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**sscrabble1****Guest**

At poker , you are dealt five cards. The probability of being dealt a flush , five cards of one suit , is 508 to 1.

But what are the odds of getting four of a flush - ie four of one suit plus one card of another suit ?

For every 39 hands that you are dealt four of a flush , should you be dealt 9 hands with all five cards of the same suit ?

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Hint: The probability of getting that last card to be of the correct suit was (13-4)/(52-4)

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Hmm. It depends on what the question means.

It could be:

"You are dealt four cards and all of them have the same suit. You are then dealt another card - what is the probability that this suit also matches?"

or

"You are dealt five cards and at least four of them have the same suit. What is the probability that five of them have the same suit?"

Why did the vector cross the road?

It wanted to be normal.

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**sscrabble****Member**- Registered: 2009-09-20
- Posts: 2

Sorry - I am not making myself clear - I am not saying you are being dealt ( for instance ) four clubs, what is the odds of the last card being a club.

I am saying you could be dealt a non-club, then four clubs - or a club, then a non-club, then three clubs - etc - So how many times will you get five clubs off the deal , rather than four plus a card of another suit ( that could be the first, 2nd,3rd,4th, or 5th card dealt.)

maybe this is a better way of putting it -

For every one occasion that you 'flop' five clubs on the deal , on how many occasions will you flop four clubs plus one card of a different suit (in any order ) on the deal ?

*Last edited by sscrabble (2009-09-21 04:58:23)*

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**Fruityloop****Member**- Registered: 2009-05-18
- Posts: 120

Hi sscrabble,

I believe you need to use the hypergeometric distribution formula. You are looking for 4 cards out of a group of 13 and 1 card from the other 39 so we have...

So the probability is 1 in 23.3.

If the odds of a flush are 508 to 1 against, the probability is 1 in 509.

So 509/23.3 gives 21.8 as the number you are looking for.

*Last edited by Fruityloop (2009-09-22 02:57:14)*

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**sscrabble****Member**- Registered: 2009-09-20
- Posts: 2

Thank you for that - I can see the logic - and also a simpler way to get to the same result - if the first four cards dealt are clubs , the odds are 9 out of 39 against the other card being a club - but because the odd card could be the first, second etc card dealt that becomes 9 out of 39 x 5 , which is 9 out of 195 , which is 21.8 to 1.

Which still seems weird , because when you have four of the suit there are 9 more of that suit in the pack, and 39 others , so you would think the ratio would be 9 to 39 , or 1 in 4.3

But then I thought if you applied that to just six cards, five of which were clubs , then applying that logic you would get a flush 1 out of two goes, whereas in fact you would only get it 1 in 6 goes as the odd card would appear in the first five cards dealt five times out of six !

Thanks for your help on here now, just glad to have got it resolved in my head !!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,106

Hi;

While having this argument on another forum. I found this, you might want to check this page out. It verifies Fruityloops's use of the hypergeometric distribution.

http://www.math.sfu.ca/~alspach/mag53/

*Last edited by bobbym (2009-09-23 19:53:31)*

**In mathematics, you don't understand things. You just get used to them.**

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**Dirtless****Guest**

Hello;

I just found this old thread, which answered my question vis a vis flush draws. What about straights?

This is a more complicated question, given that once one id dealt an initial card the seond one (in most cases) must be one of the eight that occur within 4 on either side. (Eg say you are dealt an 8. The next card must be one of 4,5,6,7 (from the "low" side), or 9,T,J,Q on the high.

But what about when one is dealt a King? The second card must be one of the 4 from the low, or an ace.

I am more interested in the result, but am likewise interested in the underlying logic/math.

Thanks

D

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,106

Hi Dirtless;

Please rephrase the question in an exact form. Are you asking the probability of being dealt a straight?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**Dirtless****Guest**

Pardon my imprecision;

I am asking essentially the same question as started this thread, except I am looking for the probability of being dealt a STRAIGHT draw, rather than a flush draw (four spades, say).

In other words, what is the probability that I will be dealt 6-7-8-9 or T-J-Q-K, and so on. A hand that gives you a one card, outside, draw to one of those 10,200 straights in the deck.

Let`s consider only hands that offers an outside straight, (6-7-8-9 rather than 5-6-8-9).

Thanks very much.

D

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,106

Hi;

Are you playing Draw Poker where you will discard one card trying to fill the straight?

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**Dirtless****Guest**

Bobby;

Yes, I am playing draw, and drawing one card to the outside straight.

