In a standard deck of 52 cards, the two red jacks face left.
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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.
There are 46 cards left (assuming no Jokers), of which 12 are red face cards, so the answer is 12/46 or 6/23 or ~26%.
The odds that the first card would be black is 50:50 or 1 out of 2, since there are 26 black cards out of 52. Now with 51 cards left, there are only 25 black cards still available, so the odds that the second card would be black is 25 out of 51. Multiply 1/2 by 25/51 and you get 25/102 or a 24.5% chance that you select 2 black cards out of a deck.
The answer is 8,648,640 in 674,274,182,400 or approximately 0.128%. To pull seven spades in a row from a deck of 52 cards may be translated as follows: In a deck of 52 cards there are 13 spades, so 13 chances to pull a spade or 13/52. After pulling one spade, there are 12 spades left and 51 cards left, so 12 chances to pull a spade or 12 / 51. After pulling the second spade, there are 11 spades left and 50 cards left, so 11 chances to pull a spade or 11 / 50. Continuing the above logic to the natural conclusion, we get the following multipliers: 13 / 52 * 12 / 51 * 11 / 50 * 10 / 49 * 9 / 48 * 8 / 47 * 7 / 46 which is equal to 8,648,640 / 674,274,182,400 or approximately 0.128%.
There are 4 aces in the deck the odds that the first card is an ace is 4/52 or 1/13. The odds the second card is an ace is 3/51 or 1/17 because there are only 3 aces and 51 cards left. The odds that both are aces are 1/13 times 1/17 which is 1/221.