Probability question

What has the highest probability of happening:

-getting a winning BJ (excluding pushes, I never insure)

-or getting ANY of the following combinations:

15 vs 10
16 vs 9-10-A

6 decks
Dealer stands on soft 17
Player may double on any first two cards
Player may double after splitting
Player may only split once
Dealer always peeks for blackjack
Late surrender available

Sorry, no BJ

The latter has a much higher probability as there are multiple ways to make up a 15 or 16 but only one way to get a blackjack.

Thanks winr winr chicken dinner! I wanted to know because the online casino where I play has a particular way of dealing with odd number bets. On an 11$ bet, surrender is as follows:

6.00$ back to player, house keeps 5.00$

On blackjack (11$):

player gets paid 16$ (house keeps .50$)

So, I figure, if surrender's are more frequent than winning a BJ, I will only make odd number bets; it should be better for me in the long run!

I assume you meant to say that surrendering an $11 bet would get you $6 back, not the $5.50 which is normal. If that is true, then you should also surrender more often than regular basic strategy says.
Your return on an $11 surrender is -5/11 or -0.4545.
You should thus surrender any hand whose EV is worse than that.

According to BJA3 you will get a winning BJ 4.603 percent of the time, while receiving a 15 vs 10, 16 vs 9, 16 vs 10, 16 vs A at total of 4.713 percent of the time. However, you can only surrender 4.309 percent of those hands due to dealer BJ's, thus you will lose 14.7 cents per 100 hands by playing odd dollar amounts.

Wow, I really appreciate the help I get on this board!

:band: :band: :band: