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The Current Blackjack Newsletter blackjack rules on your Wizard of Vegas site indicate that at the Bighorn casino a blackjack pays 2 to 1, if at least 2 or more players at the table get a blackjack. Can you tell me the effect of that rule?

teddys

I show that rule is worth 0.10% per each additional player at the table, not counting yourself. According to my blackjack house edge calculator, the house edge is 0.48% before considering that rule, or the rule allowing doubling on three cards. Doubling three or more cards is worth 0.23%. To make an educated guess, let’s assume doubling on exactly three cards is worth 0.20%, lowering the house edge to 0.28%. Considering the 2-1 blackjack rule, the following is the house edge according to the total number of players, including yourself.

### Bighorn House Edge

 Players House Edge 7 -0.32% 6 -0.22% 5 -0.12% 4 -0.02% 3 0.08% 2 0.18% 1 0.28%

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Suppose a casino had a video poker game that was over 100%, but any given player is only allowed to play it until he hits one royal. Should any strategy changes be made?

Arnold

If you want to be a perfectionist, yes. Let’s look at full pay deuces wild, for example. Normally the return is 1.00762 and a royal hits once every 45282 hands. That makes the expected profit 45282 × (1.00762 - 1) = 345.05 bet units. For a greater overall expected profit, I recommend using a less aggressive royal strategy to increase the total hands played.

In this case, the profit is maximized by following a strategy based on a royal win of 450. That will lower the actual return to 1.007534 and decrease the royal probability to 1 in 46415, resulting in an expected profit of 46415 ×(1.007534-1) = 349.68. The extra 4.6 bet units may not be worth the bother of learning a different strategy.

To find the optimal target royal value, you can use my video poker calculator, and keep lowering the pay for a royal until the overall return gets as close to 1 as possible. At that point, it is like playing for free until you hit the royal, at which point you get a bonus for the royal. In the full pay deuces wild example, the bonus is 800-450=350.

The situation is not entirely hypothetical. Slot managers have been known to prohibit advantage players from playing video poker, and usually such players get the tap on the shoulder shortly after hitting a royal.

This article from abc.net.au is about a player who manipulated the odds on a dog race in Australia. Can you explain to me how he did it?

Aussie

That is an interesting story. The betting terminology is a bit different in Australia. As I understand it, in Australia there are not separate bets for place and show, but just a place bet. The place bet will pay bettors on the first two dogs in races with seven or fewer dogs, and the first three dogs with eight or more total in the race. In the race in question, there were eight dogs, two of whom where strong favorites. Following is the general way the winning odds are calculated in a three-dog place pool in Australia, which is different from how the odds are calculated in the U.S..

1. Take out the track cut from the total pool of place bets. For the sake of argument, let’s use the usual American take-out of 17%.
2. Divide the rest into three pools.
3. Pay the winners on each dog a pro-rata basis according to the size of the pool and the amount bet on the dog. If the amount bet on the dog exceeds its share of the pool, then bettors will get a refund.

Let’s look at an example. Suppose \$100,000 is bet on place bets in an 8-dog race. Assume bets on the winning dogs total \$5,000 on dog A, \$10,000 on dog B, and \$15,000 on dog C. First, the 17% take-out would be deducted, leaving \$83,000. That would be divided by 3, leaving \$27,667 to pay the winners of each dog. Winning bets on dog A would be paid \$27,667/\$5,000 = 5.53 for 1, before any rounding (I’m not sure how they round down under). Likewise, winning bets on dog B would be paid 27667/10000=2.77 for 1 and winning bets on dog C would be paid 27667/15000 = 1.84 for 1.

The bettor in this case exploited the rules by betting such huge amounts that he pretty much controlled the odds. For the sake of simplicity, let’s assume he was the only bettor. The article said he bet \$350,000 on the two favorites and \$5,000 on each additional dog. With six underdogs (pun intended), that resulted in a total pool of 2?\$350,000 + 6?\$5,000 = \$730,000. After the take-out and split, there was \$201,997 to the winners of each dog. The rule about getting at least a push resulted in bets on the two favorites being refunded, because \$350,000 > \$201,997. However, the share of the pool on the third dog was huge compared to wagers on it. The winning odds would have been 201,997/5000 = 40.4 to 1. So, the profit on the third dog was \$5,000 ? 39.4 = \$197,000. He actually only won \$170,000, probably because of other bets on the third dog.

This technique would not work in the U.S., by the way, because in the U.S., we deduct the original wagers made on each winning dog from the total show pool and then add them back in after dividing by 3. This deduction would have caused the pools on the two favorites to be negative, resulting in just small winnings of the minimum \$0.10 per \$2 bet.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

Some gambling books say that the correct Kelly bet is advantage/variance. However, you say that is just an approximation and the correct answer is to maximize the expected log of the bankroll after the bet. My question is, how much error is there in the variance approximation?

Larry from Las Vegas

Advantage/variance is a pretty good approximation. Let’s look at full pay deuces wild, for example. The variance formula says to make a bet of 0.000295 times bankroll. Exact Kelly results in a bet of 0.000345 times bankroll.