Casino Questions and Answers – Part 3
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By John Grochowski
Q. What do you think of the new strategy of blackjack when the player who has a hard 16 with three or more cards stands when the dealer has a 10 showing? I play quite a bit of black jack and I have heard this lately from other players. Is this something new? What I don’t understand is that they say you still hit a three card hard 16 when the dealer has a 7-8-9-Ace. Do you agree with this strategy?
A. The reason we sometimes stand on 16 vs. 10 and not vs. 7s through 9s is that we have more to gain by hitting against a lower card.
Let’s say the dealer has a 7 up and a 10 down. You can push if you draw an Ace to your 16, and win by drawing a 2 or 3. But if the dealer has a 10 under a 10, you still lose on those draws.
That said, it’s not optimal to stand on any old three-card 16 against a 10. It depends on the composition of your 16. You’re better off standing instead of hitting if your hand includes at least one 4 or 5, removing from the deck a card that could give the dealer a 20 or 21.
With combinations such as 8-7-Ace or 6-7-3, you’re still better off hitting that 16 against a 10. But with 7-5-4 or 8-5-3, you stand. It’s that close a call.
Q. I don’t play the field in craps, but it has a 20 to 16 edge for the house. When I play the pass line with one come, and have a 6 and a 4 working, that gives me five ways to make point on the 6 and three ways on the 4. That’s eight ways to win and only six ways to lose (the six possible rolls totaling 7).
Can I say I have an advantage? If not, why does the 20 to 16 work for the house?
A. Even though you have eight ways to win and only six ways to lose when you have a 6 and a 4 working, you don’t have an edge. That’s because when that nasty old 7 turns up, you lose both bets at once.
Let’s run a sequence where you always have a 6 and a 4 in action, with $5 on each, and each of the 36 possible rolls turns up once. Nothing but 4, 6 and 7 matter, so we’ll just see what happens when those turn up:
On the six 7s, you lose $10 on each for a total of $70 in losses.
On the five 6s, you win $5 on each for $30 in wins
On the three 4s, you win $5 on each for $15 in wins.
So you have $45 in wins, and $70 in losses — the house has an edge.
Q. I saw a sign at a Let It Ride table that said something about an aggregate limit of $15,000. Can you tell me what that means?
A. It means that casino won’t pay more than $15,000 on one hand, even if you have a winner that would seem to call for a bigger payoff. That could come into play if you get a royal flush and have more than one bet on the table at the end.
In Let It Ride, you make three equal-sized bets at once. After you’ve seen your first three cards, you have the option of pulling one bet back. After you’ve seen the first common card that becomes part of all players’ hands, you have the option of pulling the second bet back. The third bet must stay in action.
Let’s say you’ve wagered $5 on all three spots. You’re dealt three cards to a royal, and follow basic strategy by leaving all in action as the fourth and fifth cards complete the big hand. Royals pay 1,000-1, and since you’d be getting that on three $5 wagers, your payoff would be $15,000.
But what if you were betting $10 per spot, instead of $5? Three winning $10 bets on a royal flush should bring you $30,000, but if the house has posted a $15,000 aggregate limit, then that’s all they’ll pay.
The odds against you completing a royal flush are long — only 1 in 649,740 hands are royals — so you’ll probably never run into a problem with an aggregate limit. Nonetheless, I suggest that if you see such a limit posted, you size your bets accordingly. With a $15,000 limit, don’t bet more than $5 per spot.
Q. Explain to me, in dollars and cents, why I’m better off risking $6 on a place bet on 6 instead of $5 on Big 6. Even if there’s a difference in the house edge, how can it be better for a guy on a tight bankroll to risk more money?
A. Win a $6 place bet on 6, and you’re paid $7. Win a $5 Big 6 bet, and you’re paid $5. That’s a huge difference.
Let’s say we have a statistically average session where I place the 6 11 times, and you bet Big 6 11 times. They work exactly the same way, with either bet winning on 6, but .losing if a 7 shows up first. I risk $66, and you risk $55. On the five times the 6 show up before a 7, I get back my $6 bet plus $7 in winnings, for a total of $65. You get back your $5 bet plus $5 in winnings on each of those trials, for a total of $50.
At the end, I have $65 of my $66 for $1 in losses. You have $50 of your $55 for $5 in losses. I’ve risked more, but you’ve lost five times as much as I have. Squeezing those bets down to $5 and playing Big 6 instead of placing the 6 costs you money.
John Grochowski writes a weekly syndicated newspaper column on gambling,
and is author of the “Casino Answer Book” series from Bonus Books.