I have a question for someone that can help me with maths.

What would be my advantage if I get on my first card 47% of 10 or AS instead of 5/13 ?

Thank you

What game?

If it's blackjack, what are the conditions of the game?

When you say "a 10 or Ace 47% of the time," is it still four times as likely to be a 10 as it is an Ace, and if not, how much more likely is it?

That question has many different possible answers depending on the details.

The game is blackjack with 6 decks

Yes let’s assume it’s still 4 times as likely to be a 10 as it is an ace. So let’s say something like 9.4% Ace and 37.6% 10 instead of 7.69% Ace and 30.76% 10.

Thank you very much and also if someone has the answer maybe he would be so nice to post the calculations ?

I just know getting always 10 gives 13% edge and always Ace gives 52% edge.

Thank you !

Thank you again

Quote:AZDuffmanMy question is why do some cultures use it plural "maths" vs "math" in USA?

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In the UK and Australia, since it is short for "mathematics," they use the plural.

The question that needs to be asked is, "In that case, why is it 'sports' in the USA but 'sport' in the UK?"

If you could stay focus on my subject it would be really nice from you :)

Quote:Aaron326You are very helpful :)

If you could stay focus on my subject it would be really nice from you :)

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Unfortunately, I am not well-versed in blackjack enough to know how to solve this. You have to understand that, with a change in the cards, the entire strategy changes. It's not as cut and dried as "knowing what card X will be will change the HE by N percent."

You also didn't answer one of my questions: while you say that "the probability of a 10 or Ace is 47%, is a 10 still four times as likely as an Ace? Is an Ace equally likely as a 10? Is it something else, and if so, what?

Quote:ThatDonGuyQuote:AZDuffmanMy question is why do some cultures use it plural "maths" vs "math" in USA?

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In the UK and Australia, since it is short for "mathematics," they use the plural.

The question that needs to be asked is, "In that case, why is it 'sports' in the USA but 'sport' in the UK?"

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First part, thanks. I have never had a good answer to that. I am assuming "mathematics" is a less formal word outside the USA? In the USA the full word "mathematics" is only used in the most formal ways, say a department title at a college. "Math" is all most people use.

As to the second part, well, Brits drive on the wrong side of the road as well, we can only help out so much.

On a slightly more serious note, I have been asked why the USA uses a moneyline but everyone else uses odds. I say, "we don't use the metric system, either."

Quote:Aaron326Thank you for your answer,

The game is blackjack with 6 decks

Yes let’s assume it’s still 4 times as likely to be a 10 as it is an ace. So let’s say something like 9.4% Ace and 37.6% 10 instead of 7.69% Ace and 30.76% 10.

Thank you very much and also if someone has the answer maybe he would be so nice to post the calculations ?

I just know getting always 10 gives 13% edge and always Ace gives 52% edge.

Thank you !

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Aaron326,

Since I don't know all the rules, I'll ASSUME the overall player edge based on 6 (regular) decks and the other house rules is -0.5%, or, to put it another way, the house edge is 0.5%. I'll also ASSUME that your values for 1st card X of +13% and 1st card A of +52% are also accurate.

First question: if the 1st card is 2 through 9, what is the average (negative) player edge?

For 6 regular decks (so ignoring for the moment your increased X & A probabilities), the overall player edge E is given by this equation:

E = (prob A)*(A edge) + (prob X)*(X edge) + (prob 2-9)*(2-9 edge)

Solving gives:

(2-9 edge) = {E - (prob A)*(A edge) - (prob X)*(X edge)}/(prob 2-9)

So

(2-9 edge) = {-0.5% - (1/13)*52% - (4/13)*13%}/(8/13) = -13.81...%

Let's call it -14% for a 1st card of 2-9.

Ok, now we can analyze your situation!

Given your probabilities, your edge is

E = (prob A)*(A edge) + (prob X)*(X edge) + (prob 2-9)*(2-9 edge)

E = (0.094)*52% + (0.376)*13% + (1-0.094-0.376)*(-14%) = +2.356%

Woo HOO!!! MAX BET!

Hope this helps!

Dog Hand

I assume the player uses standard basic strategy. The player advantage would be higher if he used optimum strategy for this card distribution.

