Riddle #23. Can you solve?!
Question: Riddle #23!. Can you solve!?
52 cards and a joker are placed face down on a table!. You start turning over each card one by one (without turning over each card twice)!. What is the percent chance you will turn over 4 aces before turning over the joker!?Www@Enter-QA@Com
Answers:
20%
you know im right lol
5 cards!.!.!. 4 aces and a joker, each has equal 1/5 chance of being turned lastWww@Enter-QA@Com
you know im right lol
5 cards!.!.!. 4 aces and a joker, each has equal 1/5 chance of being turned lastWww@Enter-QA@Com
I'm assuming it's a standard deck of 52 cards plus one joker, thoroughly suffled!. Simplifying things, there are only five significant cards among the 53!. The position of each card in the deck is entirely random, so each arrangement of those five cards is equally likely!. So it could be AAAAJ, AAAJA, AAJAA, AJAAA or JAAAA!. So 20%!. There's no need to include suits, since there would be the same number of variations on each of those five arrangements, so the percentage wouldn't change!. The probability is also unaffected by the number of other cards!. Provided you only have four aces and one joker, they could be in a pile of just five cards or five thousand cards and it would still be 20%!.Www@Enter-QA@Com
Man, you are going to make me haul out my probability book aren't you!?
(probability book is at home) So it goes something like this!.!.!. at the start of things you have a 1 in 53 chance that you turn over the joker, and a 4 in 53 chance that you turn over an ace!. As you turn over the cards the chances that you turn over the joker are 1 in (53-number of turned over cards) and for an ace it's 4 in (53-number of turned over cards) UNTIL, you get to the first ace (assuming you get to it before the joker)!. After you turn over the first ace the probability for the joker remains the same, but the probability for an ace is decreased to 3 in (53-number of cards turned over) and this continues, until you either turn over all the aces or the joker appears!. I can't remember the formulas off the top of my head and will have to look them up, but it is not 100% and it is not 20%, on the assumption that the deck is well shuffled!. Will try to find the book and post the answer later!.Www@Enter-QA@Com
(probability book is at home) So it goes something like this!.!.!. at the start of things you have a 1 in 53 chance that you turn over the joker, and a 4 in 53 chance that you turn over an ace!. As you turn over the cards the chances that you turn over the joker are 1 in (53-number of turned over cards) and for an ace it's 4 in (53-number of turned over cards) UNTIL, you get to the first ace (assuming you get to it before the joker)!. After you turn over the first ace the probability for the joker remains the same, but the probability for an ace is decreased to 3 in (53-number of cards turned over) and this continues, until you either turn over all the aces or the joker appears!. I can't remember the formulas off the top of my head and will have to look them up, but it is not 100% and it is not 20%, on the assumption that the deck is well shuffled!. Will try to find the book and post the answer later!.Www@Enter-QA@Com
well, in a standard 52 card deck, there are no jokers!. if there were jokers, wouldn't there be 54 cards!?
because with only 52 cards, you'd have:
4 2's
4 3's
4 4's
4 5's
4 6's
4 7's
4 8's
4 9's
4 10's
4 Jacks
4 Queens
4 Kings
and 4 aces
ergo, there is a 100% chance that you will overturn all four aces before coming across a joker, because with a standard deck of cards, there are no jokers!.Www@Enter-QA@Com
because with only 52 cards, you'd have:
4 2's
4 3's
4 4's
4 5's
4 6's
4 7's
4 8's
4 9's
4 10's
4 Jacks
4 Queens
4 Kings
and 4 aces
ergo, there is a 100% chance that you will overturn all four aces before coming across a joker, because with a standard deck of cards, there are no jokers!.Www@Enter-QA@Com
Aces do not have "a face!."
So, they cannot be "Face Down!."
Only Kings, Queens, Jacks and of course Jokers have "Faces"!.
All other cards would be turned up!.
So!.!.!.!.
My answer would be 100%,
The Aces would already be turned up before you even see a Joker (which would be "Face Down!.")
:)Www@Enter-QA@Com
So, they cannot be "Face Down!."
Only Kings, Queens, Jacks and of course Jokers have "Faces"!.
All other cards would be turned up!.
So!.!.!.!.
My answer would be 100%,
The Aces would already be turned up before you even see a Joker (which would be "Face Down!.")
:)Www@Enter-QA@Com
You didn't give enough information!. You left waaaay too much to the imagination!. What if the 52 cards were all aces!? What if the 52 cards were all jokers!? What if the 52 cards were all birthday cards!?Www@Enter-QA@Com
the jokers are at the endWww@Enter-QA@Com
One hudred % becouse the joker is the last card to be turned over as it would be on the bottom!.Www@Enter-QA@Com
you never told us if the cards were a deck
they could be all aces
or they could be all jokers
or they could be smile face cards for all we know!.!.!.!.
NOT ENOUGH INFO TO TELL!!!Www@Enter-QA@Com
they could be all aces
or they could be all jokers
or they could be smile face cards for all we know!.!.!.!.
NOT ENOUGH INFO TO TELL!!!Www@Enter-QA@Com
I got about 73% or 7!.3% cant tell :DWww@Enter-QA@Com
i dont know!.!.!.this is the hardest riddle i have ever tried to solve!.Www@Enter-QA@Com
66!.7 PERCENT I LIKE TAKIN WILD GUESSES I MIGHT WIN THE LOTTERY ONEDAY!.Www@Enter-QA@Com
51/52 if all the others are AcesWww@Enter-QA@Com
1 out of forty8Www@Enter-QA@Com
:SWww@Enter-QA@Com
99%Www@Enter-QA@Com
100%Www@Enter-QA@Com
73%Www@Enter-QA@Com
100%Www@Enter-QA@Com
100%Www@Enter-QA@Com
It depends on what the 52 cards are ^-^Www@Enter-QA@Com
Is it a normal deck of cards!?Www@Enter-QA@Com
50%Www@Enter-QA@Com
hmm are they all aces!?!?Www@Enter-QA@Com
100% cos they all aces!? idk whats the answer!?Www@Enter-QA@Com
idk!.Www@Enter-QA@Com
well we dont know what the other cards are do we :SWww@Enter-QA@Com
is it
96%Www@Enter-QA@Com
96%Www@Enter-QA@Com
um, 4 out of 52Www@Enter-QA@Com