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Ask the Wizard #208

I was playing a baccarat game in Asia, where the house paid 150 to 1 for bets on a 1 to 1 tie. What are the odds on that bet? Your site rocks, thanks for all the great work.

Jim from Las Vegas

Thank you for the kind words. My baccarat appendix shows the probability of a 1-1 tie is 0.004101. That is the most infrequent outcome in baccarat, I might add. Fair odds would be (1/p)-1 to 1, where p is the probability of winning, which comes to 242.84 to one. An easy formula for the house edge is (t-a)/(t+1), where t is the true odds, and a is the actual odds. At an actual win of only 150 to 1, the house edge is (242.84-150)/(242.84+1) = 38.1% (ouch!).

I follow your 4+ deck Blackjack basic strategy 100% of the time and I always get dirty looks whenever I hit my 12 against a dealer 2 and especially a 3. I don’t know how to explain to the other players in simple terms why what I’m doing in the best thing to do.

Everett from St. Charles, MO

I feel your pain. You can imagine how bad it gets in Spanish 21, which calls for such plays as hitting 14 against a 3. As long as it is just looks, I would let it slide. If it gets to words, I would say something like, "There are lots of other tables in here." There is no way you are going to convince simpletons like this by trying to explain the odds. The more ridiculous a belief is, the more tenaciously it tends to be held.

There is usually no sound-bite explanation anyway to why one play is better than another. To know why the correct play is what it is, one must either consider every possible way the remaining cards could fall, for both player and dealer, or play out the hand thousands of times, even millions for very borderline hands. The decision with the highest expected value is the one you should take. Only refusing insurance yields itself to being easily explained.

Hollywood Park has their new blackjack rules up on their site. I’d be interested to see an analysis if you have the time.

Justin from Redondo Beach

It seems like every casino in LA County has different blackjack rules, and they change frequently. They are often complicated to analyze. I’m afraid I have adopted an unofficial policy to spend no more time on the bizarre blackjack scene in LA.

I was playing craps at Harrah’s in St Louis, and noticed they have added place bet positions for the 2, 3, 11, and 12 to the table. I don’t remember what they paid. Do you know the odds for these bets? Thanks.

Ron from Collinsville, IL

Crapless Craps offers those two bets too. There is one way to roll a 2, and six ways to roll a 7, so the probability of winning a place bet on the 2 is 1/7. Same probability is the same for the 12. As explained in the baccarat question, if the probability of something is p, then fair odds are (1/p)-1 to 1. In this case fair odds would be 6 to 1. The house edge can be expressed as (t-a)/(t+1), where t is the true odds, and a is the actual odds. In Crapless Craps the place bet on the 2 and 12 pays 11 to 2. Using this formula, the house edge on the 2 and 12 is (6-5.5)/(6+1) = 0.5/7 = 7.14%.

In Crapless Craps the 3 and 11 pay 11 to 4. Using the same formula, t=3, and a=2.75, so the house edge is 0.25/4 = 6.25%.

There is 6/5 Double Double Bonus Poker machine with a $10,100 royal payout. It’s a $1 machine, that can take a big hit on the bankroll with only 94% paybakck. I know as the jackpot increases, so does the payback percentage. I would never even consider playing this machine otherwise. Is it worth playing? The floor manager says it’s been as high as $12,000 once before. Should I consider playing it, or just not even waste my time and money?

Nathan from Edina, MN

The return of 6/5 double double bonus is 0.946569, to be exact. My table says the probability of a royal is 0.000025. However, I like to use more significant digits that that, so let’s take the return, divided by the win, which is 0.020297/800 = 0.00002537. The return of all the wins besides the royal is 0.926273. Let’s call j the breakeven jackpot amount. Solving for j:

1 = 0.926273 + 0.00002537*j
j = (1-0.926273)/ 0.00002537 = 2,906.

The 2,906 is measured in bet units. For a $1 machine ($5 total bet) the breakeven point would be $5*2,906 = $14,530. So, $12,000 is still a long way away from break-even. Before some perfectionist writes me, as the progressive goes up, the optimal strategy will change, to be more aggressive towards playing for royals. My answer assumes the player follows the same 6/5 optimal strategy the entire time.

A simple approximation for any 52-card video poker game is to add 0.5% for every extra 1,000 coins in the meter. In the case of a $10,100 meter, that is $6,100 higher than a non-progressive. It is a dollar game, so that is 6,100 coins, so add 0.5% × (6,100/1,000) = 3.05% to the base return. The base return is 92.63%, so the total return could be approximated as 94.66% + 3.05% = 97.71%. The actual return for a $10,100 meter is 97.75%, so pretty close.