28 = 20.16%) [Tie]

When the dealer does not bust (72% of the time),

• Both players bust (7.84%) [-2]
• Both players do not bust (51.84%) [1/4 of time +2, 1/2 time tie, 1/4 of time -2] Avg = 0
• Player 1 busts, player 2 not (20.16%) [1/2 of time -2, 1/2 time tie] Avg = -1
• Player 2 busts, player 1 not (20.16%) [1/2 of time -2, 1/2 time tie] Avg = -1

Ignoring the lines that are ties, here are the results:
28% * 7.84% * -2 = -0.043904
28% * 51.84% * +2 = +0.290304
72% * 7.84% * -2 = -0.112896
72% * 20.16% * -1 = -0.145152
72% * 20.16% * -1 = -0.145152

Add up the 5 lines, and you get an expectation of -0.156800 for the two players. Divide that by the 2 units bet, and you get, predictably, the same player expectation of -7.84%.

Two players fared exactly the same as one player did. The same is true for 3 or 4 players, though the math gets tedious.

 

[London Colin]

Hi,

I'm a regular on Ken's other site, blackjacktournaments.com. I know this is an odd thread to jump into with my first post; I've been a lurker around here for some time and on this occasion the frustration of not being able to make my voice heard got the better of me.

Snipe,

Most of the answers you have received address the general, fundamental issues which make what you are investigating an ultimately pointless exercise, but I understand your frustratation that the specifics of your logic and arithmetic don't seem to be being addressed, so I thought I would have a crack at it. (Actually, I think k_c's post and one or two others have addressed the specifics.)

If playing the dealer's strategy busts 28% of the time, then the starting house advantage without any strategy is 0.28 * 0.28 (when both you and the dealer bust) = 7.84% house advantage.

There's actually an error right there, since you are neglecting the 3:2 payoff for a player natural, which makes things somewhat better for a player mimicing the dealer.

This is because if the dealer busts and 1 or 2 of you bust, the other 2 or 3 people will win -

No. You can't say that. If you want to employ your method of isolating all the variables except for dealer/player busts, then all you can say about the non-bust hands is that they will have the average EV of any non-bust hand.

With a mimic-the-dealer strategy, and continuing the ignore the benefit of the 3:2 payoff, that EV would be zero.


All of which leads to the errors in your calculation. You seem to be confusing probabilities with EVs ....

So, back to our old calculation. The house advantage now comes when the dealer and 3 people bust, and when the dealer and 4 people bust.


0.28*0.28*0.28*0.28 = 0.00614656% (the chance that the dealer and 3 people bust)

0.28*0.28*0.28*0.28*0.28 = 0.0017210368% (the chance that everyone busts)

0.00614656% + 0.0017210368% = a total house advantage of 0.7868% (rounded)

k_c alluded to the problem with the above in an earlier post. To do what you are trying to do, you need to account for every possible number of bust hands and multiply by the number of units lost in each case.

Rounding to 3dp, we have -
p(dealer + EXACTLY 1 player bust) = .28 * .72 ^3 * .28 * 4 = .117
[That's 1 bust * three non-bust * dealer bust * 4 permutations]

Similarly,
p(dealer + EXACTLY 2 players bust) = .28^2 * .72^2 * .28 * 6 = .068
p(dealer + EXACTLY 3 players bust) = .28^3 * .72 *.28 * 4 = .018
p(dealer + all 4 players bust) = .28^4 *.28 = .002

To get the overall EV for four players, multiply each probability by the number of units lost -

E(1) = .117 *1
E(2) = .068 *2
E(3) = .018 * 3
E(4) = .002 *4
---------------
.315
---------------
And divide by 4 to get the EV per player. .315/4 = .0788, or 7.88%. Without all the rounding errors, this would be the same as your original .28 *.28 figure.

And, as has been said, it's a fallacy to think that you could ever arrive at an answer for multiple hands that would be different than for a single hand, except by making errors in the increasing more complex arithmetic that becomes neccessary.

 

[KenSmith]

Thanks London Colin. Apparently we were both typing away at the same time, but between our explanations I hope we have shed enough light on this topic.

And, welcome aboard at BJInfo!

 

[snipe44]

Wow, thanks a lot for the really detailed responses Ken and London - that completely clears things up. I believe in my head I'd underestimated the combined effects of the some players bust some players don't situations.

I totally get it now - thanks guys!

Oh well, I suppose somewhere I believed it was too good to be true. All that being said, however, expecting huge returns from any gamble (even with amazing 50/50 odds) is unrealistic I guess- since in the long run, you should only break even anyways.


Thanks again everyone.