The question is the same as:
"After removing the 4 aces from a standard deck, how many different hands
of 9 cards can be dealt from the remaining 48 ?"
The calculation is
48! / (39! x 9!) = (48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 x 40) / (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2)
We get 1,677,106,640 .
If three aces have already been dealt with, there is only one ace left and 49 cards left. P(4th ace)=1/49 The odds are 1 to 49. It is here assumed that four cards cards are delat one by one and the first three were aces.
Since there are only four aces in a standard 52 card deck, the probability of being dealt five aces is zero.
There are 4 Aces in a deck of 52 cards.4
To find the probability of being dealt exactly 4 aces in a 13-card hand from a standard 52-card deck, we can use the hypergeometric distribution. The total number of ways to choose 4 aces from 4 available is ( \binom{4}{4} = 1 ), and the number of ways to choose the remaining 9 cards from the 48 non-aces is ( \binom{48}{9} ). The total number of ways to choose any 13 cards from 52 is ( \binom{52}{13} ). Thus, the probability is given by ( \frac{1 \times \binom{48}{9}}{\binom{52}{13}} ).
There are two red aces in a deck of 52 cards.
The odds are 220:1 of being dealt pocket aces.
The probability of being dealt pocket aces in a game of poker is approximately 1 in 221 hands.
Poker hands are combinations of cards (when the order does not matter, but each object can be chosen only once.)The number 52C5 of combinations of 52 cards taken 5 at a time is (52x51x50x49x48) / (5x4x3x2x1) = 2,598,960.The number of hands which contain 4 aces is 48 (the fifth card can be any of 48 other cards.)So there is 1 chance in (2,598,960 / 48) = 54,145 of being dealt 4 aces in a 5 card hand.The odds are 54,144 to 1 against. The probabilityis 1/54145 = (approx.) 0.000018469 or 0.0018469%.
If three aces have already been dealt with, there is only one ace left and 49 cards left. P(4th ace)=1/49 The odds are 1 to 49. It is here assumed that four cards cards are delat one by one and the first three were aces.
In Texas Hold'em poker, the best starting hands are typically pocket pairs like Aces, Kings, Queens, or Jacks, as well as strong high cards like Ace-King or Ace-Queen suited. These hands give players a good chance of winning the pot before the community cards are dealt.
When pocket Aces are given to a player, 99% of the time, the player will split the Aces into 2 hands. In most cases the player will push when they do this, they will win one and lose one hand.When the dealer gets the pocket aces, it's treated as a 12, unless the dealer passes 21, in which case it's made into a 2, and added to the sum of the values of cards. The dealer is not able to split these cards into 2 hands, and must play it as one hand.
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
Counting Aces as a face card, the answer is 0.0241 If Aces are not considered face cards, then the answer is 0.0181
Since there are only four aces in a standard 52 card deck, the probability of being dealt five aces is zero.
According to the Hold'em hands chart, the best starting hand to play in Texas Hold'em is pocket aces (two aces as your hole cards).
There are 4 Aces in a deck of 52 cards.4