how do you guess ace out of 3 random cards?

Question

okay.you get 3 cards out of your deck(ace,and 2 random cards).you give the cards to someone else for them to shuffle.they put it down and you can guess whats wrong and what is right.you should even know if they switched the cards.it's so hard and i saw someone on this TV show do it.if u know please tell me.i am desperate to know!thanx!!

Answer 1

rub your index finger in one stroke on the back of the card.

a high number slides easy

a low number tugs on your finger a bit

caution! only seems to be true less than 40% of the time! why? because each of the cards are exactly the same on the back.. so it's probably just my imagination.. give it a try anyway.

Answer 2

to make clean your question. the probability without understanding the cardboard isn't unavoidably 50%. in actuality it would be larger. to truly calculate it without lots artwork assuming a tie isn't a loss, locate here sum. (this is (fifty 3/fifty 3 + 40 9/fifty 3 + 40 5/fifty 3 + 40-one/fifty 3 + 37/fifty 3 + 33/fifty 3 + 29/fifty 3 + 33/fifty 3 + 37/fifty 3 + 40-one/fifty 3 + 40 5/fifty 3 + 40 9/fifty 3 + fifty 3/fifty 3) / 13 = 0.7910015 approximatly it fairly is the probability employing the optimal physique of ideas given a tie is a win of prevailing on the 1st card. that's a million/2 provided which you wager RANDOMLY. in case you wager randomly then the probability of prevailing is mostly a million/2. then you definately can basically locate the probabiliyt of prevailing the game as a million - (a million/2)^(n-a million) ([n-a million] by using fact the 1st card is desperate you in straightforward terms play n-a million circumstances) in spite of the shown fact that I advise simulation... if truth be told in case you had a honest fifty two card deck and used n of them, then the probability could in straightforward terms count on n. you should get an extremely precise answer with little actual artwork. you may make a table that gave you the probability for values of n. in case you assume that a tie is a loss... then the guessing probability is smaller than a million/2 (except all ties are out already i.e., all last card numbers are unique) EDIT analyzing up on the game from wikipedia a tie is a loss. we are able to nonetheless locate the 1st probability with some replace (fifty 3/fifty 3 + 40 9/fifty 3 + 40 5/fifty 3 + 40-one/fifty 3 + 37/fifty 3 + 33/fifty 3 + 29/fifty 3 + 33/fifty 3 + 37/fifty 3 + 40-one/fifty 3 + 40 5/fifty 3 + 40 9/fifty 3 + fifty 3/fifty 3) / 13 - 4/fifty 3 = 0.7155298 (in straightforward terms somewhat decrease) the game became into performed with 5 taking part in cards.... no longer 15... and because im fascinated i'm going to write code now to calculate the proability. EDIT2 i basically found out the optimal answer demands greater coding than i want to do :P in spite of the shown fact that that's conceivable..

Answer 3

It looks like you're describing a game called "3-card monte". The idea is to pick out one specific card (in your case, an ace) out of three cards. Another person shuffles the cards by moving them around very quickly, and that's where the trick is played. The person shuffling the cards will have some skill at sleight of hand, so if you try to follow the ace, you'll probably guess incorrectly most of the time.

I'd guess that if you really want to learn how to pick out the correct card, you should learn how to shuffle them and learn the tricks that the dealer uses. By the way, the dealer knows that if you lose track of the ace, you only have a 1 in 3 chance of choosing it correctly, so the odds are greatly in his favor.

Answer 4

Also note the the dealer may also palm the card in his hand and replace it with a other card so that the ace may not even be on the table when you go to chose your card

Answer 5

go on youtube and search for card magic and then tutorual. it shows how to do all card tricks

People so this with marked cards or with card skill to keep track of a card.
on three card monte, the card you are looking for isnt even in the three, it is removed and then put back after