yes i understand your point of view. but why dont you try shuffling a deck until the top card is the king of diamonds, the second card is the 9 of hearts and the 3rd card is the 4 of spades.
Originally posted by zjones
Hopefully when it finally shows up you'll realize that it wasn't a 50/50 shot.
Originally posted by badgerer
yes i understand your point of view. but why dont you try shuffling a deck until the top card is the king of diamonds, the second card is the 9 of hearts and the 3rd card is the 4 of spades.Spoiler
If you don't understand combinations versus permutations, and why a certain arrangement of cards will have a statistically higher probability of occurring than other arrangements of cards, then I'll explain that in greater detail to you.
If you're just trying to troll me, then I'll go ahead and leave the forum since people are obviously way better at the game than me and don't need my idiot advice.
the 50/50 thing was a joke, but i do have a serious point.
what i'm saying is if you shuffle a deck of cards the chance of getting 2 :spade:3 :spade:4 
at the top is the same as getting any other unique 3 card combo like for example k :diamond:9 :heart:4 (i don't know what the odds are exactly). the odds do not change just because its a 3 to a straight flush. do you see what i mean?
every poker deal is unique and therefore has the same chance of occurring as any other deal. same goes for the hand in the OP. i don't think the probability changes because 5 villains got dealt pairs and hit the board.
am i wrong?
Originally posted by badgerer
the 50/50 thing was a joke, but i do have a serious point.what i'm saying is if you shuffle a deck of cards the chance of getting 2 :spade:3 :spade:4
every poker deal is unique and therefore has the same chance of occurring as any other deal. same goes for the hand in the OP. i don't think the probability changes because 5 villains got dealt pairs and hit the board.
am i wrong?
No, you're not wrong at all. Any card has the same probability of showing up as any other card. Every single shuffle is going to have the same probability of 52 cards being in a certain order as other shuffles will.
So you're right when you say that getting 2♠ 3♠ 4♠ is just as likely as getting K♦ 9♥ 4♠. The problem that people want to figure out is how often certain things will occur over other things. As you may or may not be aware, there is somehwere in the range of 80 unvigintillion ways to arrange a deck of 52 cards. It's a statistical improbability that anyone will ever see the same arrangement of 52 cards twice.
However, it's not a statistcal improbability that certain cards will show up among multiple hands, given how specific those cards are. The more specific the arrangement, the less likely it is to occur. That's because the number of possible arrangements gets reduced the more specific it is. If all 52 cards have to be in an exact order, then there is only 1 possible way to arrange the cards. If only 1 card has to show up in a certain order, on the other hand, then there are trillions, upon trillions, upon trillions, upon trillions, upon trillions of ways to arrange the other 51 cards in the deck.
Therefore, we can say that the 2♦ will show up in a particular spot much more often than 2 specific cards will show up next to each other, and that 2 specific cards will show up more often than 3 specific cards. The 2♠ 3♠ 4♠ has the same probability of showing up as K♦ 9♥ 4♠, but they are both more likely to show up than 5♦ 8♥ 7♣ J♣, because there are less possible ways to arrange 48 cards in a 52 card deck than there are ways to arrange 49.
So, when we ask the probability of 5 pocket pairs making 4 sets with exactly 1 card that's different from them all, we're asking how many different ways 15 cards can be arranged that will yield this result. Even if the 15 cards show up, but they're not in a special order, then the conditions won't be met. So we're narrowing down the possible ways for those 15 cards to be arranged. Will 15 cards always have the same probability of showing up as any other 15 cards? Yes, but only approximately 1 in 7.75 billion will have an arrangement of 15 cards that contains 5 pocket pairs, 4 sets, and 1 card that's different from all the other cards.
i get what you are saying, you obviously know your maths better that i do. i think our disagreement is more about interpretation of the question. ie is the question "how likely is five pocket pairs making four sets which all lose to a 4 card straight" or something to that effect or is it "how likely is this specific hand".
in hindsight, your response is probably what the OP was looking for, but i dont think its a very helpful way to think about these kinds of hands.
on another note, i was staring at the word unvigintillion in your post for a while. at first i thought it was some kind of mathematical jargon, then i was sure it was a made up word, then curiosity got the better of me and i googled it. you learn something new everyday!
The thing that struck me is not the cards setup preflop, but the way everybody got a set before the river got em all with the only player without a set.
sick
Originally posted by WinStrat
The thing that struck me is not the cards setup preflop, but the way everybody got a set before the river got em all with the only player without a set.sick
There is an old card trick that looks a lot like this.
♦ Using a brand-new -- or any deck sorted into suits in order
♦ Deal 7 5-card hands -- except that the dealer deals every 2nd round to himself off the bottom.
♦ When done, 6 players have a full house except the dealer, who has a straight flush.
Although such a setup seems fantastically improbable there are a lot of different combinations that give the same result.
There are 8.06582E+67 [*] distinct ways to order a 52-card deck, but many of them give rise to exactly the same poker hands, since the order of cards in a poker hand doesn't matter, nor does the suit.
To use the card trick as an example,
Once you have sorted the deck, you can pile the suits up 24 different ways,
Each suit can be ordered 13 different ways (one for each starting card -- although some orders won't result in a straight flush to the dealer)
All will not result in exactly the same hands, but all will result in the same scenario: 6 boats and a straight flush (except as noted).
Cheers,
--VS
[*] I used Excel's permut(52,52)
it would be interesting to calculate the probability of the event that a cooler of 5 hands of set or better happens.
