consecutive aces in a deck of cards

**george12345** · January 10th 2011, 07:02 AM

I have come across a problem relating to playing cards, and am having some difficulty figuring out the solution.

Assume you have 1 deck of shuffled cards. You proceed to deal the cards out one by one. I would like to know the probability of getting 2 consecutive aces anywhere in the sequence of 52 cards.

Thus far I have the following:
The chance of getting four consecutive aces anywhere in the deck.

I have calculated as follows:
A sequence can start on Card 1, 2, 3, .... up to the 49th spot. Thus there are 49 possible start positions for the sequence. Further, since there are 4 aces, each sequence can be arranged in 4x3x2x1 ways (4! ways). Thus there are 49x24 possibilities. With these 1,176 possibilites the remaining cards of the deck can be arranged in 48! ways for EACH of the 1,176 possibilities. Thus in total there are 48! x 1,176 ways in which a deck of cards can contain a 4-ace sequence.
There are 52! in which a deck of cards can be arranged.
Thus the probability that a particular card arrangement contains a 4-ace sequence is

(1,176 x 48!) / 52! = 1,176 x 48! / (52x51x50x49x48!)
= 1176 / (52x51x50) = 0.018% (or roughly 1 in 5,525)

I then looked at the likelihood of getting a 3-ace sequence

A sequence can start on Card 1, 2, 3, .... up to the 50th spot. Thus there are 50 possible start positions for the sequence. Further, since there are 4 aces, each sequence can be arranged in 4x3x2 ways. Thus there are 50x24 possibilities. With these 1,200 possibilites the remaining cards of the deck can be arranged in 49! ways for EACH of the 1,200 possibilities. Thus in total there are 49! x 1,200 ways in which a deck of cards can contain a 3-ace sequence.
However this includes the 4-ace sequences which I want to exclude.
49! x 1200 = 7.29 x 10^65
48! x 1176 = 1.46 x 10^64

After the subtraction we are left with 7.16 x 10^65

Thus the probability that a particular card arrangement contains a 3-ace sequence is

(7.16 x 10^65) / 52! = 0.904% (or roughly 1 in 110.5)

Is my approach correct thus far?

**e^(i*pi)** · January 10th 2011, 07:07 AM

Isn't the chance of getting 4 aces, in turn, simply $\dfrac{4}{52} \cdot \dfrac{3}{51} \cdot \dfrac{2}{50} \cdot \dfrac{1}{49} = \dfrac{4! \cdot 48!}{52!}$

**snowtea** · January 10th 2011, 07:10 AM

Originally Posted by e^(i*pi)

Isn't the chance of getting 4 aces, in turn, simply $\dfrac{4}{52} \cdot \dfrac{3}{51} \cdot \dfrac{2}{50} \cdot \dfrac{1}{49} = \dfrac{4! \cdot 48!}{52!}$

I think he means getting 4 consecutive aces anywhere when dealing all the cards (not just in the beginning).

**snowtea** · January 10th 2011, 07:11 AM

Originally Posted by george12345

Assume you have 1 deck of shuffled cards. You proceed to deal the cards out one by one. I would like to know the probability of getting 2 consecutive aces anywhere in the sequence of 104 cards.

Since when does a deck have 104 cards?

**george12345** · January 10th 2011, 07:26 AM

Originally Posted by e^(i*pi)

Isn't the chance of getting 4 aces, in turn, simply $\dfrac{4}{52} \cdot \dfrac{3}{51} \cdot \dfrac{2}{50} \cdot \dfrac{1}{49} = \dfrac{4! \cdot 48!}{52!}$

Hi Donor, thanks for your quick reply! Your answer - i think that's the probability of getting four aces in a row when drawing four cards out of deck without replacement. I am looking for the probability of getting a sequence of 4 aces anywhere in the deck of 52 cards. and then the probability of getting a sequence of 3 aces anywhere in the deck, and so on. Does that make sense?

**Soroban** · January 10th 2011, 08:23 AM

Hello, george12345!

You have a deck of shuffled cards and you deal the cards out one by one.
Find the probability of getting 2 consecutive aces in the sequence of 52 cards.

There are: $52!$ possible orderings of the cards.

We have: .4 aces and 48 Others.

Place the 48 Others in row, leaving a space before, after, and between them.

. . $\_\,O\,\_\,O\,\_\,O\,\_\,O\,\hdots \,\_\,O\,\_$

There are 49 spaces.

Select 4 of the spaces and insert the four Aces (in some order).

. . There are: . $_{49}P_4$ ways.

Hence: . $P(\text{no consecutive Aces}) \;=\;\dfrac{_{49}P_4}{52!}$

Therefore: . $P(\text{some consecutive Aces}) \;=\;1 - \dfrac{_{49}P_4}{52!}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If the problem meant exactly one pair of consecutive Aces,
. . we can modify the above procedure.

We have the 48 Others with spaces before, after, and between them.

