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What are the odds of this happening?

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What are the odds of this happening? - Sat Apr 28, 2012, 02:43 AM
(#1)
Aus-Redda's Avatar
Since: Jun 2011
Posts: 30
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I was dealt pocket 10’ in middle pos. The table folded to me, so I raised, I had about 9G in chips and early stage of tourney. Blinds are 20/40. Armanisergio 2nd spot behind the button pushed all in. I took a chance thinking coin flip against 2 live over cards. He had (edit he had AA not A4) A4 and flopped an A. I lose a proportion of my stack. I was prepared to gamble here, if I lose the hand at least I’m not out of the tourney.

Now “Armanisergio” has doubled up.

Bang! I am dealt pocket 10’s again, back to back. I rasied again to about 320, “Macmarttigan” shoves all and “Armanisergio” pushes over the top all in. Again I have a decision to make with my 10’s. I figured this time that Armanisergio had a decent hand like J’s or Q’s. I certainly did not put him on AA again.

If you zoom into the screen shot, you will see that I asked, “Hmmm are my 10’s good again”? I decided not to call him this time.

I have some questions that I would like to ask.

Q1) What are the odds of a player being dealt a pocket pair on a 9 handed table in a tournament?

Q2) What are the odds of 3 players being dealt a pocket pair on a 9 handed table in a tournament?

Q3) What are the odds of 3 players being dealt a pocket pair pre-flop and each player making a set (3 of a kind) on the flop?

Q4) What are the odds of the middle pair (in this case 10’s), hitting quads?

I would like to know statistically what the odds of this type of hand occurring. I would deeply appreciate some feedback.

Thank you.

Last edited by Aus-Redda; Sat Apr 28, 2012 at 02:56 AM..
 
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Sat Apr 28, 2012, 06:23 AM
(#2)
TrumpinJoe's Avatar
Since: Jun 2010
Posts: 4,557
Smart ass answer #1: It happened so the probability is 100%.

Smart ass answer #2: It happened so get over it.

Actual answers:
Q1 – 1.4:1
Q2 – 14:1
Q3 – 350:1
Q4 – Flop 400:1; Turn 100:1; River 18:1

Last edited by TrumpinJoe; Sun Apr 29, 2012 at 12:09 PM.. Reason: correction
 
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Sat Apr 28, 2012, 07:28 AM
(#3)
Ovalman's Avatar
Since: Feb 2011
Posts: 1,778
TrumpinJoe, can you tell me how you arrived at those odds?
 
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his odds are way out lol! - Wed May 02, 2012, 11:00 AM
(#4)
Aus-Redda's Avatar
Since: Jun 2011
Posts: 30
Quote:
Originally Posted by Ovalman View Post
TrumpinJoe, can you tell me how you arrived at those odds?
Thank you all for replying to this post. I did some revision on my mathematics (I love mathematics….I just hate thinking, ha-ha). I think that I may have figured it out. I will try to answer my own questions. If I am incorrect, please feel free to comment.

Q1) What are the odds of a player being dealt a pocket pair on a 9 handed table in a tournament?
Q2) What are the odds of 3 players being dealt a pocket pair on a 9 handed table in a tournament?
Q3) What are the odds of 3 players being dealt a pocket pair pre-flop and each player making a set (3 of a kind) on the flop?
Q4) What are the odds of the middle pair (in this case 10’s), hitting quads?

Read more: What are the odds of this happening? - PokerSchoolOnline Forum http://www.pokerschoolonline.com/for...#ixzz1ti7ESWX0

Assuming the following is true and correct.
Courtesy (http://people.richland.edu/james/lec.../ch04-not.html)
(The only difference in the definition of a permutation and a combination is whether order is important.)
A combination of n objects, arranged in groups of size r, without repetition, and order being important is:
nCr = C(n,r) = n! / ( (n-r)! * r! )


Answers to questions:
(I apologize for the confusion relating to number of players seated and table size, mathematically, this is irrelevant, I was referring to specific outcomes).

Q1) What are the odds of a player being dealt a pocket pair?

Answer:

52! / ( (52-2)! * 2!) = 1326

There are 1326 ways of being dealt 2 cards out of a field of 52 cards. There are 6 ways (considering suits where c=clubs, h=hearts, d=diamonds and s=spades) that a player can be dealt a pair out of those 1326 combinations.
Example: AA can be dealt as: Ac Ah or Ac Ad or Ac As or Ah Ad or Ah As or Ad As.

Therefore 1326/6 = 221. The odds of being dealt any pair pre-flop are 1 in 221.

Q2) What are the odds of 3 players being dealt a pocket pair?

