Calcul of edge

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

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.

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).

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?

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

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)."

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

People come here mainly to say hello to each other, but they waited for many days for you to say thank you.

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?