[bonanova]

Commemorating the post that matches my member number,

let's try a bit of useless [this side of Vegas at least] calculations.

In a standard deck of cards there are 52 cards: 1-10, J Q K of four suits S H D C.

In draw poker, 5 cards are dealt to each player. Hands are valued in increasing degree as follows.

Edit: Examples of each type of hand shown in red.

0 - High card: Highest ranking card. This hand is none of the following types. - 3D 6H 8C JH KD

1 - Pair: 2 cards of same rank; 3 cards of different ranks - 4C 4D x x x

2 - Two Pair: 2 cards of one rank, 2 cards of another rank, 1 card of different rank. - 6H 6S 9D 9H x

3 - Three of a kind: 3 cards of same rank; 2 cards of different ranks. - KC KD KS x x

4 - Straight: 5 cards in rank sequence, but not all of same suit. - 3D 4D 5C 6H 7S

5 - Flush: 5 cards all of the same suit, but not all in sequence. - 3H 6H 7H 9H JH

6 - Full House: 3 cards of one rank; 2 cards of another rank. - AH AC AS 8S 8H

7 - Four of a kind: 4 cards of same rank. 7S 7H 7D 7C x

8 - Straight flush: 5 cards in sequence and all of same suit. - 4H 5H 6H 7H 8H.

Prove that the valuation of 0-8 is reasonable by computing the probabilities of being dealt each type of hand.

Express the probabilities as fractions where the denominator is the number of possible unique 5-card hands dealt from a single deck.

Easy check - they add up to unity.

The prize for being first with all correct answers is 10 "attaboy"s.

The current value of an "attaboy" is that you may combine 100 of them with $0.85 for a cup of coffee at Dunkin Donuts.

[Guest]

Commemorating the post that matches my member number,

let's try a bit of useless [this side of Vegas at least] calculations.

In a standard deck of cards there are 52 cards: 1-10, J Q K of four suits S H D C.

In draw poker, 5 cards are dealt to each player. Hands are valued in increasing degree as follows.

0 - High card: Highest ranking card. This hand is none of the following types. - e.g. 3D 6H 8C JH KD

1 - Pair: 2 cards of same rank; 3 cards of different ranks - 4C 4D x x x

2 - Two Pair: 2 cards of one rank, 2 cards of another rank, 1 card of different rank. - 6H 6S 9D 9H x

3 - Three of a kind: 3 cards of same rank; 2 cards of different ranks. - KC KD KS x x

4 - Straight: 5 cards in rank sequence, but not all of same suit. - 3D 4D 5C 6H 7S

5 - Flush: 5 cards all of the same suit, but not all in sequence. - 3H 6H 7H 9H JH

6 - Full House: 3 cards of one rank; 2 cards of another rank. - AH AC AS 8S 8H

7 - Four of a kind: 4 cards of same rank. 7S 7H 7D 7C x

8 - Straight flush: 5 cards in sequence and all of same suit. - 4H 5H 6H 7H 8H.

Prove that the valuation of 0-8 is reasonable by computing the probabilities of being dealt each type of hand.

Express the probabilities as fractions where the denominator is the number of possible unique 5-card hands dealt from a single deck.

Easy check - they add up to unity.

The prize for being first with all correct answers is 10 "attaboy"s.

The current value of an "attaboy" is that you may combine 100 of them with $0.85 for a cup of coffee at Dunkin Donuts.

What are you asking us? the probability of getting your proposed hands?

How I'm understanding it, the answer to what you're asking for 0 would be:

5/52 * 4/51 * 3/50 * 2/49 * 1/48

Those are the chances of getting 3D 6H 8C JH KD

Edited March 28, 2008 by bonanova

[Guest]

Commemorating the post that matches my member number,

let's try a bit of useless [this side of Vegas at least] calculations.

In a standard deck of cards there are 52 cards: 1-10, J Q K of four suits S H D C.

In draw poker, 5 cards are dealt to each player. Hands are valued in increasing degree as follows.

Edit: Examples of each type of hand shown in red.

0 - High card: Highest ranking card. This hand is none of the following types. - 3D 6H 8C JH KD

1 - Pair: 2 cards of same rank; 3 cards of different ranks - 4C 4D x x x

2 - Two Pair: 2 cards of one rank, 2 cards of another rank, 1 card of different rank. - 6H 6S 9D 9H x

3 - Three of a kind: 3 cards of same rank; 2 cards of different ranks. - KC KD KS x x

4 - Straight: 5 cards in rank sequence, but not all of same suit. - 3D 4D 5C 6H 7S

5 - Flush: 5 cards all of the same suit, but not all in sequence. - 3H 6H 7H 9H JH

6 - Full House: 3 cards of one rank; 2 cards of another rank. - AH AC AS 8S 8H

7 - Four of a kind: 4 cards of same rank. 7S 7H 7D 7C x

8 - Straight flush: 5 cards in sequence and all of same suit. - 4H 5H 6H 7H 8H.

Prove that the valuation of 0-8 is reasonable by computing the probabilities of being dealt each type of hand.

Express the probabilities as fractions where the denominator is the number of possible unique 5-card hands dealt from a single deck.

Easy check - they add up to unity.

The prize for being first with all correct answers is 10 "attaboy"s.