. . $\_\,O\,\_\,O\,\_\,O\,\_\,O\,\hdots\,\_\,O\,\_$ . (49 spaces)

We have 4 aces.
Select two of them to be together; there are $_4C_2 \,=\,6$ choices.

Duct-tape the pair together.
Then we have 3 "cards" to distribute: . $\boxed{AA}\,,\:A,\:A$
And they can be placed in: . $_{49}P_3$ ways.

Hence, there are: . $6(_{49}P_3)$ ways to have one pair of consecutive Aces.

The probability is: . $\dfrac{6(_{49}C_3)}{52!}$

**Soroban** · January 10th 2011, 08:45 AM

Your two cases are even easier.

There are $52!$ possible orderings of the cards.

4 consecutive Aces

Duct-tape the four Aces together.
. . They can be in: $4! = 24$ orders.

Then we have 49 "cards" to arrange: . $\boxed{AAAA}\,,\,O,\,O,\,O\,\hdots\,O$
There are: $49!$ arrangements.

Therefore: . $P(\text{4 consecutive Ace}) \;=\;\dfrac{24\cdot49!}{52!}$

3 consecutive Aces

Choose 3 of the Aces to be together; there are $_4C_3 = 4$ choices.
. . and they can be in $3! = 6$ orders.

Duct-tape the 3 Aces together.
So we have two "cards": . $\boxed{AAA}\,,\,A$ .to distribute among the 49 spaces.
. . There are: . $_{49}P_2$ ways.

Hence, there are: . $4\cdot6\cdot(_{49}P_2)$ ways to have 3 consecutive Aces.

Therefore: . $P(\text{3 consecutive Aces}) \;=\;\dfrac{24\cdot(_{49}P_2)}{52!}$

**george12345** · January 10th 2011, 08:54 AM

hi soroban
you have used a completely approach then what I have been trying to do- and your approach seems much simpler.
So basically you are determining in how many ways you can mix 4 aces into the 48 other cards without letting them be in a consecutive order, because you are only allowing one space in between each O? And the difference between 52! and the number that you determine must be the number of ways in which you can have two consecutive aces.

Presumbaly that number would also include instances where there are 3 or 4 aces in succession?

Can that approach also be used for determining the likelihood of getting 3 aces or 4 aces?

**Opalg** · January 10th 2011, 10:03 AM

My approach is similar to Soroban's, but I think the numbers come out different and I'm not sure which of us is right.

Edit. My method was very wrong!

**george12345** · January 10th 2011, 03:16 PM

Originally Posted by Soroban

Your two cases are even easier.

There are $52!$ possible orderings of the cards.

4 consecutive Aces

Duct-tape the four Aces together.
. . They can be in: $4! = 24$ orders.

Then we have 49 "cards" to arrange: . $\boxed{AAAA}\,,\,O,\,O,\,O\,\hdots\,O$
There are: $49!$ arrangements.

Therefore: . $P(\text{4 consecutive Ace}) \;=\;\dfrac{24\cdot49!}{52!}$

3 consecutive Aces

Choose 3 of the Aces to be together; there are $_4C_3 = 4$ choices.
. . and they can be in $3! = 6$ orders.

Duct-tape the 3 Aces together.
So we have two "cards": . $\boxed{AAA}\,,\,A$ .to distribute among the 49 spaces.
. . There are: . $_{49}P_2$ ways.

Hence, there are: . $4\cdot6\cdot(_{49}P_2)$ ways to have 3 consecutive Aces.

Therefore: . $P(\text{3 consecutive Aces}) \;=\;\dfrac{24\cdot(_{49}P_2)}{52!}$

Thanks for your reply. The way you explained this seems logical to me - thank you so much! I got the same answer as you for the case of 4 consecutive aces with a probability of 0.0181% (or 1 in 5525). For the 3 consecutive aces I got a different answer - however I think I know why.

Originally Posted by Soroban

Therefore: . $P(\text{3 consecutive Aces}) \;=\;\dfrac{24\cdot(_{49}P_2)}{52!}$

This results still needs to be multiplied by 48! right? As there are 48! ways of ordering the remaining 48 (all non-ace) cards, for each of the successful permutations that you have identified.

If this is multiplied out I arrive at a probability of 0.8688% (or about 1 in 113). The formula you stated above on its own would give a probability of 0.0000%, which doesn't seem right - since the probability of getting a 3-ace sequence should be higher than getting a 4-ace sequence. Can you confirm whether you agree?
In particular this figure of 0.8688% does not include 3-ace sequences that actually form part of a 4-ace sequence - am I understanding correctly?

**george12345** · January 10th 2011, 04:04 PM

Originally Posted by Soroban

Hello, george12345!

There are: $52!$ possible orderings of the cards.

We have: .4 aces and 48 Others.

Place the 48 Others in row, leaving a space before, after, and between them.

. . $\_\,O\,\_\,O\,\_\,O\,\_\,O\,\hdots \,\_\,O\,\_$

There are 49 spaces.

