I have been looking at the way people calculate the odds of poker hands for a few hours and i think there is a HUGE flaw. In poker a player is never dealt two CONSECUTIVE cards. This in my view changes the odds, while also making the number of people at the table as well as the players position a material factor. For example

2 handed: first player dealt; odds of getting pocket aces = (4/52)x(3/50) = .46% or 1/217

2 handed: second player dealt; odds of getting pocket aces = (4/51)x(3/49) = ~.48% or 1/209

correct me if I'm wrong but this infers that the greater the number of players and the further in the action you are (closer to the button) the higher the probability of being dealt aces. Assuming that no other players have been.

or to be exact

The odds of being dealt pocket aces at a ten handed table on the button = (4/43)x(3/39) = .72% or 1/139

Where as the odds of being dealt pocket aces at a ten handed table on the small blind =(4/52)x(3/42) = .55% or 1/182

for this reason i believe the answer to "what is the probability of being dealt pocket aces" can not possibly be greater then 1/217 and should from here forward be described as the least likely ratio. that is in a heads up match being dealt the first card or 1 in 217 times.

  • I'm not into that math, but the probability to dealt rockets is around 0.004%, to dealt either aces and kings is about 0.009% and to dealt any pair in general is about 6% – user1165 Jun 10 '15 at 22:42
  • 1
    0.004% is 1 in 25000. That's not correct. There are only 52 * 51 = 2652 unique starting hands, so the odds of getting any two specific cards in a specific order is exactly 1 in 2652. There are 12 unique ways of getting dealt AA (Ac first As second, AcAd, AcAd, AsAc, AsAd, AsAh, AdAc, AdAh, AdAs, AhAc, AhAd, AhAs). 12 in 2652 is equivalent to 1 in 221, so there is a 1/221 = 0.4525% chance to be dealt pocket aces in any form in holdem. – Chris Farmer Jun 10 '15 at 22:57
  • ah i see, never got into the ratio, thanx for the explanation – user1165 Jun 11 '15 at 2:18
  • Your calculations only make sense if the game is rigged so that none of your opponents can ever get an ace. – Paulpro Jun 11 '15 at 3:59
  • Also, regarding "" "what is the probability of being dealt pocket aces" can not possibly be greater then 1/217"", 1/217 is the smallest number in your answer not the greatest. All the other numbers like 1/139 and 1/182 are greater and represent a higher probability. – Paulpro Jun 11 '15 at 3:59

The odds of getting aces do not at all depend on the number of cards remaining in the deck. They depend solely on the number of cards in the deck (52), how many aces are in the deck (4), and how many cards you receive from that deck (2 in holdem).

You have a 4 in 52 (or 1 in 13) chance to get an initial ace. If you get that first ace, you then have a 3 in 51 (or 1 in 17) chance to get a second ace. That's 1 in (13 * 17) or 1/221 chance of getting pocket aces.

Your math implies that you know that the other players at your table are getting cards but are not being dealt any of the aces. For example, in your first example which says "(4/52)x(3/50) = .46% or 1/217" you are saying (with your 3/50 value) that a non-ace card has somehow been removed from the deck and been given to your heads-up opponent. In fact there are 51 cards whose identities are unknown to you, and you are getting one of those 51 cards. The fact that one of those unknown cards is now in the possession of your opponent isn't important. Using 50 in the denominator there is incorrect. If you have information about what kinds of cards your opponents hold, then it would be appropriate to include that in your odds calculation, but since you don't, you can't.

If you know that the other players aren't getting any aces but you are, then I recommend you tip your dealer.

To be brief,

In poker a player is never dealt two CONSECUTIVE cards. this in my view changes the odds,

No.

while also making the number of people at the table as well as the players position a material factor.

No.

Since the desk is randomly ordered, the order of dealing does not in fact change the likelihood of receiving any two cards. In fact, if you burn 50 cards and hand the remaining two to a player, the odds are exactly the same for him receiving any particular hand.

For example: 2 handed: first player dealt; odds of getting pocket aces = (4/52)x(3/50) = .46% or 1/217

2 handed: second player dealt; odds of getting pocket aces = (4/51)x(3/49) = ~.48% or 1/209

This is wrongly formulated; it assumes that the first player receives no Aces (single or pair). Once you make an adjustment for the (small probability) event that he does, the probability is exactly equal. See the following calculation:

(4/52)(3/51) = 0.00452488687 , from google

(48/52)(47/51)(4/50)(3/49) + (4/52)(48/51)(3/50)(2/49) + (48/52)(4/51)(3/50)(2/49) + (4/52)(3/51)(2/50)(1/49) = 0.00452488687 , from google

The second formula above can be derived from http://en.wikipedia.org/wiki/Law_of_total_probability#Statement

Refer to http://en.wikipedia.org/wiki/Hypergeometric_distribution for the appropriate probability distribution to use in calculating poker odds.

You have to make a clear distinction about what you know up front and what not.

If player #1 and player #2 get two cards, they have exactly the same chance that it's pocket aces.

Now player #1 looks at his cards, but not player #2 and player #1 sees that he doesn't have pocket aces. Now he has knowledge about the deck and what was left: 50 cards, all aces still in there and player #2 has 2 of those cards, so now he knows player #2 has a bit higher chance of having pocket aces. If however player #1 draws his cards and sees an ace, he also know that player #2 will have a much lower chance to have pocket aces, because there's 50 cards left with only 3 aces.

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