# You are dealt 4 cards from a single (initially complete) deck. likelihood that you receive 3 aces?

52 initially

4/52 = 7.69%

3/51 = 5.88%

2/50 = 4%

do i add this? or multiply this? am i even on the right track?

Relevance

3 aces can be selected from 4 in 4C3 ways and balance one card from balance 48 in 48 ways

=> there are 48C3 x 48 ways of getting 3 aces out of 52C4 total n umber of getting any 4 cards.

=> required probability

= 4C3 * 48/52C4

= (4x48x24)/(52x51x50x49)

= 0.0007092.

I have rechecked my answer and it is correct. To explain it differently,

Suppose the first card selected is ace. Its probability is 4/52 as you rightly thought. Similarly 2nd ace and 3rd ace probabilities are 3/51 and 2/50 Now fourth non-ace probability is 48/49. But one more thing to be kept in mind is that the nonace card may be picked up as first, second, third or fourth attempt which makes the final asnwer as

(4/52)*(3/51)*(2/50)*(48/49)*4

which is the same as what I worked out above.

• Login to reply the answers
• Leave them as fractions and multiply them out.

For example if the question was 3 cards dealt and 3 aces

the the probability is as you said 4/52 * 3/51 * 2/50 = 24/132,600 =.0001809. Not too likely a chance.

Then you have the added complication of having four cards dealt and only 3 of them required. You have to consider abc (already done), but also abd,acd and bcd in the same way as above and ADD the probabilities together.

Source(s): self learning for my degree course . hey I'm 51 give me a SMALL break.....lol
• Login to reply the answers
• The are 4 aces in a customary deck of 50 two enjoying cards. So on a single draw (if the deck is done) you have a 4 in fifty two danger of pulling an ace, which reduces to a million in 13. precis: a million in 13 (or 4 in fifty two) that's a 7.6923% danger.

• Login to reply the answers
• Anonymous
• Login to reply the answers