you have three cards face down, one of them is the ace that you want. you pick one card, your friend turns over one of the other two cards that (s)he knows is not the ace, and puts it back in the deck. before you look at the card you chose, (s)he offers you the option of swapping your card with the one card left on the table.
apparently, you can increase your chances by swapping to 2/3 if you swap.
why?
this really is true - maths fun facts book (called "How Long is a Piece of String) - great read actually.
2007-05-25 05:37:39 · 7 answers · asked by randomperson-randomplace 2 in Science & Mathematics ➔ Mathematics
If you have one card in your hand, there is a 1/3 chance that you hold the ace and a 2/3 chance that the ace is still on the table. Your friend removes one card that is definitely not the ace. Your card is still only 33% likely to be the ace. The remaining card is now 66% likely. Swap!
2007-05-25 05:42:53 · answer #1 · answered by TychaBrahe 7 · 2⤊ 0⤋
After you choose a card, the probability is 1/3 that the ace is the card you've chosen and 2/3 that it is in the other group of two cards. Removing a non-ace (which you know exists) from the group of two doesn't change the probability that the group of two contains the ace. And if the group of two contained the ace, the ace is the remaining card.
To show this mathematically requires conditional probability:
Let A be the event that the card remaining from the group of two is the ace. Let B be the event that the ace was in the group of two before a card was removed and B-comp be the complement of B, i.e., that the ace was not in the group of two before a card was removed. We're interested in P(A).
P(A) = P(A intersect B) + P(A intersect B-comp)
= P(A|B)P(B) + P(A|B-comp)P(B-comp)
= (1)(2/3) + (0)(1/3)
= 2/3
2007-05-25 13:07:40 · answer #2 · answered by Chet 1 · 0⤊ 0⤋
If you type "Monty Hall problem" into your favorite search engine, you'll find explanations, and even little games you can play, that are just like this, mathematically.
On Let's Make a Deal, there were three doors: one with a real prize, the other two with bogus prizes. The contestent would pick a door, Monty would open another door, that was a bogus prize, and ask the contestant whether they wanted to switch.
Exactly the same situation.
You can find sites were you can try swapping over and over, and see how often you get the prize, how often you get the "goat" then use the "stick" strategy over and over.
2007-05-25 20:07:31 · answer #3 · answered by tehabwa 7 · 0⤊ 0⤋
think about it. 1st, your odds are 1/3 of getting the ace you want. your friend then removes a card she knows is not the ace. with a card you don't want gone, now your chances are 2/3 you'll get the ace you want. do the probability formula. 1/3 x 2/3 = 2/3 if you swap.
2007-05-25 12:45:37 · answer #4 · answered by jkidd 2 · 0⤊ 1⤋
First answer is correct: TychaBrahe.
This is the famous Monty Hall Problem as popularized on "Let's make a deal."
See link below for full explanation.
2007-05-25 13:02:45 · answer #5 · answered by Will 4 · 2⤊ 0⤋
no, it is 50% you only have two cards left on the table. you started out with one in three, now you have one in two.
2007-05-25 12:46:37 · answer #6 · answered by dadforfive 6 · 0⤊ 3⤋
dadforfive, all I have to say is, "Try it." Really.
2007-05-25 13:07:23 · answer #7 · answered by ryanker1 4 · 0⤊ 0⤋