Two poker enthusiasts asked me about something they heard while watching a hold 'em tournament on TV. They said they heard the color announcer say, "If you hold an ace, then there's a 70 percent chance that you have the only ace dealt out among the players." They asked me if that was correct. I told them they misheard the statement; they insisted they heard correctly. If so, then the announcer erred, big time.

If the dealer dealt you an ace and a kicker, then 50 cards remained from which to deal nine hands to your opponents. The number of ways to do that equals 18,053,528,883,775 [C(50,18)].

The number of ways to deal 18 cards from 47 non-ace cards equals 4,568,648,125,690 [C(47,18)]. We divide to find the probability, 0.25306.

So, either they each misheard the announcer, or the announcer had it almost exactly backwards. If you have an ace, then the chance that someone else has an ace equals ~75 percent.

"Well, perhaps he meant when an ace appeared on the flop," they wondered.

Nope. When an ace and two other cards appear on the flop the probability that the dealer didn't distribute any aces obtains from [1-C(45,18)/C(47,18)], or 0.62442. An ace on the flop improves your chances of having the only ace to ~38 percent, hardly 70 percent.

Parenthetically, that fact shows the importance of your kicker. More than 62 percent of the time you need to overcome an opponent's kicker in order to win the pot.

"Well, perhaps he meant in a nine-handed game, not a ten-handed game," they wondered.

In a nine-handed game, if the dealer dealt you an ace and a kicker, then that left 50 cards from which to deal nine hands to your opponents. The number of ways to do that equals 4,923,689,695,575 [C(50,16)]. We divide that number into 1,503,232,609,098 to find the probability of no one else having an ace equals 0.30531.

Thus, you have a 69.5 percent chance of running into an ace among your eight opponents before the flop. If the dealer flops an ace, you still have a 57 percent chance of encountering another ace among your eight opponents.

"Well, what if there were two aces on the board after the dealer laid out all five community cards," they asked.

We can do that easily enough. C(44,18)/C(45,18) obtains the probability in a ten-handed game, 0.6. Ina nine-handed game, C(44,16)/C(45,16) yields the probability, 0.644. When two aces lie on the table after the river, you have a 60 percent chance of having the only ace in a ten-handed game and a 64 percent chance in a nine-handed game. (Those numbers don't give you the nuts by any means: you still need to have a better kicker much of the time.)

"Of course those numbers hold true whenever you hold trips with two on the table. Look at this deal:

Kh-9h
Ad-9d

9s-Qc-3d-2h-9s

"You have Kh-9h. The nine on the river makes open trips (three nines) for you. Lying low until she raised you on the river, an opponent stayed with her suited ace-nine!

"How many times do you have to lose to a bigger kicker when holding open trips before you get the idea that the chance of someone's holding the case card is far from small," we asked.

They didn't answer. They walked off saying, "64 percent is pretty close to 70 percent-that's probably what he meant-he just made a rounding error."

Riiight.

Mr. Burke is the author of Flop: The Art of Winning at Low-Limit Hold 'Em, on sale at amazon.com & kokopellipress.com. E-mail your Hold 'Em questions to richardburke@comcast.net