Probability calculation : 5 cards ...?

Question

What is the chance of having no pair of cards of the same rank next to each other in a hand of five cards (from a deck of 52 cards) ? According to me it's impossible to calculate unless you check all the possibilities.

Answer 1

First, break it into cases:

1) No pairs.
2) One pair.
3) Three of a kind.
4) Four of a kind.
5) Two pairs.
6) Full house (pair + 3 of a kind).

First we find the probability of each of those cases. Then find the probability, in each case, that there is no adjacent cards of the same rank.

STEP 1: Probability of each case

First of all, we need to know how many hands total there are. The answer is, of course:

52C5 = 2598960

Case 1) None of a kind:
(4^5) × (13C5)

There are 13C5 ways to pick the rank, and for each card, 4 different suits it could be.

Probability: P1 = 4^5 × 13C5 / 52C5
P1 = 2112 / 4165 = 0.507083

Note that P1 is not the probability of having a "worthless," "no pair," "high card only," or "nothing" poker hand. P1 includes the chances of getting a straight, for example, because a straight has no pairs.

Case 2) One pair:

(4C2 4^3) ×(13 × 12C3)

There are 12C4 ways to choose the ranks (excluding the pair's rank), and 13 ways to choose which rank will have the pair. Among them, there are 4 choices for each suit EXCEPT for the pair - it's 4C2 ways to choose a suit for them.

P2 = (4C2 × 4^3) ×(13 × 12C3) / 52C5
P2 = 352 / 833 = 0.422569

(For the rest, I'll spare the explanation.)

Case 3) Three of a kind:

P3 = (4C3 × 4^2) × (13 × 12C3) / 52C5
P3 = 88 / 4165 = 0.0211285

Case 4) Four of a kind:

P4 = (4C4 × 4) × (13 × 12C1) / 52C5
P4 = 1 / 4165 = 0.000240096

Case 5) Two pairs:

P5 = (4C2 × 4C2 × 4) × (13C2 × 11C3) / 52C5
P5 = 18 / 4165 = 0.00432173

Case 6) Full house (pair + 3 of a kind):

P6 = (4C2 × 4C3) × (13×12) / 52C5
P6 = 6 / 4165 = 0.00144058

Now we have to calculate the probabilities, in each of those cases, of having no adjacent cards of the same rank:

Case 1) No pairs:

A1 = 1 (clearly - there are no pairs)

Case 2) One pair:

A2 = 1 - 4! / 5!
A2 = 4 / 5 = 0.8

4! / 5! is the number of arrangements with pair / total number of arrangements.

Case 3) Three of a kind:

There's only one way to split them up here, so we get:

A3 = 1/5!
A3 = 1 / 120 = 0.00833333

The one way to arrange with no pair is (if P is the one in the triplet) PxPyP

Case 4) Four of a kind:

A4 = 0

There's no way to avoid it here.

Case 5) Two pairs:

There are 5! total arrangements.

There are 4! arrangements with pair 1 together.

There are 4! arrangements with pair 2 together.

There are 3! with both.

Thus the probability of having no pair together is:

A5 = (5! - 4! - 4! + 3!) / 5!
A5 = 13 / 20 = 0.65

Case 6) Full house (pair + 3 of a kind):

A6 = 1/5!
A6 = 1 / 120 = 0.00833333

There's one way to do it here, same as case 3.

FINALLY you have to compute the probability by multiplying:

P1 × A1 + P2 × A2 + P3 × A3 + P4 × A4 + P5 × A5

If you work it out, you get:

P = 211949 / 249900 = 0.848546

I will not duplicate the algebra necessary, to simplify it. Just use your calculator to verify (you can check some of my computations on the sites listed in my see-alsos).

It's about 84.9% likely that you will not have a pair in your hand that is adjacent. Notice that it cannot be 50% like another answerer asserted. The odds of getting no pairs at all is more than 50%, so it's 50% + all the possibilities where you have a pair, but it isn't adjacent in your hand.

Answer 2

to find the probability of having a hand with no pairs is challenging but it can be done.

first example, lets assume that you can't have any pairs, three of a kind or four of a kinds (only because for a three of a kind and four of a kind you have a pair of cards of equal rank included in the hand)

how many hands are like this? Well there are 52 different cards that can be the first card drawn. for example let's say it is a king. to make sure that there is no possible pairs, the other three kings have to be removed from the deck leaving 48 possible cards to be drawn for the second card, say a jack. same logic here, remove the other three jacks and drawn again. Total number of five card hands (drawn in a certain order) is:
52 * 48 * 44 * 40 * 36 = 158146560

now you need to know how many possible hands there are in general. Normally that would be 52 choose 5. 52! / (5! (52 - 5)!) = 2598960. but here i've allowed for order to matter so I need to look at the number of permutations of 52 choose 5 which is: 52! / ( (52 - 5)!) = 311875200

so the probability of having and hand with no pairs, and for that matter no card of repeating rank you have 158146560 / 311875200 = 2112 / 4165 = 0.507083

Answer 3

I can give you a close approximation -- of course in poker you don't care if they are "next to each other" but ill give it a shot


catching a pair
(4p2)(48p3)/(52p2) = 0.003992982

however there are thirteen different cards for which this can happen so I would mulitply this number by 13

0.051908764

and you want the probability that there are no pairs so 1 minus this which is 0.9481

after reading the poster below he is correct

1- (4c2)(48c2)/(52C2) *13 is about .50

if you mean that the pairs aren't exactly next to one another. but if you mean that they are exactly next to one another in your hand there is only about a 5% chance of that happening

Answer 4

Here are the probabilities for being dealt a five card pat hand.

. . . . . . . . . . . . . . . . .Hands
Hand . . . . . . . . . . .Possible . . . Probability

Royal Flush . . . . . . . . . . . . 4 . . . . . 0.0002%
Straight Flush . . . . . . . . . 36 . . . . . 0.0014%
Four of a Kind . . . . . . . .624 . . . . . 0.0240%
Full House . . . . . . . . . 3,744 . . . . . .0.1441%
Flush . . . . . . . . . . . . . .5,108 . . . . . .0.1965%
Straight . . . . . . . . . . .10,200 . . . . . .0.3925%
Three of a Kind . . . . 54,912 . . . . . .2.1128%
Two Pairs . . . . . . . .123,552 . . . . . .4.7539%
One Pair . . . . . . .1,098,240 . . . . .42.2569%
No Pairs. . . . . . . 1,302,540 . . . . .50.1177%

Total . . . . . . . . . . .2,598,960 . . . . 100.0000%

Answer 5

you're impressive in case you ought to get carry of P( RRRBB playing cards.) i.e., if the order is accompanied. yet interior the question purely P(3 purple and 2 black playing cards) is to be acquired. The order isn't substantial. consequently the possibility acquired is to be more desirable with the form of techniques of choosing 3 purple playing cards out of 5. The form of techniques of choosing 3 purple out of 5 = 5C3 = 10 P(3red and 2 black playing cards) = 10*26/fifty two*25/fifty one*24/50*26/40 9*25/40 8 = 0.3251 <<<<<<<<<<<<<<<<<<<<<<<<<<<<< The question could be responded by means of ability of applying the combinations P(3 purple and 2 black playing cards) = 26C3*26C2 / 52C5 = 0.3251