A casino I frequent is offering side bets in blackjack. A regular game is $5 or $10 minimum bet. The side bets are black or red pays 3 to 1 and point total of 20 pays 8 to 1. Bets are resolved based on the player's original two cards ($50 max 8 decks). My questions are what is the casino edge for each bet and is there a card counting system for these bets?
Thanks
Judy
> A casino I frequent is offering side bets in
> blackjack. A regular game is $5 or $10
> minimum bet. The side bets are black or red
> pays 3 to 1
What is "black or red"? The dealer's upcard? I analyze this completely, including a count to beat the wager, on pp. 32-33 of BJA3. House edge is 3.85% for the non-counter.
> and point total of 20 pays 8 to
> 1. Bets are resolved based on the player's
> original two cards ($50 max 8 decks).
A player gets a 20 (T-T) or (A-9), almost exactly 10% of the time in a 6-deck game. So, the correct payoff should be 9 to 1. For every ten bets the non-counter makes, he's one unit short, being paid 8 to 1. So, the house edge is a whopping 10%.
Obviously, in high counts, we get more T-T and even A-9, but I don't know what count we need to push the frequency to over 11.11%, so that we have the edge. Perhaps someone can sim it. I can also go somewhere to look that up, but it will take a few minutes.
Don
"What is "black or red"? The dealer's upcard? I analyze this completely, including a count to beat the wager, on pp. 32-33 of BJA3. House edge is 3.85% for the non-counter"
No, it's the side bet that your first two cards will be red, at 3-1, then another side bet that your first two cards will be black at 3-1. It's a big house edge unless you're counting.
The other bet on the first two card 20, is $100 max, not $50, if it's the same game I was talking about a little while ago.
> What is "black or red"? The
> dealer's upcard? I analyze this completely,
> including a count to beat the wager, on pp.
> 32-33 of BJA3. House edge is 3.85% for the
> non-counter.
> A player gets a 20 (T-T) or (A-9), almost
> exactly 10% of the time in a 6-deck game.
> So, the correct payoff should be 9 to 1. For
> every ten bets the non-counter makes, he's
> one unit short, being paid 8 to 1. So, the
> house edge is a whopping 10%.
> Obviously, in high counts, we get more T-T
> and even A-9, but I don't know what count we
> need to push the frequency to over 11.11%,
> so that we have the edge. Perhaps someone
> can sim it. I can also go somewhere to look
> that up, but it will take a few minutes.
> Don
> No, it's the side bet that your first two
> cards will be red, at 3-1, then another side
> bet that your first two cards will be black
> at 3-1. It's a big house edge unless you're
> counting.
That's 3 TO 1, not 3 FOR 1.
Don
> That's 3 TO 1, not 3 FOR 1.
> Don
Don
The side bets are:
2 red: The player's first two cards are red. pays 3 to 1.
2 black: The player's first two cards are black. pays 3 to 1.
What is the casino edge? Is there a card counting system to get an advantage?
Sorry I didn't previously explain it clearly.
Thanks
Judy
> The side bets are:
> 2 red: The player's first two cards are red.
> pays 3 to 1.
> 2 black: The player's first two cards are
> black. pays 3 to 1.
> What is the casino edge?
0.241% for the non-counter.
> Is there a card
> counting system to get an advantage?
Most certainly. Count one color minus the other. If the count isn't zero, you have an edge! :-)
Don
That can't be right, can it?? I thought the deck would have to be 66% in favor of one color to have an edge on a 3 to 1 bet, no? Aren't the true odds of getting 2 red or 2 black on your first two cards 4 to 1, or am I all mixed up?
> 0.241% for the non-counter.
> Most certainly. Count one color minus the
> other. If the count isn't zero, you have an
> edge! :-)
> Don
> That can't be right, can it??
Er, yes.
> I thought the
> deck would have to be 66% in favor of one
> color to have an edge on a 3 to 1 bet, no?
No.
> Aren't the true odds of getting 2 red or 2
> black on your first two cards 4 to 1, or am
> I all mixed up?
All mixed up. Roughly one chance in two each time. 1/2 x 1/2 = 1/4. One chance in four is odds of 3 to 1 against.
Don
Ahhh, I see the light now. Thanks for straightening me out. I should be making this side bet a lot more often than I have been.
You do know this same game also has the over/under 13 side bet. That one I do have down cold. ;-)
> Er, yes.
> No.
> All mixed up. Roughly one chance in two each
> time. 1/2 x 1/2 = 1/4. One chance in four is
> odds of 3 to 1 against.
> Don
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