question involving a deck of cards

I have searched this area of the forum, as this is where I believe it would be, and saw nothing. If this question has been posted, please direct me to where it was as I did not find it.

The question is this. Given a deck of cards, what is the probability of any two chosen ranks are next to each other within the entirety of the deck.

To give an example, if I were to choose 3 and 8, as I fan through a full deck of 52 cards, what is the probability of both cards appearing next to each other at least once within the deck?

I have had conflicting information about how to properly estimate these odds. I believe that I may just be doing this wrong with my calculation. Please enlighten me to the correct way to determine this, if I am wrong.

For the first 8's position, there are 40 "safe spots" out of 48 total spots where it is not touching a 3. 40/48 odds of not touching. Second 8 safe odds: 39/47. Third: 38/46. Fourth: 37/45. 40*39*38*37/48*47*46*45 = 0.47

Quote: freeko

I have searched this area of the forum, as this is where I believe it would be, and saw nothing. If this question has been posted, please direct me to where it was as I did not find it.

The question is this. Given a deck of cards, what is the probability of any two chosen ranks are next to each other within the entirety of the deck.

To give an example, if I were to choose 3 and 8, as I fan through a full deck of 52 cards, what is the probability of both cards appearing next to each other at least once within the deck?

I have had conflicting information about how to properly estimate these odds. I believe that I may just be doing this wrong with my calculation. Please enlighten me to the correct way to determine this, if I am wrong.

For the first 8's position, there are 40 "safe spots" out of 48 total spots where it is not touching a 3. 40/48 odds of not touching. Second 8 safe odds: 39/47. Third: 38/46. Fourth: 37/45. 40*39*38*37/48*47*46*45 = 0.47

If I understand your question correctly, the probability of finding any two ranked cards adjacent is around 70%. Remember that the order you find them in doesn't matter, and it also doesn't matter if there are more than one "match" in the deck.

Edit: Still digging, but some have calculated ~51%. Apparently the math is quite daunting.

To accurately answer this question, you would either need to run through all 52! combinations or run a simulation. However, I submit the following estimate:

1. The probability of any given pair of consecutive cards being one of two chosen ranks is (8/52)*(3/51)
2. There are 51 sets of consecutive cards.
3. Thus, there 51*(8/52)*(3/51) = 0.4615 expected sets of consecutive cards.
4. Using the exponential distribution, the probability of 0 consecutive sets, when 0.4615 are expected is e^-0.4615 = 63.03%. So, the probability of one or more consecutive set is 36.97%.

Note to Aye: Remember that only two chosen ranks count, like kings and queens.

https://wizardofvegas.com/forum/questions-and-answers/math/9312-the-two-card-trick/ :+)

Quote: Wizard

To accurately answer this question, you would either need to run through all 52! combinations or run a simulation. However, I submit the following estimate:

1. The probability of any given pair of consecutive cards being one of two chosen ranks is (8/52)*(3/51)
2. There are 51 sets of consecutive cards.
3. Thus, there 51*(8/52)*(3/51) = 0.4615 expected sets of consecutive cards.
4. Using the exponential distribution, the probability of 0 consecutive sets, when 0.4615 are expected is e^-0.4615 = 63.03%. So, the probability of one or more consecutive set is 36.97%.

Note to Aye: Remember that only two chosen ranks count, like kings and queens.

I could be wrong, but wouldn't the probability of a given pair of pre-determined (or chosen) cards with different ranks be (8/52)*(4/51)?

Following the calculations on out, the probability of one or more consecutive set is then 45.96%?

Quote: Wizard

Quote: MidwestAP

I could be wrong, but wouldn't the probability of a given pair of pre-determined (or chosen) cards with different ranks be (8/52)*(4/51)?

Following the calculations on out, the probability of one or more consecutive set is then 45.96%?

The OP said TWO chosen ranks.

Maybe I'm misunderstanding the question, but two chosen ranks reads to me that the first card can be either of the ranks, thus an 8 of 52 chance. Then by default, the second card would have to be the other chosen rank, with probability of 4 of 51.

Quote: Wizard

The OP said TWO chosen ranks.

Yes, but given the example of 8 and 3... you have 8/52 for the first of the two consecutive cards, then given the only way the series counts as a 'match' is if the OTHER chosen card is draw... which there should be 0 of them drawn thus far (assuming first hit in the deck), so 4/51?

Edit: MidwestAP worded it a lot better than me above =p.

Quote: MidwestAP

Quote: Wizard

Quote: MidwestAP

I could be wrong, but wouldn't the probability of a given pair of pre-determined (or chosen) cards with different ranks be (8/52)*(4/51)?

Following the calculations on out, the probability of one or more consecutive set is then 45.96%?

The OP said TWO chosen ranks.

Maybe I'm misunderstanding the question, but two chosen ranks reads to me that the first card can be either of the ranks, thus an 8 of 52 chance. Then by default, the second card would have to be the other chosen rank, with probability of 4 of 51.

It seems to me the difference is the way it is being interpreted.

If the two chosen ranks were 8 and 3 then it looks like the Wizard is basing his calculations on whether there are either two consecutive 8's or two consecutive 3's.

It seems like MidwestAP assumption was the same as mine that if the chosen cards were 8 and 3, whether there is an 8 following a 3 or a 3 following an 8.

I deleted my former post because I misunderstood the question.

So, to rehprase it again, somebody picks any two ranks, say kings and queens. What is the probability any such two ranks appear consecutively in the deck, including a consecutive king and queen.

My estimate would be 1-e^(51*(8/52)*(7/51)) = 1-e^-1.0769 = 65.94%.

From the 2012 thread...

Quote: ThatDonGuy

Er...it's 45.12953% if there are only 51 cards in the deck.

With a 52-card deck, it's 48.6279%.

I stand corrected.

Makes me think my estimate isn't appropriate, or I misunderstood the problem again.

Maybe the OP can clarify the question.

Given two pre-determined ranks (like 8 and 3) , there are three solutions being considered.

1) Either of the ranks followed by the same rank (3,3 or 8,8)
2) Either of the ranks followed by the other rank (3,8 or 8,3)
3) Either of the ranks followed by either of the ranks (3,3 or 8,8, or 3,8 or 8,3)

My interpretation is he meant door #3.