OK smart guys...What are the odds of guessing 5 cards in a row (not suit) in a deck of playing cards?

Assuming you don't put the cards back in the deck, the odds are 311,875,200 (or 52x51x50x49x48) to 1. If you put the cards back in the deck, the odds are 380,204,032 to 1, or 52 to an exponent of 5.

OK smart guys...What are the odds of guessing 5 cards in a row (not suit) in a deck of playing cards?
one in 371,293 The math is simple. The chances of guessing one card is one out of 13. The second time is one out of the one out of 13. The third is one out of the one out of the one out of the 13, and so on. So you end up calculating 13*13*13*13*13=371,293
Reply:It's actually completely dependent upon which cards you pick. The chance of picking the first card correctly is 4/52; there are 4 of each card in a deck. This fraction can be reduced to 1/13.





This is where it becomes tricky. If the first guess is correct, the chance of guessing that same particular card for the second guess would be 3/51, which can be reduced to 1/17. In this case, the chance of guessing both right would be (1/13)*(1/17) = 1/221, or 0.5%.





f the card guessed was DIFFERENT from the first card guessed, the odds of picking the next card correctly would become 4/51. In this case, the chance becomes (1/13)*(4/51) = 4/663, or 0.6%.





The best case odds (in which a person guesses a DIFFERENT card value each time) would be (1/13)*(4/51)*(2/25)*(4/49)*(1/12) = 32/9746100, or 0.0003% (I reduced the fractions in the equation).





The worst case odds (in which a person guesses the same card


and guesses it 4 times in a row, then has to choose another value) are (1/13)*(1/17)*(1/25)*(1/49)*(1/12) = 1/3248700, or 0.00003% (notice this percentage is only 1/10 of the best case odds).





Once again, this is completely dependent upon the cards guessed and which are left; your range of percentage for guessing them all right would range from 3 in a million to 3 in 10 million.
Reply:52 to 1 odds.........
Reply:A deck has 52 cards. If a person guesses the first card, that is 1 out of 52 possibilities. If he then guesses the second card, that is 1 out of 51, but the odds of him being able to guess BOTH of those correctly is 1 out of 52 multilplied by 1 out of 51 which becomes 1 out of 2,652. If he goes on to guess the third card correctly, then this becomes 1 out of 132,600. Guessing the fourth card would be 1 out of 6,497,400. So ... to answer your question, the odds of guessing the fifth card correctly, too, become 1 out of 311,875,200. I'm betting not a lot of people are going to be able to pull that off!!
Reply:5/52 = 0.09615





Approximately 9-10% of the time.
Reply:Pretty damn big.
Reply:It depends on if you reveal what the card is before I guess the next one. Because if I guess all five of them at once and then you reveal them the odds are different than if i guess one and then you show it to me and then I guess the second one, since I obviously wouldn't try guessing that the second one is the same as the first.





DUH!
Reply:50%......lol i dunno!!!!
Reply:Do you put the cards back in after you draw one, or do you draw five out at once? If you draw five out at once, is it completely random what card you will guess next, or is there some sort of logic involved? And when you say "not suit" does that mean that the suit doesn't matter? These are important questions to figure this problem out, I'm really not sure how people have gotten answers, but since most of them are very distinct, I assume they are also not sure :)








First of all, you can add details by using the link to the right of the question: What he told me is that you choose 3 in a row out of one deck, and then choose 2 in a row (I assume that you shuffle the deck).





Here are some numbers (again I assume suit is not important):


Odds if choice is completely random:


Choosing 1st: 1/13


Choosing 2nd: (48/51)•(4/51)+(3/51)•(3/51)=67/867


Choosing 3rd: (3/51)•(2/50)^2 + (48/51)•[(6/50)•(3/50) + (44/50)•(4/50)]= 777/10625





Therefore the odds of choosing 3 in a row are (1/13)•(67/867)•(777/10625) =17353/39918125 and the odds of choosing 2 in a row are (1/13)•(67/867)=67/11271. Since the deck is shuffled in between they are mutually exclusive, thus the odds of choosing 2 in a row then 3 in a row are:


(17353/39918125)•(67/11271) = 1162651/449917186875 ≈ 2.584•10^(-4)%





If you do this intelligently though (that means you choose so that no two cards have the same value) you can increase your odds:





Choosing the 1st: 1/13


Choosing the 2nd: 4/51


Choosing the 3rd: 4/50





Odds of choosing 2 in a row are (1/13)•(4/51)=4/663 and the odds of choosing 3 in a row are (1/13)•(4/51)•(4/50)=8/16575. Therefore the overall odds are (4/663)•(8/16575) = 32/10989225 ≈ 2.912•10^(-4)%





Choosing intelligently greatly increases your odds.
Reply:.000000487%





(1/48)*(1/47)*(1/46)*(1/45)*1/44)





Results may vary depending on if you pull out any duplicate card numbers.
Reply:Is it 1 in 311875200? Unless you are David Blaine, then it would be 1 in 1 (after editing).

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