Quote:gordonm888I get a player advantage of 3.215% for a case where the tens and aces are 46.67% of the cards in the shoe. Specifically, 300 cards in the shoe, 140 are Ten,Ace and 160 are 2-9.

I assume the player uses standard basic strategy. The player advantage would be higher if he used optimum strategy for this card distribution.

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Oh I don’t think OP meant to change the deck composition but rather somehow knew with some probability the first card of the shoe only was 47% ten/face or A.

Last thing, I guess I have to substract HE from the Edge you found ? So E = 2.356 - 0.5 = 1.856 ?

Or 2.356 is already with HE substracted ?

Quote:Aaron326I really thank you very much for your calculations doghand !!!

Last thing, I guess I have to substract HE from the Edge you found ? So E = 2.356 - 0.5 = 1.856 ?

Or 2.356 is already with HE substracted ?

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Aaron326,

Thank you for your kind words.

The 2.356% I quoted includes the house edge.

Note, though, that you should re-do the calculations yourself with the actual house edge and the actual player edge for each upcard. One way to get these values is to use this website:

https://www.bjstrat.net/cgi-bin/cdca.cgi

Enter the correct rule set, then click "compute" to get the overall player edge (it will be negative, of course). Then set the player hand to have just an Ace and compute to get the A edge. Repeat for 2, 3, ... up to 9, then average those eight results to get the 2-9 edge. Next, set a T to get the X edge. Finally, use the equation I gave to find your actual edge.

Hope this helps!

Dog Hand

I thought about this again. A more reasonable question to ask is this, what is the new EV when player knows the next card for him is NOT a face card? This happens when player had a glimpse of this card.

Let me follow Dog Hand calculation to find out.

EV = (prob A)*(A edge) + (prob X)*(X edge) + (prob 2-9)*(2-9 edge)

= (1/10)*52% + (1/10)*13% + (8/10)*(-14%) = +0.18%

Bet minimum!

Why can’t I just do 2.356% + 0.5 = 2.856% to find my edge without HE or if HE = 0

I thought we always could do this way like knowing our Edge and then subtract the HE to have our real expectation.

I will have a look of your link thank you :)

I meant why not calculate my player edge independently than the HE and then just substract the HE from the edge.

So imagine player edge 2.856% -0.5 (=HE) = 2.356%

I always thought doing this way was the right way but maybe I was wrong …

We cannot do it this way at all ? (Ignore HE for calculations and then substract the HE at the end)

Thanks !

Why bet minimum with a +EV game?Quote:acesideYou do not need to substrate the ordinary EV from the edge Dog Hand found, so the 2.356% is the EV.

I thought about this again. A more reasonable question to ask is this, what is the new EV when player knows the next card for him is NOT a face card? This happens when player had a glimpse of this card.

Let me follow Dog Hand calculation to find out.

EV = (prob A)*(A edge) + (prob X)*(X edge) + (prob 2-9)*(2-9 edge)

= (1/10)*52% + (1/10)*13% + (8/10)*(-14%) = +0.18%

Bet minimum!

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Quote:MentalWhy bet minimum with a +EV game?Quote:acesideYou do not need to substrate the ordinary EV from the edge Dog Hand found, so the 2.356% is the EV.

I thought about this again. A more reasonable question to ask is this, what is the new EV when player knows the next card for him is NOT a face card? This happens when player had a glimpse of this card.

Let me follow Dog Hand calculation to find out.

EV = (prob A)*(A edge) + (prob X)*(X edge) + (prob 2-9)*(2-9 edge)

= (1/10)*52% + (1/10)*13% + (8/10)*(-14%) = +0.18%

Bet minimum!

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Good catch! Actually you are right. Player may bet maximum.

Quote:acesideYou can just say, player gains an edge of 2.856% = 0.5% +2.356%. This is correct.

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This I know but I was just thinking if we could do the calculation for player edge without HE first and then just subtract the HE.

It would be easier in case I would like to do same calculation just removing surrender rules etc …

Here Doghand (that I respect for his answer) ask me to do calculation again and maybe he has a good reason.

I was just wondering if we could do it the way I asked.

