What s the probability of getting AN ace in 5 consecutive draws from a deck of 52 cards? Do we need to take into account of the arrangement?

Method 1 : Not to consider the arrangement

A : an ace (there are 4 cards that are A)
N : not an ace (there are 52 - 4 = 48 cards that are N)

When 5 cards are drawn from 52 cards :
Total number of combinations without restriction = C(52,5)

To fulfill the criteria, 1 A is drawn from 4 A cards [C(4,1)], and 4 N cards are drawn for 48 N cards [C(48, 4)].
Number of combinations of getting an ace = C(4,1) × C(48,4)

P(getting an ace in 5 consecutive draws from a deck 52 cards)
= C(4,1) × C(48,4) / C(52,5)
= 4 × 194580 / 2598960
= 3243/10829


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Method 2 : To consider the arrangement

A : an ace (there are 4 cards that are A)
N : not an ace (there are 52 - 4 = 48 cards that are N)

When 5 cards are drawn from 52 cards and arranged in order:
Total number of arrangements without restriction = P(52,5)

To fulfill the criteria, 1 A is drawn from 4 A cards [C(4,1)], and 4 N cards are drawn for 48 N cards [C(48, 4)]. Then, arrange the 5 cards drawn [P(5,5)].
Number of combinations of getting an ace = C(4,1) × C(48,4) × P(5,5)

P(getting an ace in 5 consecutive draws from a deck 52 cards)
= C(4,1) × C(48,4) × P(5,5) / P(52,5)
= 4 × 194580 × 120 / 311875200
= 3243/10829

Are you looking for the probability of *exactly one* ace? Then yes, there are different ways to get exactly 1. It could be any of the 5 cards drawn.

You could calculate the probability that you draw 1 ace and then 4 non-aces, but then you'd have to multiply by 5 for the ways to pick which of the 5 cards is the ace vs. the non-ace.

P(first card is ace) = 4/52
P(second card isn't an ace) = 48/51
P(third card isn't an ace) = 47/50
P(fourth card isn't an ace) = 46/49
P(fifth card isn't an ace) = 45/48
C(5,1) = ways to pick the position of the chosen ace = 5
5 * 4/52 * 48/51 * 47/50 * 46/49 * 45/48
= 3243/10829
≈ 29.9%

Or you could use combinations.
There are C(4,1) ways to pick the suit of the ace in the 5 cards drawn.
There are C(48,4) ways to pick the cards for the non-aces.
There are C(52,5) ways to pick *any* five cards.

C(4,1) * C(48,4) / C(52,5)
= 3243/10829
≈ 29.9%

P.S. If you are looking for the probability of *at least* one ace, then it's even easier. Calculate the opposite case of drawing *no aces* and then subtract from 1. --> 1 - (48/52 * 47/51 * 46/50 * 45/49 * 44/48)

Do you mean "at least one ace" or "just one ace" (in the five cards drawn)?
What "arrangement" do you mean? Of the deck before drawing? Of the five drawn cards?
Is each card replaced before drawing the next?

Assuming the draw is random, you don't need to take into account the arrangement. You DO need to take into account whether or not the draws are with replacement. Once you draw the first card, do you put it back? Or leave it out?

(1) With replacement, i.e. putting the card back each time:

The draws are independent, meaning each draw of an ace doesn't affect the probability of the next draw being an ace. The odds of an ace on each draw is 4/52 (4 aces out of 52 cards), which reduces to 1/13.

The odds of drawing an ace five times with replacement would be
1/13 * 1/13 * 1/13 * 1/13 * 1/13, or 1/(13^5), or 1 / 371,293.

(2) Without replacement, i.e. taking each card and leaving it out...the odds are zero. You don't even need to do the math.

How do we know? Because there are only 4 aces! Once you have drawn 4, there is no fifth ace you can draw.