Wouldn't 8% be great! I'd go all in every time with a single A paired to the board, kicker be dammed!

However, it is not 8% it would be 62.44%

Waggho’s method is the correct way to calculate the chance of someone else having an Ace. As he says, there are 18 cards in the hands of other players. 47 cards remain unseen and of these 2 are Aces. On the first of the 18 cards that makes the chance of no ace 45/47. Then we multiple the chance of an A not showing another 17 times. So we have:
45/47 x 44/46 x 43/45 ….. on to the 18th one which would be 28/30. So this works out to:
(45*44*43*42*41…..*29*28) / (47*46*45…*31*30). We would calculate this by using factorials. Just in case you don’t know what a factorial is (if you care?), it is simply the value of a series of numbers multiplied, i.e. 5 factorial (which is written 5!) is 1x2x3x4x5=120. To answer the above question the equation would be:
(45!/27!) / (47!/29!). This gives us the percentage chance that no Ace was drawn in the 18 cards as 37.56%. So the chance that at least one was drawn is 62.44%

Thanks,
BreakfastMan