Assuming blackjack values, there are seven cards per suit that can be dealt first that could lead to a total of 19. Each of these has one card per suit to be dealt next to make 19, except an 8 has two. If it's a standard 4-deck situation, the odds are:
(6/13)(16/207) + (1/13)(32/207) = 128/2691
The probability that the first four cards are face cards is (16/52)*(15/51)*(14/50)*(13/49) = 43680/6497400 = approx 0.0067
If you are drawing two cards from a full deck of cards (without jokers) then the probability will depend upon whether the the first card is replaced before the second is drawn, but the probability will also be different to being dealt a hand whilst playing Bridge (or Whist), which will again be different to being dealt a hand at Canasta. Without the SPECIFIC context of the two cards being got, I cannot give you a more specific answer.
There are 52 cards of which 26 (a half) are black. So he probability that the first card is black is 26/52= 1/2
If the card is drawn randomly, the probability is 1/4.
In Texas Hold'em, each player is dealt two cards face down as their starting hand. These are called the "hole cards." The first three community cards dealt face up on the table are called the "flop."
There is one chance in four of the first card being a diamond (since there are four suits). There is one chance in four of the second card being a diamond. The chances of both being a diamond is 1/4*1/4 = 1/16, or thus one chance in 16
The minimum number of cards that must be dealt, from an arbitrarily shuffled deck of 52 cards, to guarantee that three cards are from some same suit is 9.The basis for 9 is that the first four cards could be from four different suits, the next four cards could be from four different suits, and the ninth card is guaranteed to match the suit of two of the previously dealt cards. The minimum number, without the guarantee, is 3, but the probability of that is only 0.052, or about 1 in 20.
There are 36 number cards, 12 face cards, and 4 aces. So the probability of your first card from a full pack being a number card is 36:16, or 9:4.
It is (8/52)*(7/51) = 56/2652 = 0.0211 (approx).
The probability of drawing a king followed by a queen is (4 in 52) times (4 in 51), or 16 in 2652, or 4 in 663, or 0.006033.
The odds of being dealt AK in Texas Hold'em is computed as follows:The first card can be either an A or a K, a total of 8 possible cards out of 52 cards in the deck. So, the probability is 8/52 = .153846.If you get an A on your first card, there are 4 Ks; if you get a K on your first card, there are 4 As. So in either case, if you get an A or a K on your first cards, there are 4 possible cards out of the remaining 51 cards that will make the AK hand, which is a probability of 8/51 = .078431Now, multiply two probabilities and you have .0121.To convert from probabilities into odds, divide the probability by (1 - probability). So, .0121 / (1 - .0121) = about 82 to 1.
The probability that two cards drawn from a deck of cards being an Ace followed by a King is 1 in 13 (for the Ace) times 4 in 51 (for the King) which is equal to 4 in 663.