Anthony92600
Anthony92600
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November 30th, 2024 at 9:33:53 AM permalink
Hello everyone I’m new here and I have a pure theory question :)

Let’s imagine a 6 decks BJ game with an house edge of 0.5% for someone playing Basic strategy

What would be the EDGE of that player if :

1) every time the dealer upcard is a 9 10 or AS, the chance that the dealer hit a 9 or 10 on his third card is 7/13 instead of 5/13
2) every time the dealer upcard is a card from 2 to 8, the chance that the dealer hit a 9 or 10 on his second card (the hidden one) is 7/13 instead of 5/13

Calculations are welcome :)
And thank you a lot guys
Anthony
aceside
aceside
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November 30th, 2024 at 10:03:05 AM permalink
I don’t quite understand this question, but let me give an example. If the dealer upcard is an Ace, what is the probability of a dealer hit?

Hit-17 and Stand-17 are different.
Romes
Romes
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December 7th, 2024 at 8:08:27 PM permalink
So this sounds like 2 extra big cards out of every 13... which would mean 4 extra big cards per 52... which would mean a +4 TC, and on a "generic" game each TC is worth about .5% to the player... so -.5 + 2 = +1.5%.

Just rough napkin math, but sounds easy enough to get the gist of what you're referring to, no? For the differences between 1) and 2) simply weight them accordingly (6/13 vs 7/13).
Playing it correctly means you've already won.
aceside
aceside
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camaplRomes
December 8th, 2024 at 4:59:24 AM permalink
Quote: Romes

So this sounds like 2 extra big cards out of every 13... which would mean 4 extra big cards per 52...
link to original post


Do you mean, which would mean 8 extra big cards per 52?
DogHand
DogHand
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December 8th, 2024 at 10:16:56 AM permalink
Quote: aceside

Quote: Romes

So this sounds like 2 extra big cards out of every 13... which would mean 4 extra big cards per 52...
link to original post


Do you mean, which would mean 8 extra big cards per 52?
link to original post


Somebody tore the napkin in half while Romes was calculating 😉

Dog Hand
Romes
Romes
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December 8th, 2024 at 6:55:31 PM permalink
Quote: aceside

Quote: Romes

So this sounds like 2 extra big cards out of every 13... which would mean 4 extra big cards per 52...
link to original post


Do you mean, which would mean 8 extra big cards per 52?
link to original post

This is what I get for napkinning at night XD.

Fine... CORRECTED:

"So this sounds like 2 extra big cards out of every 13... which would mean 8 extra big cards per 52... which would mean a +8 TC, and on a "generic" game each TC is worth about .5% to the player... so -.5 + 4 = +3.5%.

Just rough napkin math, but sounds easy enough to get the gist of what you're referring to, no? For the differences between 1) and 2) simply weight them accordingly (6/13 vs 7/13)."
Playing it correctly means you've already won.
Anthony92600
Anthony92600
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December 17th, 2024 at 9:38:27 AM permalink
Hello guys,

I think you misunderstood my question.
I don’t mean that the deck is high.
I was just imagining a game which would be biased in a way that the dealer will have an increased chance of hitting a 9 or 10 on either his second card either his third card.

You can reply to my initial post again if you know the math :)

Also it would be appreciate to know what changes in basic strategy would be profitable in this case.

Thank you
aceside
aceside
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December 17th, 2024 at 9:48:08 AM permalink
People come here mainly to say hello to each other, but they waited for many days for you to say thank you.
Romes
Romes
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December 17th, 2024 at 8:38:52 PM permalink
Quote: Anthony92600

Hello guys,

I think you misunderstood my question...
...
I was just imagining a game which would be biased in a way that the dealer will have an increased chance of hitting a 9 or 10 on either his second card either his third card...

We're saying the same thing. Maybe a miscommunication/translation? How else would the dealer have an 'increased chance' at getting a high card?

If the dealer were to have an 'increased chance' then this would only happen naturally (aka no cheating) when the deck is in a state in which there is a higher concentration of high cards. The math above is accurate and from what you just said I think we understood what you were originally asking, no?
Playing it correctly means you've already won.
tyler498
tyler498
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January 26th, 2025 at 9:03:34 PM permalink
Quote: Anthony92600

Hello everyone I’m new here and I have a pure theory question :)

Let’s imagine a 6 decks BJ game with an house edge of 0.5% for someone playing Basic strategy

What would be the EDGE of that player if :

1) every time the dealer upcard is a 9 10 or AS, the chance that the dealer hit a 9 or 10 on his third card is 7/13 instead of 5/13
2) every time the dealer upcard is a card from 2 to 8, the chance that the dealer hit a 9 or 10 on his second card (the hidden one) is 7/13 instead of 5/13

Calculations are welcome :)
And thank you a lot guys
Anthony
link to original post



If I understand your rephrasing of the question correctly, and everything is normal other than 1) and 2), then this is likely a negative change for the player.
Player first 2 cards => standard odds
Dealer first card => standard odds
Dealer 2nd card => high count
It is like the player is taking cards from a normal deck, dealer's first card from normal deck. Then the dealer is drawing cards 2 and 3 from a hot deck. Exact result can only be calculated by SIM, but it seems to me like this would make the house edge way worse.
Edit: since it is the drawing card for dealeer that is changing, you can use deviations for high count to reduce the edge (surrender more, stand more or dealer stiff) if you're playing this for some reason
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