I'm terrible at math but how often does AA, KK, QQ, AKs, come up?
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I'm terrible at math but how often does AA, KK, QQ, AKs, come up?
A particular rank of card has a 1 in 13 chance of coming up.
So an unpaired hand, such as AK, has a 1 in 13x13 chance of coming up, or 1 in 169 (not quite, as there is one less card in the deck after the first card, but this is close enough).
With respect to a paired hand, the first card is 1 in 13, and the second card is 3 in 51, or 1 in 17 (because there is one less card in the deck and one less card of that rank). So it is 1 in 13x17, or 1 in 221.
With suited cards, the formula is as follows. We know there is a 1 in 169 chance of any combination of AK. Of the 16 combinations of AK, 4 are suited. So 1 out of 4 AK's are suited.
That means you have a 1 in 676 chance of AK suited.
The chance of being dealt one of these hands is 1.66%
AA-6, KK-6, QQ-6, AKs-4
Total 22 Combinations / 1326 Starting Hands
(Edit)
Hand...........Every # Hands
AKs------------ 331.5
AA ------------- 221
AA,KK,QQ,AKs - 60.27
So every 221 hands, you get AA. Right?
yes, on average. Check out this post in the beginners digest, it's exactly what you're talking about.Quote:
Originally Posted by ArcadianRock
http://www.flopturnriver.com/phpBB2/...tc-t75711.html
from that post: there's 6 combinations of AA and 1326 total possible card combinations in the deck. so 6 in 1326 chance or 1 in 221
AA: (Or any pocket pair)
4 ways to draw the first card, 3 ways to draw the second card.
Since we are calculating the probability of being drawn two aces, we multiply the chance of being drawn the first ace by the chance of being drawn the second. So we have:
4/52 * 3/51 (which reduces to) = 1/13 * 1/17 = 1/221
Therefore your chance of being dealt AA is 1/221. This is actually the same result for being dealt any specific pocket pair, because the probability of being dealt each card is calculated the same way. So if you wanted to find the probability of being dealt say, pocket 7's, it's the same method.
AKs: (Ace Dealt First) -see tyrn's correction for combinations, this example was created using permutations, which assumes order matter, when in practice it does not.
This is slightly different, but the main thing is to keep in mind the theory - the theory does not change. Simply count how many cards are left that make the hand you're trying to find the probability for. Ask yourself how many cards are left in the deck that make my hand after I am dealt my first card?
There's 4 ways to choose an Ace out of the deck, as we know. However, there's only one way to choose the King for your hand to be suited, therefore:
4/52 * 1/51 = 1/13 * 1/51 = 1/663.
So as you can see, when you want to find the probability of being dealt AK suited (when being dealt an Ace first), you must remember that there is only one way to choose the second card to match your suit.
Now you know the theory, you should be able to figure out any other cases.
(8 ways to choose first card for AKs for combinations, 1 way to choose second)
You have to remember you can be dealt AKs or KAsQuote:
Originally Posted by Micro2Macro
So your first card can be any ace or king, which makes it 8/52 * 1/51 = 1/331.5
Oh shit you're right. I was thinking along the lines of 'after being dealt an ace' rather than 'being dealt 'AKs''. My brain was using permutations but they don't matter in practice because the AK or KA is the same (god damn stats class teaching me useless permutation garbage).Quote:
Originally Posted by tyrn