# Probability of obtaining Ace High Straight

Consider a standard 52 card deck of playing cards. Suppose that we reduce the number of cards in the deck by

• removing one of the Aces

• removing one of the Queens

• removing one of the Jacks

• removing one of the tens

• removing one of the threes.

The cards that are removed are discarded and are not used for the remainder of this question. As such we now have a deck that consists of just 47 cards. Two cards are selected but are not kings leaving 45 cards in total.

Question: P(at least one of the cards is a King) if the player was obtain an Ace-High straight in poker with the 2 cards selected before?

Attempt: There is a 4/45 chance that the card can be a king however, i do not know what to do with this information to find the answer. Please help!

• do you know how to calculate an ace high straight with all the cards? Ignoring suits, there is one set of cards that is an ace high straight , A,K,Q,J,10 - now normally that leads to $4^5$ different possibilities when introducing suits. (that includes straight flushes) , but now, each card only has 3 possible suits, apart from the king. The loss of the 3 reduces the number of possible hands, but not the number of ways of getting the straight
– Cato
May 17, 2017 at 15:03
• I think it is $(3 \times 4 \times 3 \times 3 \times 3) / (^{47}_5C)$ - Are you familiar with those sorts of terms? (counts straight flush as a straight - otherwise more info on missing cards needed)
– Cato
May 17, 2017 at 15:11

$$\frac { 3 * 4 * 3 * 3 * 3 } { \binom{47}{5} } = 0.00021122 = 1 / 42856$$