When I read first the calculations of doghand it didn’t look like the HE of 0.5 was in the equation ? Unless I didn’t read properly all his calculation.

So I need to get those new numbers for the HE I will want

That is like asking where the HE of baccarat appears in the calculation. The game you describe is a completely different game compared to ordinary BJ. You don't need to use the ordinary game HE as a starting point for your HE calculations. You might be used to people stating that such-and-such a rule change adds X % to the HE. But this is normally computed by doing a completely new calculation of the house edge and subtracting.Quote:Aaron326Where does the HE of 0.5 appears in the calculations of doghand ?

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EV = (prob A)*(A edge) + (prob X)*(X edge) + (prob 2-9)*(2-9 edge)

= (1/10)*52% + (1/10)*13% + (8/10)*(-14%) = -4.7%

Bet minimum!

In the first table of data you can notice the "COMBINED" column. This is where if you KNOW you're getting an A or 10 as your first card but nothing else, you have ~21.33% advantage in the game (Assumes 6D H17 DA2 DAS, ~.61% HE). This assumes 50% split, whereas the OP said this was 47% of the time.

This should be sufficient for the OP to answer his own question to the level of granularity of his choosing =).

Quoting Eliot:

“Consider the following example. Assume an AP is playing a heads-up game against a sloppy dealer. Suppose the dealer has a Ten as his up card. Next, suppose the player is holding the cards 4/3 for a total of 7. His correct play is, obviously, to hit. But, assume he sees the corner of the card (as in the picture above) is the Ten of Spades. He then stands on his hard 7. That way he preserves the valuable Ten as his first card for the next round. When the next round comes, if the dealer didn’t take a hit, the AP makes a large wager knowing he will get that Ten as his first card. Depending on the blackjack rules at the table, the AP may have more than a 14% edge over the house on that hand.”

Eliot is a mathematician but probably didn’t calculate this situation. Suppose this is a 6-deck game for flat betting all the way. What should a player do upon an exposed next card of 10 according to math?

If you are forced to flat bet, then it still might be worth it to make the play. Giving up the 17 you could have had doesn't cost you much EV against a 10.Quote:acesideI just read this article but found it confusing on this part.

Quoting Eliot:

“Consider the following example. Assume an AP is playing a heads-up game against a sloppy dealer. Suppose the dealer has a Ten as his up card. Next, suppose the player is holding the cards 4/3 for a total of 7. His correct play is, obviously, to hit. But, assume he sees the corner of the card (as in the picture above) is the Ten of Spades. He then stands on his hard 7. That way he preserves the valuable Ten as his first card for the next round. When the next round comes, if the dealer didn’t take a hit, the AP makes a large wager knowing he will get that Ten as his first card. Depending on the blackjack rules at the table, the AP may have more than a 14% edge over the house on that hand.”

Eliot is a mathematician but probably didn’t calculate this situation. Suppose this is a 6-deck game for flat betting all the way. What should a player do upon an exposed next card of 10 according to math?

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What is not explained fully is that you are not guaranteed to see the 10 in the next hand. The dealer could have 2-6 as a hole card, so the 10 will be used to bust the dealer's current hand. This is a decent outcome compared to taking your chances with a 17 total.

Are there many situations where you are forced to flat bet, though? Bumping up the bet to catch a 10 on the next round is where the real value lies.

If you use up the 10 for your 17, then there is no more knowledge to exploit. The Wizard's probability charts will give you the numbers:Quote:acesideThere are almost no such situations forcing you to flat bet; however, oftentimes you are already max betting when you hold 4/3 for a total of 7. I’d like to see dealer’s bust rate and final 17 tie rate to be more confident in this specific situation where the dealer has an upcard of 10 and the next coming card is a sure 10.

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https://wizardofodds.com/games/blackjack/dealer-odds-blackjack-us-rules/ for S17

Tie at 17 = 0.124156 Dealer bust = 0.232499

The dealer wins 64% of the time.

EV = (5/13-8/13) = -0.23;

If the player hits on 4/3 with this exposed Ten, the

EV= 0.23-0.64 = -0.41.

This means that standing on player 4/3 gives a much better EV, so the strategy decision is Stand regardless of what amount the player has betted.