Stats question

[Sir_Bee]

I was curious if anyone here would be able to help me. Here is the scenario I am trying to figure out:

I have 3 cards, the King of Spades, King of Clubs, and the Ace of Hearts. I put them all face down on the table and mix them up. You then pick up a card at random. The chance you draw the Ace is 1/3. Now lets say I took whatever you drew and put it back in the pile, mixed the cards up again, and you then drew another card. The odds of you picking the Ace at least one of those two times now becomes 5/9. This was calculated the long way. There are 9 possible draws of 2 cards, 5 of them have an Ace. I then did it the long way again to find out what would happen if I added another draw. I got to 19 out of 27 that had at least one A. Order is not important, and repetitions are allowed.

I am pretty sure there is an easier way to do this, I just cannot remember what the formula is I need to use. If anyone knows it, could they please let me know. Also, if I have made a mistake in my math, please correct me.

[BeautifulTorment]

You just keep multiplying 1/3*1/3*1/3 depending on how many times you want to get the same card in a row.

Edit: Actually, that is the chance of repeatedly picking the same card. Give me a minute to think about what you actually asked for!

Edited January 14, 2013 by BeautifulTorment

[SitusKiss]

I'm not sure I follow what you are looking to calculate here. It seems like 2 different probabilities are in play.

Let me see if I can get this straight...

1) You know the 3 cards (K, K, A) you start with and place them face down. Your odds are currently 1/3 (or 33.333...%) of pulling the Ace.

2) You now place that card you pulled back into the deck and draw a random (potentially even the King you just put back) card to get 3 back on the table. I see a fluctuation possibility here depending on the 1st card you picked, from the 3, originally on the table:

a. If you drew the Ace, your odds will now be... (4%)

b. If you drew one of the Kings, your odds will now be...(56%) - I came up with this by combining two probabilities 6% for drawing an Ace from the remaining 50 cards and 50% for drawing the Ace from the 2 cards left, of the original 3.

I'm not even sure this ^ is accurate, haha. It makes my head hurt a little.

[Sir_Bee]

I'm not sure I follow what you are looking to calculate here. It seems like 2 different probabilities are in play.

Let me see if I can get this straight...

1) You know the 3 cards (K, K, A) you start with and place them face down. Your odds are currently 1/3 (or 33.333...%) of pulling the Ace.

2) You now place that card you pulled back into the deck and draw a random (potentially even the King you just put back) card to get 3 back on the table. I see a fluctuation possibility here depending on the 1st card you picked, from the 3, originally on the table:

a. If you drew the Ace, your odds will now be... (4%)

b. If you drew one of the Kings, your odds will now be...(56%) - I came up with this by combining two probabilities 6% for drawing an Ace from the remaining 50 cards and 50% for drawing the Ace from the 2 cards left, of the original 3.

I'm not even sure this ^ is accurate, haha. It makes my head hurt a little.

The only 3 cards being considered are the two kings and the Ace. So the second draw is identical to the first, it has the same odds of pulling an Ace (1/3). My question is how to simply calculate the odds of getting at least one Ace over both draws. Same goes for 3 draws.