Riddle- No one has solved it yet!?!
Question: The Mad Hatter is holding four cards, an ace of clubs, an ace of hearts, an ace of diamonds, and an ace of spades. Alice draws two of the cards. What is the probability that Alice will draw at least one red ace.
don't just assume. i will tell you it is not 50/50
or 1/4 or 3/4 or 1/8 or 1/50
Answers: The Mad Hatter is holding four cards, an ace of clubs, an ace of hearts, an ace of diamonds, and an ace of spades. Alice draws two of the cards. What is the probability that Alice will draw at least one red ace.
don't just assume. i will tell you it is not 50/50
or 1/4 or 3/4 or 1/8 or 1/50
is it 5/6?
100%
None because it didnt say what color it was it just said shape it was!=)
(did i get it right?)
0% since shes puling 2 cards...best answer?
it is TREE!
? I would still say 50/50
One out of two.
(C(2,1)*C(2,1))/C(4,2) + C(2,2)/C(4,2)
Assuming you have learned combinations and permutations in math, this would be the correct answer
i dont know tell us
1/2 or 50%
15/28
Is it 1/3?
None because the Mad Hatter has all the aces so Alice will draw something different in the deck!
5/6
There are 6 possible combos for her to draw and 5 of them have at least 1 red ace.
Mad has all the aces, so there has to be two decks. So that 104- 4 so 100 she has 2 in 100 or 1/50 shot.
there is no red ace so it is 0
That's not quite a riddle. It's more of a math question. I'll use alternative occurrences of probability theory to answer.
There are two events in your question - the drawing of the cards. Let's call them Draw1 and Draw2 and apply them to the formula for alternative occurrence of two events:
P(Draw1 v Draw2) = P(Draw1) + P(Draw2) - P(Draw1 x Draw2)
That is, the probability that one or the other or both of two events will occur is equal to the probability that the first will occur, plus the probability that the second will occur, minus the probability that they both occur.
Your question's phrasing is ambiguous though. If the Hatter is holding 4 cards and 4 aces, that's 8 cards total.
So that's 2/8 + 2/8 - (2/8 * 1/7) =
14/56 + 14/56 - 2/56 =
13/28
When we ignore the twist of the additional four cards, that leaves only the four aces. If that's the case, then it's as follows:
2/4 + 2/4 - (2/4 * 1/3) =
6/12 + 6/12 - 2/12 =
10/12 =
5/6