The current value of an "attaboy" is that you may combine 100 of them with $0.85 for a cup of coffee at Dunkin Donuts.

I see a trick to this that might catch a few people, but computing probabilities like this might take me a while.

[Guest]

Long term reader, first time poster...

The number of ways of picking k unordered outcomes from n possibilities is n! / (k!(n-k)!)

So the number of possible hands of 5 cards from a pack of 52 is 52! / (5! * (52-5)!) = 2,598,960, so that's my denominator

Some hands are easier than others, might take me a while to calculate them all, I think the key challenge here is

Not to double count (e.g. A hand containing a full-house should not also count as a hand containing either a pair or two pairs)

[Guest]

Long term reader, first time poster...

The number of ways of picking k unordered outcomes from n possibilities is n! / (k!(n-k)!)

So the number of possible hands of 5 cards from a pack of 52 is 52! / (5! * (52-5)!) = 2,598,960, so that's my denominator

Some hands are easier than others, might take me a while to calculate them all, I think the key challenge here is

Not to double count (e.g. A hand containing a full-house should not also count as a hand containing either a pair or two pairs)

but has that been wroked out not including the order, e.g. you could pick 1h, 5s, 3c, jd, 9d but you could also pick these 5 cards in the order, 5s, jd, 1h, 9d, 3c?

[bonanova]

Dackombe has the denominator. Welcome to the Den!

The numerators are the challenge, for the reasons he mentioned.

Hint: you can discuss the probabilities one at a time, as if it were

eight puzzles. You might be surprised which one is the most

complicated to describe and calculate. Have fun with this.

To me, the interesting part is that it's fairly easy to look at five

cards and not confuse 4 of a kind with 2 pairs for example.

But if I'm a little bit sloppy with my math, I might count it that way.

Or even worse, count 4 of a kind simply as a pair.

What's needed is to describe what all 5 cards must or must not be.

[Guest]

but has that been wroked out not including the order, e.g. you could pick 1h, 5s, 3c, jd, 9d but you could also pick these 5 cards in the order, 5s, jd, 1h, 9d, 3c?

I would think that the probability of getting the hand would be the same, regardless of order

[Guest]

I think the hands with the fewest possibilites are the easiest to work out so I'll start with the straight flush and simply count them...

A|K|Q|J|10, K|Q|J|10|9, ... , 6|5|4|3|2, 5|4|3|2|A.

There are 10 possible straight flushes in each suit (from Ace high down to 5 high) so that's 40 possible straight flushes. Probability 40/2598960

Also I'll just count the Four of a kinds...

There are 13 ranks so that's 13 different four of a kinds, each one needs a fifth card drawn from the remaining 52-4=48. So that's 13*48=624 possible four-of kind hands. Probability 624/2598960

[Guest]

Now it gets a little more complicated...

Three of one kind, two of another.

There are 13 kinds that we could have 3 of, say 3 aces. Each of these must have one of the four suits missing so thats 13*4=52 possible three-of-a-kinds. Now we need we two more cards.

There are 12 possible kinds left (since we already have three of one kind) and 6 posible combinations of each of these (either Club/Diamond, Club/Heart, Club/Spade, Diamond/Heart, Diamond/Spade or Heart/Spade). So thats 12*6=72 possible pairs from the remaining cards.

52 combinations of three * 72 combinations of two gives a total of 3744 possible Full House hands. Probability 3744/2598960

[Guest]

sorry i've been working on this all day:P cba to post working although if required ill try and remember them

ok i agree with the ones that dackombe has already posted first of all

3 of a kind 54912/2598960

two pairs 123552/2598960

one pair 1098240/2598960

highest card 1317888/2598960

[Guest]

You have missed an assumption, one that is always made, and is usually wrong. For these probabilities to work that way that you intend, you must be the ONLY player. Otherwise, you must have the number of other players as well, as they will also get cards. This will change the combinatorics.

[Guest]

Dackombe has the denominator. Welcome to the Den!

The numerators are the challenge, for the reasons he mentioned.

Hint: you can discuss the probabilities one at a time, as if it were

eight puzzles. You might be surprised which one is the most

complicated to describe and calculate. Have fun with this.

To me, the interesting part is that it's fairly easy to look at five

cards and not confuse 4 of a kind with 2 pairs for example.

But if I'm a little bit sloppy with my math, I might count it that way.

Or even worse, count 4 of a kind simply as a pair.

What's needed is to describe what all 5 cards must or must not be.

Is the most complicated one the high card one? Because really, that one is the probability that you don't get ANY of the other possible hands. So it's 1-(probability of getting any other hand). And of course, you would need to find out all of the other probabilities first to get this one.

Edited March 28, 2008 by Seventh Sage

[bonanova]

Great job, all ... !

Hands calculated so far:

By Dackombe:

Total hands - 2 598 960

Straight Flush - 40

Four of a Kind - 624

Full House - 3 744

By andreay:

Three of a Kind - 54 912

Two Pair - 123 552

Pair - 1 098 240

High Card - 1 317 888 <- very close, but slightly off.

Seventh Sage is correct about the most difficult one to calculate.

High Card.

Three numbers left to find.

Flush - fairly easy

Straight - easier

High Card - hardest

What we know is that these three must total

2598960 - [40+624+3744+54912+123552+1098240] = 1 417 848.