Select 4 of the spaces and insert the four Aces (in some order).

. . There are: . $_{49}P_4$ ways.

Hence: . $P(\text{no consecutive Aces}) \;=\;\dfrac{_{49}P_4}{52!}$

Therefore: . $P(\text{some consecutive Aces}) \;=\;1 - \dfrac{_{49}P_4}{52!}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If the problem meant exactly one pair of consecutive Aces,
. . we can modify the above procedure.

We have the 48 Others with spaces before, after, and between them.

. . $\_\,O\,\_\,O\,\_\,O\,\_\,O\,\hdots\,\_\,O\,\_$ . (49 spaces)

We have 4 aces.
Select two of them to be together; there are $_4C_2 \,=\,6$ choices.

Duct-tape the pair together.
Then we have 3 "cards" to distribute: . $\boxed{AA}\,,\:A,\:A$
And they can be placed in: . $_{49}P_3$ ways.

Hence, there are: . $6(_{49}P_3)$ ways to have one pair of consecutive Aces.

The probability is: . $\dfrac{6(_{49}C_3)}{52!}$

I'd like to try and also follow this through.

The probability of having two consecutive aces in the deck comprises situations of AA, A, A (single instance) as well as AA, AA (double instance).

The single instance occurs as follows:
49 spaces
Three clumps AA, A, A
49 P 3 ways of placing the 3 clumps into 49 spaces = 110,544
Select 2 aces from 4 => 6 possibilities (AB, AC, AD, BC, BD, CD)
48! possibilities of ordering the remaining (non-ace) 48 cards = 48!

The product of these three is 8.2337 x10^66 - which out of 52! total possible combinations equates to a 10.21% probability - that seems right since 1/13th of the time an ace will be followed by another ace, or can be preceded by an ace etc.

The double instance occurs as follows:
49 spaces
Two clumps AA, AA
49 P 2 ways of placing the clumps into 49 spaces = 2,352
Select 2 lots of 2 aces to be chosen from 4 => 24 possibilities (4x3x2x1)
48! possibilities of ordering the remaining (non-ace) 48 cards = 48!

The product of these three is 7.007 x10^65 - which out of 52! possible combinations for the deck equates to a 0.8687% probability. This is the same probability as having a 3-ace sequence - which seems logical since both cases deal with the placing of two clumps of cards, AAA & A in one situation and AA & AA in other situation.

The sum of the two proabilities is 10.21%+0.87% = 11.08%

Thus in summary I have the following probabilities:
P(four consecutive aces anywhere in the deck) = 0.0181%
P(three consecutive aces anywhere in the deck) = 0.8869% (excludes situations where a 3-ace sequence forms part of a 4-ace sequence)
P(two consecutive aces anywhere in the deck) = 11.0769% (excludes situations where a 2-ace sequence forms part of a a 3-ace or 4-ace sequence)

It would be great if you could let me know whether I have followed this through correctly or not. Many thanks in advance!

**Wilmer** · January 10th 2011, 06:54 PM

I ran a simulation program for 4 concecutives (a few million tries) and got same as Soroban: ~.000181

**george12345** · January 11th 2011, 01:58 AM

Originally Posted by Wilmer

I ran a simulation program for 4 concecutives (a few million tries) and got same as Soroban: ~.000181

Hi Wilmer, that's reassuring - thank you!
Does your simulation program also work for multiple decks?

**Wilmer** · January 11th 2011, 03:22 AM

Originally Posted by george12345

Does your simulation program also work for multiple decks?

It would, but it's set up for 52 cards and I'm not interested in making it larger.
As I said, results matched Soroban's, so I suggest you use his formula...

**awkward** · January 14th 2011, 02:00 PM

Originally Posted by Soroban

Hello, george12345!

There are: $52!$ possible orderings of the cards.

We have: .4 aces and 48 Others.

Place the 48 Others in row, leaving a space before, after, and between them.

. . $\_\,O\,\_\,O\,\_\,O\,\_\,O\,\hdots \,\_\,O\,\_$

There are 49 spaces.

Select 4 of the spaces and insert the four Aces (in some order).

. . There are: . $_{49}P_4$ ways.

Hence: . $P(\text{no consecutive Aces}) \;=\;\dfrac{_{49}P_4}{52!}$

Therefore: . $P(\text{some consecutive Aces}) \;=\;1 - \dfrac{_{49}P_4}{52!}$

[snip]

Soroban,

If you're going to count all the permutations of the cards in which there are no consecutive aces, it's not enough to account for the positions of the aces; you must also count the arrangements of the other cards. Once you have placed the aces, which can be done in P(49, 4) ways, the remaining cards can be arranged in 48! ways.

So the probability of no consecutive aces is

$\frac{P(49,4) \times 48!}{52!}$

and the probability of at least two consecutive aces is

$1- \frac{P(49,4) \times 48!}{52!} \approx 0.2174$

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