Answer:
To answer this question, I have determined that each pair is not related to any other pair.

Therefore, I arrived at 3*221 = 663. The odds of 3 players each being dealt a different pocket pair pre-flop are 1 in 663.

Q3) What are the odds of 3 players being dealt a pocket pair pre-flop and each player making a set (3 of a kind) on the flop?

Answer:
It gets a bit tricky here and I hope my mathematics will hold up.

Mathematically 52-6 =46
Now we have to calculate (using the nCr formula) combinations of 1 of 2 cards being dealt from the remaining field of 46 cards and then multiply that by 3 in order for each player to hit 1 out of 2 cards to make a set (3 of a kind).

46! / ( (46-2)! * 2!) = 1035

Therefore: I arrived at 663*1035 = 686,205.

The odds of three pairs pre-flop, each seeing the flop and flopping a set are 1 in 686,205


Q4) What are the odds of the middle pair (in this case 10’s), hitting quads?

There are 43 cards left. There is only 1 (10) remaining in the deck.

The odds that trip 10’s makes quads = 686,205 * 43 = (29506815)

1 in 29,506,815

Please feel free to comment on this or correct my mathematics.
 
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Wed May 02, 2012, 01:43 PM
(#5)
JDean's Avatar
Since: Aug 2010
Posts: 3,145
BronzeStar
Quote:
Originally Posted by Aus-Redda View Post
Answers to questions:
(I apologize for the confusion relating to number of players seated and table size, mathematically, this is irrelevant, I was referring to specific outcomes).

Q1) What are the odds of a player being dealt a pocket pair?

Answer:

52! / ( (52-2)! * 2!) = 1326

There are 1326 ways of being dealt 2 cards out of a field of 52 cards. There are 6 ways (considering suits where c=clubs, h=hearts, d=diamonds and s=spades) that a player can be dealt a pair out of those 1326 combinations.
Example: AA can be dealt as: Ac Ah or Ac Ad or Ac As or Ah Ad or Ah As or Ad As.

Therefore 1326/6 = 221. The odds of being dealt any pair pre-flop are 1 in 221.
Those are the odds of being dealt a SPECIFIC pocket pair pre flop. You have 13 potential pp you can be dealt to fulfill your criteria (what are the chances of a player being dealt [ANY] pp pre flop).

You must then adjust all your numbers after this...


Double Bracelet Winner
 
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Thu Sep 20, 2012, 09:03 AM
(#6)
Aus-Redda's Avatar
Since: Jun 2011
Posts: 30
Quote:
Originally Posted by JDean View Post
Those are the odds of being dealt a SPECIFIC pocket pair pre flop. You have 13 potential pp you can be dealt to fulfill your criteria (what are the chances of a player being dealt [ANY] pp pre flop).

You must then adjust all your numbers after this...
I see your point and it is valid.
I appreciate the correction.

The odds I stated were based on being dealt a specific pair.

The odds on being dealt a random pair are 1 in 17.

I will try to recalculate this on the odds of 3 players being dealt a random pair pre-flop. Then based on the flop, calculate the odds of this hand occurring.

Last edited by Aus-Redda; Thu Sep 20, 2012 at 10:05 AM..
 
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Thu Sep 20, 2012, 09:41 AM
(#7)
Aus-Redda's Avatar
Since: Jun 2011
Posts: 30
Quote:
Originally Posted by JDean View Post
Those are the odds of being dealt a SPECIFIC pocket pair pre flop. You have 13 potential pp you can be dealt to fulfill your criteria (what are the chances of a player being dealt [ANY] pp pre flop).

You must then adjust all your numbers after this...
I appreciate your comment and I hope other readers of this post will appreciate the difference.

example 1) If I sat down at a table and said....come on dealer how about AA this time the odds are 1 in 221 that I will get that hand. The reason for this is, the player is asking for a specific outcome.

As you correctly pointed out. there are numereous ways a player can be dealt "ANY PAIR".

Therefor doing the maths....the odds of being dealt a random pair are now reduced to 1 in 17. this can be explained as follows.

The odds of being dealt a specific pair 221 divided by the number of ranks 13 that could make a pair. 221/13 = 17.

I hope this makes sense.

Last edited by Aus-Redda; Thu Sep 20, 2012 at 09:48 AM..
 
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Thu Sep 20, 2012, 09:55 AM
(#8)
Aus-Redda's Avatar
Since: Jun 2011
Posts: 30
This also rases another question about raising with middle order pairs!

Last edited by Aus-Redda; Thu Sep 20, 2012 at 09:57 AM..
 

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