Some blackjack shifts and craps high-low bets

QUESTION: Counting cards seems too much like work to me. I’m there to play and have fun. But I like talking with my card-counting brother-in-law.

Last time we talked about blackjack, it came up that he always stands on 16 when the dealer has a 10 card up. He said it doesn’t cost very much, that it’s almost the right play anyway, and that it looks less suspicious than if he hit sometimes but stood whenever the count was even slightly in the player’s favor.

He made it sound good and I know he knows what he’s talking about. I was wondering if you could give me the numbers on how close to the right play it is. Could someone who doesn’t count cards gain from standing?

ANSWER: The numbers differ slightly depending on the composition of your hard 16, but let’s look at 9-7 vs. 10 in a six-deck game in which the dealer hits soft 17.

If you hit, your average loss per $1 wagered is 53.54 cents; if you stand, the loss rises a tiny amount to 53.68 cents.

With 10-6, it’s 53.47 cents with a hit, or 54.10 if you stand.

With multiple cards, the better play can swing in favor of standing even if you don’t keep count. With 2-7-7, the average loss is 53.77 cents if you hit and slightly lower at 53.66 cents if you stand.

A single low card, even with multiple decks remaining to be played, swings the odds in favor of standing by a tiny amount.

Card counters use “index numbers” to tell them the counts at which it’s advisable to play opposite basic strategy. For 16 vs. 10, the index number is 0 – it swings even if a positive count is not enough to make up an entire +1 in the true count that divides a running count by the number of unseen decks remaining.

If there’s a way for non-counters to show a very, very small gain, it would be to stand on hard 16 vs. 10 if your hand includes three or more cards, but to follow basic strategy and hit on 9-7 or 10-6 while splitting 8-8.


QUESTION: I saw something at craps I’d never seen before. They had a bet called low dice, high dice, and you could bet on whether the total of the next roll would be 6 or under or 8 or higher. Sevens lose.

Are these decent bets?

ANSWER: I had never heard of this bet until I received your e-mail, so I searched and found a description from Michael Shackleford at wizardofodds.com.

On the low bet, payoffs are even money on 3 through 6, but 2 is worth a big 5-1 pay. Similarly, the high bet pays even money on 8 through 11 but 5-1 on 12.

Either bet has 15 possible winning rolls, with 14 combinations paying even money and one paying 5-1. There are 21 possible losing rolls.

If you wagered $5 per roll for 36 rolls, with each possible combo turning up once, you would risk $180. On the 14 even-money payoffs, you would keep $70 in wagers and get $70 in winnings for a total of $140. On the one 5-1 payoff, you’d keep the $5 bet and get $25 in winnings.

At the end of the trial, you’d have a total of $170 and the house would have $10 of your original $180.

Divide the house’s $10 profit by your $180 risk, then multiply by 100 to convert to percent, and you get a 5.56 percent house edge. That’s not a bet I’d make when there are better options such as pass, don’t pass, come, don’t come and place bets on 6 or 8.

This article is provided by the Frank Scoblete Network. Melissa A. Kaplan is the network's managing editor. If you would like to use this article on your website, please contact Casino City Press, the exclusive web syndication outlet for the Frank Scoblete Network. To contact Frank, please e-mail him at fscobe@optonline.net.