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I heard there was a "perfect" game of bridge recently, where each player got 13 cards of one of the four suits. What are the odds of that?

Rob from Las Vegas

There are also rumors of that happening in a game of whist in Warwickshire in 2011. For the benefit of other readers, this is asking what is the probability of dividing a 52-card deck into four groups of 13 cards each, where each set is composed entirely of the 13 cards of one of the suits.

There are combin(52,13)*combin(39,13)*combin(26,13) = 53,644,737,765,488,800,000,000,000,000 possible ways to arrange 52 cards into four 13 card sets. 4! = 24 would be the number of winning combinations, since you could distribute the four suits to the four players any way you wish. So the odds are equal to 1 in 2,235,197,406,895,370,000,000,000,000. To give that number some perspective, if all 7.5 billion people on earth dealt bridge hands at a rate of one per second, the probability of someone dealing a so-called perfect hand before the sun explodes in 5 billion years is 1 in 16,558.

However, some define a "perfect" hand as just one player getting all 13 cards of any suit. I show the probability of this is 1 in 39,688,347,497 games. This probably does happen once in a while somewhere on earth.

Wiz, I know you're a wet blanket when it comes to the lottery, but can you suggest an easier way to turn \$2 into a \$1,000,000?

"Anonymous" .

Yes. The house advantage in most lotto-based lottery games is close to 50%. So, a hypothetical \$2 game with a \$1,000,000 jackpot, with no smaller prizes, would need to have a probability of winning of 0.5*(2/1000000) = 1 in one million to retain a 50% house edge.

Here is my strategy to turn \$2 into \$1,000,000 with better odds than that.

• First bet \$2 on any number in double-zero roulette. You can find \$2 minimums at some Vegas casinos, like the El Cortez and South Point. If you win, you'll be up to \$72.
• Next, parlay your \$72 onto another single-number bet. If you win, you'll be at \$2,592.
• Next, carry that \$2,592 into one of the high-end casinos on the Strip, like the Wynn, Venetian, or Bellagio. Bet your \$2,592 from roulette on the Banker bet in baccarat. Do this a total of nine times, letting everything ride each time. After your ninth win, you'll be up to \$1,056,687. Your ninth bet would be for \$541,891, which I'm sure any of these casinos would take if they saw you win it in front of their noses.

The probability of winning a single-number bet in double-zero roulette is 1/38. The probability of winning the Banker bet in baccarat is 50.6825%, not counting ties. So, the probability of two roulette wins and nine Banker wins is (1/38)^2 × 0.506825^9 = 1 in 654,404. Those are better odds than the 1 in a million you would get with a lottery and you'll have a little more than a million dollars too.

I disagree with your assertion that Popeye had one anchor and one tank tattoo. Here is a still from the cartoon Blow Me Down, showing an anchor tattoo on both arms. Please issue a correction.

MustangSally

Okay, point taken. Here is my evidence that Popeye has a tank tattoo on is left arm. I submit for your consideration that Popeye had the anchor on that arm removed and replaced it with a tank. Click on the image for a larger version.

This question in raised and discussed in my forum at .

An alien has abducted ten logicians and placed them in a room. He explains to them that he first order them in a line, from tallest to shortest, with each person facing the direction of the next shortest person in front, so that each person may see all shorter logicians, but no taller ones. He then explains that he will place a black or white hat on each person, however nobody will be able to see the color of his own hat, only the hats of the shorter logicians. The distribution of black and white hats could be anything, not necessarily five and five.

The alien then explains that he will ask each logician, starting with the tallest, and moving down the line, the color of his hat. The logicians may hear the response of those acting before them. Other than the black/white responses, they may not communicate in any way once the game has started. If more than one logician is wrong, then they will all be eaten. If at least nine answers are correct, they will be returned safely to earth. The alien then gives them some time to strategize. What should their strategy be?

"Anonymous" .

Following is one possible strategy. Have the first logician to act say "black" if he sees an even number of black hats and "white" for an odd number on the other nine logicians. He will have a 50% chance of matching the color of his own hat, so he is the only one allowed to be wrong. Whatever he says, define that as the "running color."

Next, the second logician to act will take a count of the black hats on the shorter eight logicians and match it to a color by the same odd and even method as the first logician. If this agrees with the color stated by the first logician, then he must have a white hat, and he should state white. If it disagrees, he must have a black hat, and should state black. If he says "white," then the running color will stay the same. If he says "black," then the running color will flip to the opposite color.

Next, the third logician to act will do exactly as the second one, but taking a count of the black hats on the seven shorter logicians. Likewise, if he agrees with the running color, he will say "white" and the running color stays the same. If he disagrees with the running color, he will say "black" and the running color will flip.

Each of the other logicians will do exactly the same thing.

Of course, they could just as easily associate black to an odd total and white to even. The important thing is they all agree which color means which odd/even parity. Following either strategy will result in the second to tenth logicians to act all being correct and the first one having a 50% chance, so they will all live. This strategy will work for any number of logicians.