Deck of Cards Probability Question

Suppose 5 cards of a poker hand are dealt one at a time without replace­ment. If the first two cards drawn are aces what is the probability two of the last three will be aces.

this is a conditional probability problem.
using simplified form of bayes' th,
P[2 of the last 3 are aces | 1st 2 are aces]
= P[1st 2 are aces & 2 of the last 3 are aces] / P[1st 2 are aces]
= (4/52 * 3/51 ) * (3 * 2/50 * 1/49 * 48/48) / (4/52 * 3/51)
= 3/1225 or 0.2449%
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someone answered 1 out of 25, which is not correct. There are 52 cards in the deck, your question was if all 5 cards were dealt at the same time. The prior answer is discounting the first 2 cards which is a common mistake. the probability does not change just because 2 of the cards revealed are aces, because in reality their value has NO bearing what so ever on what the last 3 cards are. 4 of the 52 cards are aces (1 of 13 cards is an ace, but not necessarily 1 of EVERY 13 will be, but that is the actual probability of "any 4 of a kind in a 5 card hand is


The probability is 0.000240..

2 out of 50 or 1 out of 25 cards could be an ace.