The answer will depend on the exact situation.If you are dealt a single card, the probability of that single card not being a queen is 12/13 - assuming you have no knowledge about the other cards.Here is another example. If you already hold three queens in your hand (and no other cards have been dealt), the probability of the next card being dealt being a queen is 1/49, so the probability of NOT getting a queen is 48/49 - higher than in the previous example.
The answer will depend on how many cards you are dealt.
If the pack is well shuffled, the probability is 1/52.
1/13
The probability, if the cards are dealt often enough, is 1.On a single deal, the prob is 3.69379*10^-6
Probability = Chance of Success / Total Chances (Chance of Success + Chance of Failure) There are 4 aces in a 52 card deck and 48 cards that are not aces. Probability of being dealt an ace = 4 / (4 + 48) = 4/52 = .0769 or about 7.7 percent
1/26 under the assumption that one is using a standard 52 card deck with no jokers. 1/27 if the jokers are included.
If only one card is dealt randomly from a deck of cards, the probability is 1/52.
You have four chances out of 52, therefore 4/52=1/13.
The probability is 0. One card cannot be a club and a spade!
If the pack is well shuffled, the probability is 1/52.
Probability of 2 of clubs = 1/52 or 0.0192.
Since there are only four aces in a standard 52 card deck, the probability of being dealt five aces is zero.
1/13
The probability, if the cards are dealt often enough, is 1.On a single deal, the prob is 3.69379*10^-6
Probability = Chance of Success / Total Chances (Chance of Success + Chance of Failure) There are 4 aces in a 52 card deck and 48 cards that are not aces. Probability of being dealt an ace = 4 / (4 + 48) = 4/52 = .0769 or about 7.7 percent
It is 28/52 = 7/13
1/26 under the assumption that one is using a standard 52 card deck with no jokers. 1/27 if the jokers are included.
It is 0.00111 approx.