In Wikipedia, it shows that the Pair probability for 7-card poker hand is 43.8%.
After that 2 card is dealt to me, lets say Ace of Spades and 7 of Diamonds, does knowing these 2 cards effect the probability of Pair? If so how? If not, why?
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Probability is a measure of certainty based on information, so yes, every piece of information you get changes the probabilities.– Lee Daniel CrockerOct 7, 2021 at 19:41
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2 Answers
Intuition says yes...
For any random seven cards, there are 21 ways to combine them in forming a pair: pick one of them, then pick another, and divide the total count by 2 because you can change the order.
Knowing that your first two cards are not a pair, you have lost one of the potential ways to form at least one pair, reducing your chances by a small amount, since you only have 20 ways remaining to achieve it (see below for further explanation).
... and the math agrees.
The probability that you end up with no pairs at the end, with zero cards already in hand, is 135168/643195 or just under 21%.
Note that this means you get at least a pair in 79% of hands, but some of those hands are more than a pair (two pair, trips, quads, full house) and some without a pair are ranked higher (flush, straight).
With two cards in hand, and no pair yet, that increases to 8448/37835 or just over 22%.
The odds of getting no pair at all went up by 1.3% once you miss the first possible pair.
This reasoning applies to some extent with more cards being pulled, but other factors begin to grow and balance out, such as the likelihood that you match an existing in-hand card, which grows as you add more cards to the hand.
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Are you counting flushes and straights as well? Pair or better should be 82%+ of all 7 card combos? Oct 7, 2021 at 8:24
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Ranking of hands is irrelevant to the question, and therefore the answer. You can expect two or more cards of the same rank(s) in 79% of seven-card hands. I have no idea where you got 82% from.– NijOct 7, 2021 at 9:27
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1I see what you're saying. From the OP's 43.8, I assumed that higher-ranked hands should be excluded. Oct 7, 2021 at 15:44
Note that your figure of 43.8% for a pair means "chance for a hand that is worth exactly one pair". So it excludes both better and worse hands.
Some hands with a pair in them aren't even counted here (for instance if the seven cards have a pair and a flush, it's just a flush). But if you make the flush less likely, then some of those pairs that weren't counted before become "just a pair".
So the overall strength of your hands goes down (from 82% pair or better to only 80%), but the probability of a hand having exactly one pair goes up.
| Hands | one pair exactly | one pair or better |
|---|---|---|
| All combinations 7 cards | 43.8% | 82.5% |
| A♠ 7⋄ hole cards | 45.8% | 80.5% |
