Randomly Shuffled - 3 Ace's in a row

[ItalianFX]

I was just playing around with some numbers and wanted to see what a graph looked like for the probability of pulling out a selected card from a standard deck of cards that were randomly shuffled, without putting each selected card back into the deck.

Since we are selecting 1 card, the first time it is 1/52, and then 1/51, 1/50...etc. Suprisingly, maybe not, it is an exponential curve.

I decided to pick out the Ace/images/graemlins/diamond.gif, which happened on the 18th card (2.86% chance). However, when doing this, the 16th, 17th, and 18th card were all Aces.

My question is: What is the probability of pulling out 3 Aces in a row from a random place in the deck after being randomly shuffled each time?

[fiskebent]

C(4,3)/C(52,3) or 1 in 5525.

[ItalianFX]

[ QUOTE ]
C(4,3)/C(52,3) or 1 in 5525.

[/ QUOTE ]

Ok, I quit. I don't want to try it again. /images/graemlins/shocked.gif

Could you show me in a longer, written out form, how that is calculated? i'm not familiar with the C(x,x) thing.

[BruceZ]

[ QUOTE ]
I was just playing around with some numbers and wanted to see what a graph looked like for the probability of pulling out a selected card from a standard deck of cards that were randomly shuffled, without putting each selected card back into the deck.

Since we are selecting 1 card, the first time it is 1/52, and then 1/51, 1/50...etc. Suprisingly, maybe not, it is an exponential curve.

I decided to pick out the Ace/images/graemlins/diamond.gif, which happened on the 18th card (2.86% chance). However, when doing this, the 16th, 17th, and 18th card were all Aces.

My question is: What is the probability of pulling out 3 Aces in a row from a random place in the deck after being randomly shuffled each time?

[/ QUOTE ]

The probability of 3 consecutive aces somewhere in the deck is:

(49 + 48*47 + 2*48) / C(52,4) =~ 1 in 112.8.

There are C(52,4) total ways to select places for aces, irrespective of which ace goes in which place. Of these, 49 will have 4 consecutive aces since this is the number of places they can start. If there are exactly 3 consecutive, there are 50 places they can start, but for 48 of these, there willl be 47 places remaining for the final ace so that it doesn't come immediately before or after the 3 consecutive. If the 3 consecutive start with either of the 2 places 1 or 49, then there will be 48 places left for the remaining aces, since only 1 place will be excluded.

[BruceZ]

[ QUOTE ]
Ok, I quit. I don't want to try it again. /images/graemlins/shocked.gif

Could you show me in a longer, written out form, how that is calculated? i'm not familiar with the C(x,x) thing.

[/ QUOTE ]

Here is a way without combinatorics. The probability of 3 aces in a row starting with the first card is 4/52 * 3/51 * 2/50. Since 3 consecutive can start at any of 50 positions, we can multiply this by 50 to get very close to the final answer, but that would double count the times that we get 4 in a row, so to be exact we must subtract that off. The probability of 4 in a row starting with the first card is 4/52 * 3/51 * 2/50 * 1/49, and we can multiply this by the 49 places it can start to get exactly the probability of 4 in a row. Subtracting this from the previous result gives the final answer:

50*(4/52 * 3/51 * 2/50) - 49*(4/52 * 3/51 * 2/50 * 1/49) =~ 1 in 112.8.

This is in agreement with my first result.

[BruceZ]

Note, fiskebent's answer of 1 in 5525 is the probability of getting 3 in a row starting with a particular card, and this is 4/52 * 3/51 * 2/50. Maybe that is what you were actually asking. My answer is for the whole deck.

[fiskebent]

I simplified the question to "What are the odds of drawing 3 aces when I draw 3 cards from a deck of cards". I think that's the same scenario. Doesn't really matter how you draw the cards.

C(4,3): How many 3-card combinations can you make with 4 aces? The result is 4 combinations.
C(52,3): You many 3-card combinations can you make with 52 cards? Answer is 22100.

So 4 out of 22100 three-card combinations contain 3 aces. That's the same as 1 in 5525.

Bruce, your number is the probability that the aces are lined up. Not that we also start drawing at one where we can draw two more.

[ItalianFX]

Just to clarify, maybe you both understood what I was saying:

Draw a card: 7s
Shuffle
Draw a card: 2d
Shuffle
Draw a card: Th
Shuffle
Draw a card: Ad
Shuffle
Draw a card: Ah
Shuffle
Draw a card: Ac

And, I was selecting from about the middle of the deck each time.

It doesn't matter which card you use. In my example, I was looking for the Ad, and it just happened that 3 Aces came. The probability has to be the same for any 3 cards.

[fiskebent]

OK. Then we need to figure out what the odds are that 2 aces come after the ace of diamonds.

Odds that the first card after [Ad] is an ace is 3/51. Odds that the second card is an ace too is 2/50.

3/51 * 2/50 = 1 in 425.

With the C function it's C(3,2) / C(51,2)

[AaronBrown]

The shuffles and drawing from the middle of the deck don't matter, the probability would be the same if you just shuffled once and dealt from the top, or shuffled after every third card and drew from the bottom, or any other scheme.

BruceZ's answer is correct for the chance of getting three Aces in a row at some point in the deck. This was the situation you described. Fishbent's is correct for the first three cards being Aces.

The other possible thing you could have meant, is what is the probability of getting two Aces immediately after getting the Ace of diamonds. This is (2/51)*(1/50) = 1/1,275.

Your statement:

[ QUOTE ]
It doesn't matter which card you use. In my example, I was looking for the Ad, and it just happened that 3 Aces came. The probability has to be the same for any 3 cards.

[/ QUOTE ]
Is not quite true. If the "any three cards" means "A/images/graemlins/diamond.gif A/images/graemlins/heart.gif A/images/graemlins/club.gif" then the chances are about 1/4 BruceZ's answer. If you asked the probability of getting, say, three diamonds in a row, the answer is much higher.

To be precise, you should say the probability has to be the same for "three of any four cards."

[ItalianFX]

[ QUOTE ]

Your statement:

[ QUOTE ]
It doesn't matter which card you use. In my example, I was looking for the Ad, and it just happened that 3 Aces came. The probability has to be the same for any 3 cards.

[/ QUOTE ]
Is not quite true. If the "any three cards" means "A/images/graemlins/diamond.gif A/images/graemlins/heart.gif A/images/graemlins/club.gif" then the chances are about 1/4 BruceZ's answer. If you asked the probability of getting, say, three diamonds in a row, the answer is much higher.

To be precise, you should say the probability has to be the same for "three of any four cards."

[/ QUOTE ]

You're right. That was more of an implied statement since we were talkng about 3 of the same 4 rather than 3 of the same suit.

[fiskebent]

[ QUOTE ]
The other possible thing you could have meant, is what is the probability of getting two Aces immediately after getting the Ace of diamonds. This is (2/51)*(1/50) = 1/1,275.


[/ QUOTE ]

Isn't it (3/51)*(2/50)? There are 3 aces left. Not 2.

[AaronBrown]

You're right. I'm wrong.