30 cards are selected from a deck 1) without replacement 2) with replacement.

We wish to know the chance of getting at least one ace.

1) Each time we select a card, it is either an ace or not. The simplest way to approach the problem is to realise that the probability of gettting at least one ace is equal to 1 - the probability of getting no aces.

At the nth turn there are 53-n cards left out of the 52 in the deck, and 53-n-4 of them are not aces. The probability of not getting an ace at each of the n=1 to 30 turns is then (49-n)/(53-n). We write this as

Pr(no aces) =

`prod_(n=1)^30 ((49-n)/(53-n)) = (48/52).(47/51)...(20/24).(19/23) = 0.027`

Therefore the probability of getting at least one ace is 1 minus this, ie

Pr(>=1 ace) = 1 - Pr(no aces) = 1 - 0.027 = 0.973 to 3dp.

The probability of not getting an ace starts out quite high (48/52) = 0.923. The longer you go without choosing an ace, however, the higher the chance of choosing one (the pack is getting smaller, but all the aces are still there). The probability of not getting an ace at the 30th turn has gone down to (19/23) = 0.826. On top of this, the current probability is multiplied by the probability of not having chosen an ace so far (which gets smaller and smaller the more turns you have). If left with only four cards, given an ace hasn't been chosen already, the next card chosen is an ace with probabilitiy 1, by which point the probability of getting no aces is 0.

2) If choosing the cards with replacement, the pack remains the same size at each turn and the probabilities remain constant. The events are now independent as opposed to dependent as the current result of choosing a card doesn't depend on what has gone before.

Again it is simpler to realise that the probability of choosing at least one ace is equality to 1 - the probability of choosing no aces. The probability of not choosing an ace at each turn, independently, is 48/52. The probability of doing this 30 times in a row is

`(48/52)^30 = 0.091 ` to 3dp

Now this is higher than choosing an ace 30 times in a row ((4/52)^30, approx 0), but the sheer number of times we require the card not to be an ace, of any suit, makes the probability small.

The probability of choosing a least one ace on any of these 30 independent occasions is then

Pr(at least one ace) = 1 - Pr(no aces) = `1-(48/52)^30 = 0.909 ` to 3dp

Answer:

The probability WITHOUT replacement is 0.973 (3dp), whereas WITH replacement it is 0.909 (3dp). The probability is higher when we don't replace the cards, as, as the pack dwindles in size, the chance of getting an ace (if one hasn't already been picked) increases as the choice of alternative cards reduces.