# We have a 52 cards pack, we draw 2, what`s the probability that both are aces?

Ok so i dunno if my answer is correct:

the favorable cases are: combination of 4 taken 2

the total cases: combination of 52 taken 3

Update:

the total cases: combination of 52 taken 2

Relevance

Odds of first being an ace are 4 in 52, odd of second being an ace (assuming first really was one) are 3 in 51. To get the overall odds of drawing 2 aces you just multiply these together 4/52 * 3/51 = 12/2652 = 1/221... so the odds are 1 in 221 of drawing two aces in a row.

• For the first card you have 4 favorable cases and 52 total cases. Then you have a probability p_1 = 4/52 = 1/13 for obtain an ace here. If you draw an ace, do you have a probability p_2 = 3/51 = 1/17 for obtain a second ace (rest 3 aces and 51 cards). Then, the probability that you have both aces is (p_1)(p_2) = (4/52)(3/51) = (1/13)(1/17) = 1/221 = (combination of 4 taken 2)/(combination of 52 taken 2), the same you calculate.

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• There are 4 Aces.

When you draw a first card probability that it is an ace = 4/52=1/13

When you draw a second card probability that it is an ace = 3/51=1/17

Hence probability that 2 drawn cards are aces= (1/13)×(1/17)=3/221

Even otherwise favourable cases of selecting

two aces out of four= C(4,2) =(4!)/{(2!)×(4−2)!}=6

Total cases of selecting

two cards out of 52= C(52,2)=1326

Hence probability that 2 drawn cards are aces= C(4,2)/ C(52,2)

=6/1326=1/221

• There are 4 card communities, each and each with 13 playing cards. in case you do no longer placed the cardboard you drew back interior the %., the risk is: the two hearts: 13/fifty two * 12/fifty one= a million/17 2 aces: 4/fifty two * 3/fifty one=a million/221 King and queen: 4/fifty two * 4/fifty one= 4/663 wager you may circulate on from there