A shuffle through the gaming mailbag

Q. I read in the news about that roulette wheel that hit number 19 seven times in a row. What are the odds, and how much would you have won if you'd bet it and just let your winnings ride for all seven?

A. For those who haven't seen the story, it started when a poker player took a picture of a lighted display that showed the last 16 numbers at a roulette table at the Rio in Las Vegas. Seven in a row were No. 19.

After the story first appeared, it was reported that there weren't actually seven 19s in a row. The Detroit News followed up with a quote from Caesars Entertainment spokesman Gary Thompson, who said, "There was no one playing at the table, it was just a diagnostic test being done."

OK, so the streak of seven 19s didn't actually happen. What if it had? The odds are 1 in 38 -- the total number of spaces on a double-zero wheel -- raised to the seventh power. That's 1 in 114,415,582,592, or 1 in more than 114 billion.

As to how much a bettor would have won, let's assume a $500 maximum bet. I've seen maximums on inside bets as low as $50, but let's give our bettor a chance at a good, profitable run. On the first win, that 19 would bring a 36-1 payoff, leading to a $37 wager on the second. The 36-1 payoff on that would bring $1,332 in winnings. Each of the remaining five wagers in the sequence would be limited to $500, and each would bring $18,000 in winnings.

So the total win would be five times $18,000, or $90,000, plus $1,332, plus $36, for a total of $91,368. And the player would still have the initial $1 wager.

One other thing. The odds of No. 19 turning up are 1 in 38 at every step. After one win, it's 1 in 38 that the next number will be 19, and after two wins it's 1 in 38 that the third spin will be 19. Assuming a balanced wheel in good working order, the odds don't change. The wheel has no memory. It doesn't know what numbers have turned up in the past, and it doesn't go into any makeup mode after a streak. In the long run, the streak just fades into statistical insignificance.

Q. My girlfriend and I are about to go on an overnight casino trip, and we both figure to play some video poker. The game we like is Double Double Bonus Poker. She's had a couple of royal flushes. I've never had one, but I've been pretty lucky in getting the four aces with the low card.

If we want to give ourselves some playing time on Double Double Bonus, how much money do you think we need? We'll mostly be playing quarter games.

A. Double Double Bonus Poker is one of the most popular video poker games, and the 2,000-coin bonanza on four aces with a 2, 3 or 4 is a big reason people like it so much. That's half the value of a royal flush on a hand that comes up about 2.5 times as often -- once per 40,799 hands for a royal, once per 16,236 for four aces and a kicker.

However, Double Double Bonus Poker is a high-volatility game with much of its return concentrated in relatively rare hands. The two big paying hands -- royals and four aces with kickers -- account for less than a hundredth of a percent of all hands, but account for about 4.4% of total payback.

I ran the numbers through the Video Poker for Winners software, and found that if you were planning about 10 hours of play at 500 hands an hour on 25-cent, single-hand games, and you wanted to keep your risk of ruin to about 5%, it would require a starting bankroll of about $885.

If the casino doesn't have the full-pay version and you wind up playing 8/5 Double Double Bonus, then it takes a starting bankroll of $1,010 to hold your chances of going bust within 10 hours to about 5%.

Double Double Bonus Poker is just that kind of game. There's a chance of big, fast rewards, but it also can extract rapid losses. It takes more cash to keep you going than do lower volatility games such as Jacks or Better or Bonus Poker.