In a stack of 52 shuffled cards, what is the probability of getting all 4 kings before the ace of spade? I know there are lot of smart people here, please explain to me how to find this probability.
thinking about it briefly, would it not be 1/5? All the other cards are irrelevant, so you just toss them out. So you have the four kings and the ace. Shuffle those up, and the probability of the ace being last is 1/5.
The probability of getting each individual king before the Ace of Spades is 1/2. Since you need that to happen 4 times the odds are 1/16. The number of cards in the deck are irrelevant.
This is correct. All other cards are irrelevant. So, dealing with these 5 cards, there are only 5 possible ways you can draw them (because it doesn't matter if the King is a club, spade, heart, or diamond). Possible Draws: A K K K K K A K K K K K A K K K K K A K K K K K A Only 5 possibilities, thus the answer is 20%.
If the only cards in the pot were 4 kings and an ace of spades, then I get the 20% thing, but the OP specifically states that they are in a full deck.
Doesn't matter. So what do you if you come across one of the other 47 cards? Nothing. You toss it aside. They're irrelevant. There are still only 5 ways those 5 specific cards can appear (with the type of king being a non-factor).
Ohh, ok, I gotcha. I was thinking that the cards had to be the only 5 chosen in the right order out of the 52 cards.
Just to clarify, what exactly is the question asking? "Drawing five cards total, what is the probability of drawing all four kings and then the ace of spades?" or "Drawing all 52 cards one-by-one, what is the probability of seeing all four kings before the ace of spades?"
i'm good at math but often miss the easy way to look at things when it comes to probability. the other cards are irrelevant and it is 1/5 like others have said. and i did it the long way anyway and also and got 1/5 so hey, at least the long way worked.
Statistics and Probability. I hate this subject. The only class that I got nothing out of in College.
What is your long way? I am quite curious to know. A following up to this question is "what is the probability of getting all four kings before getting ANY ace?"
(52/4)*(51/3)*(50/2)*(49/1)*(48/4) For stats, I think it'd be 52C4* (48/4) Or 1:3,248,700 draws. You want a king for the first draw. 4 kings in a deck. Next draw there are 3... Final draw is ace but there are 4 aces. Since the draws are dependent upon each other because of order, you multiply the probabilities.
The question is being asked backwards. It should say: What is the probability that, during a one-at-a-time drawing of cards from a shuffled 52-card deck that the Ace of Spades will be drawn LAST after first seeing all four Kings drawn before that? Therefore, the specific probability of any one card being drawn on any one pull is not relevant. The only issue at hand is the likelihood that the Ace will be drawn after the 4 Kings. The entire 52-card draw is considered one entire act, and the only relevant question is how likely the Ace will be the last of those particular 5 cards. There are only 5 possible outcomes: The Ace will either be first, second, third, fourth, or fifth. The question is concerned specifically with the odds of one particular outcome: fifth position. Those who have said 20% are correct.
i just went through each position the ace could be picked (1 thru 52) and did the probability you'd get all 4 kings before that position. so if the ace was picked 3rd, there would be 0 chance to get all 4 kings. if it was picked 10th, then of the 9 cards before, there would be 51C9 ways to pick those 9 cards, and and 47C5 ways that included 4 kings (47 b/c you're taking the ace and 4 kings out of the equation and 5 b/c there are 5 other cards in the 9 cards besides the 4 kings). then i just put that in excel for all 52 positions and added up the individual probabilities and it came about to 0.2 or 1/5. it sounded a lot more complicated when i just wrote it than it really was, but it still got me to 1/5, just about 10 minutes after the easy way would have.
