Supposing you are playing on the first box and you see tht the first card that is coming on the next hand is a ten ( a situation that happens quite often),does anybody know what your edge over the house is?
Go over to a site like bjmath.com or get a book with EV tables.
You will have 10 possible hands: T,2 T,3 T,4 ... T,T T,A
with T,T being about 4 times more likely than any other.
The dealer will have 10 possible upcards: 2, 3, 4, ... T, A
with T upcard being about 4 times more likely than any other.
Look up and add these 100 possible expectation numbers together (remembering to multiply the combinations with another T in them by 4)
The sum is a pretty good estimate of your advantage. I think is in the neighborhood of 12% advantage
You have an EV of about .55 on the most common hand: T,T vs T
So, it means that I should bet as if true count is about 25, right?
This is fundamentally different to a high true count. Although the advantage is roughly the same the variance is different, there is no kelly divisor. That means the kelly bettor bets proportional to his exact advantage because of the impossibility of splits, dd's etc.
This is fundamentally different to a high true count. Although the advantage is roughly the same the variance is different, there is no kelly divisor. That means the kelly bettor bets proportional to his exact advantage because of the impossibility of splits, dd's etc.
If surrender is allowed, the variance will be less than 1. As a result, the kelly bet is a little larger.