D

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 97,106

Hi;

I have this in my notes:

Hmmmm. Computer count yields an answer close to that guy's formula but slightly lower.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**OzMark****Member**- Registered: 2013-06-10
- Posts: 9

Say the first card must be the middle card in the straight, 3 4 5 6 7 8 9 10 J Q, these are the 10 types of straights

it can be any of 4 suits so 10 x 4 = 40, use 40/52

The second card must be from within cards 2 places lower or 2 places higher (2+2=4) side of the middle card,

eg. either slot A 2 4 or 5 for a 3 middle card

it can be any of 4 suits so 4 x 4 = 16, use 16/51

The third card must be from one of the 4 slots outside the middle card, but 1 has been used; 4-1 =3

it can be any of 4 suits so 3 x 4 = 12, use 12/50

The fourth card must be from one of the 4 slots outside the middle card, but 2 have been used; 4-2 =2

it can be any of 4 suits so 2 x 4 = 8, use 8/49

The fourth card must be from one of the 4 slots outside the middle card, but 3 have been used; 4-3 =1

it can be any of 4 suits so 1 x 4 = 4, use 4/48

But the middle card doesnt have to be the first, it can be dealt in any of the 5 positions

either 1st 2nd 3rd 4th or 5th, so multiply by 5

The order of the other cards does matter

5 x 40 x 16 x 12 x 8 x 4 = 10240 = 0.3940 %

52 51 50 49 48 2598960

ANSWER

Chance of a straight in the first deal of 5 cards = 0.3940 % (1 in 253.8)

This is the figure given in wiki for a straight

en.wikipedia.org/wiki/Poker_probability

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Also notice that poker tables list Royal Flushs and Straight Flushs seperate to the other flushes

4 Royal Flush + 36 Straight Flush + 10200 Other Straights = 10240 All Straights

The guy at the first post made a similar mistake for For Flushes, its 1 in 504.8 for any flush in the first hand,

not 1 in 508 as he says, which is for other flushes but not the royal or straight flushes

*Last edited by OzMark (2013-06-14 14:11:42)*

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**OzMark****Member**- Registered: 2013-06-10
- Posts: 9

OK the redraw situation

If you get to throw out a card and get a fresh one, for the second chance straight probability

multiply the 0.3940 % by

(48/4 * 44/48 * 4/47) = 0.936

= 93.6%

almost same chance of getting a straight on the 6th card as on the 5th card

48/4 cancels the 5th card prob, because we no longer deal with a wanted card there

44/48 (= [48-4]/48) is the prob that the 5th card is unwanted for a straight

so you discard the incorrect 5th card and draw a 6th card, the chance the 6th card is what you want is 4/47

0.3940% (chance of straight on a 6 card deal, on the 5th card)

0.3940% x 0.936 = 0.3689 % (chance of straight on a 6 card deal, on the 6th card)

0.3940% x 1.936 = 0.7629 % (chance of straight on a 6 card deal, on the 5th or 6th card)

ANSWER:

Chance of a straight with a 1 card redraw = 0.7628 % (1 in 131.1)

(on a deal of either 5 or 6 cards)

Im happy to expand on the information in this post or the above post if anyone who wants to understand is still stuck

----------------------------------

Here is the probability of straight on a 6 card deal, on the 6th card in an expanded form

5 x 40 x 16 x 12 x 8 x 44 x 4 = 112640 x 4 = 0.3689 %

52 51 50 49 48 47 2598960 47

*Last edited by OzMark (2013-06-14 20:43:42)*

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**OzMark****Member**- Registered: 2013-06-10
- Posts: 9

Fifth card makes first card of straight, 4C1 = 4

4 x 40 x 12 x 8 x 4 x 4 = 2048 = 0.0788 %

52 51 50 49 48 2598960

Fifth card makes last card of straight, 4C1 = 4

4 x 40 x 12 x 8 x 4 x 4 = 2048 = 0.0788 %

52 51 50 49 48 2598960

Fifth card makes an inside card of straight, 4 x 3 = 12 (4 places lead card of straight, 3 places end card)

12 x 40 x 12 x 8 x 4 x 4 = 6144 = 0.2364 %

52 51 50 49 48 2598960

Sums to 0.3940%, so is correct

Remove the fifth card (last 4/48 term) to see the picture with 4 cards dealt

= Chance of getting the first 4 cards of a straight (fifth card inside or outside)

0.3940% *48/4 = 4.728 %

For outside straights only

(0.0788% + 0.0788%) *48/4 = 1.8912 %

ANSWER

Chance of getting the first 4 cards of a straight deal, where a fifth card can make a straight on the start or end = 1.8912 %

*Last edited by OzMark (2013-06-16 18:21:16)*

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