How many 5-card poker hands have all 5 cards ranked 7 or lower (that is, every card is a 2, 3, 4, 5, 6, or 7)?
Choose the answer below that would lead, after calculation, to the correct answer to this counting problem: How many 5-card poker hands have all 5 cards ranked 7 or lower (that is, every card is a 2, 3, 4, 5, 6, or 7)?
e. 4* (6P5)
Since the order does not matter, we’re dealing with combinations, not permutations.
There are 4 suits and 6 cards included in consideration per suit (2, 3, 4, 5, 6 and 7).
So 4×6=24 considered for a combination of 5 at a time, therefore : a. 24C5
In poker, when the cards are dealt to find out who will start, what suits are ranked the highest?
Typically, in a card game… Highest to Lowest:
By the way, Zak n is correct. In poker, the person to the left of the dealer always goes first.
In Bridge, the highest to lowest:
a poker hand consists of 5 cards selected at random from an ordinary deck.find the probability that a poker ha?
a poker hand consists of 5 cards selected at random from an ordinary deck.find the probability that a poker hand contains:(a)4 aces(b)4 of a kind(4 cards of the same rank)(c)5 consecutive cards of the same suit
find the probability, of the letters mentioned, using what’s given. (a)4 aces(b)4 of a kind(4 cards of the same rank)(c)5 consecutive cards of the same suit
I don’t need to find the probability. It’s already been done for me:
Your highest poker losing hand by card rank?
I had 4 of a kind and was beaten by straigh flush (texas hold’em) but luckely only in videogame 😀
I lost with like a King high straight-flush to a Royal. I wasn’t surprised though because 4/5 of our hands were on the board, so I didn’t lose much.
Probability of Getting Nothing/High Card/Garbage In 4 Card Poker?
Determine the probabilities of all the possible hands which can be dealt in the game of 4-card poker. The deck used in 4-card poker is similar to a standard 52-card deck except for the following:
All spades are removed, all jacks are removed, and all 2s are removed. So the deck consists of 3 suits with cards ranking from 3 to 10 plus queens, kings and aces (meaning that getting a 9 ,10, Q, K is still considered a straight) in each suit. Aces are always treated as being one rank above kings. The total number of cards in the deck is 3 x 11 = 33 cards.
Round all probabilities to 6 decimal places. The sum of all your probabilities should be 100%.
I want to know the probability of getting nothing. Nothing: no two cards have the same rank, the four cards are not in sequence, and the four cards are not all the same suit.
The probability of getting nothing is 25116/40920. How ever I want to know how you can get it without adding up all the probabilities of the 3 of a kind, 2 pair, etc. and subtracting them of from the total possibilities. I know it’s a bit wordy but does it make sense?
Unfortunately calculating the complement really is significantly easier than what you want to be able to do. Even if I wanted to do it manually by writing out all 40,920 possible hands I would still be thinking in terms of crossing out all the straights, flushes and hands that contained a pair. The alternative would be to count all the various types of hands that might occur – a task almost as daunting as writing out all possible hands.
Here is what would be involved:
Step ONE: You would start by counting all the hands you would include a “3”: There are 30 choices for your second card (can have anything other than another three) and a choice of 27 for your third card. Your number of choices for a fourth card need to be counted separately depending on what your first three cards had been. For the 45 hands where you had picked all the cards out of the same suit you could only pick a card of a different suit that did not have the same rank as one of your first three ( 16 choices ). For the 9 hands where you had picked a 4 and 5, you could only pick something other than a 6 ( 21 choices )
You would multiply these numbers together and add them (don’t forget to subtract hands that fall into both of these categories) and then divide by 4 factorial since order does not matter.
Step TWO: Then you would count all the hands that started with a “4” and did not have a “3” – same concept, different number of choices at each step.
Etc. For steps THREE through SEVEN. (You can’t have a hand where the lowest rank was a ten )
I really don’t think you want to go through with it, but I see your point, it would be nice to have some kind of way of checking your work – it is not morally satisfying to count up all the non garbage hands and subtract that from 40920 and say that you haven’t miscounted anything.
Whenever I am faced with this kind of concern and am unable to convince myself not to worry about it, I write myself a short program and have the computer create a database of the 40,920 hands. Then I have the program identify each of the hands that have a value, then I visually inspect a random sample of the remaining 25,116 hands to see if I had missed anything. Usually I get glassy eyed and stop looking after I’ve checked a couple of hundred.
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