Probability of someone guessing a card whilst its still being shuffled

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rishrush

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Jun 5, 2016
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Hi

while someone is shuffling, and someone says a card, what is the probability that the card will be the same as the one that is chosen? ( this will be better explained below)

Could someone please explain if my reasoning is wrong?


because if I hadn't shuffled the cards yet, but my friend says "5 of clubs" , then the card that will be on top will be determined by the shuffle. so the probability of the shuffle ending up with the "5 of clubs is on top" is 1/52.


then if i say


ive already shuffled and theres a card on top, lets say its the 5 of clubs, then the probability of my friend choosing the 5 of clubs would be 1/52


so theres an event that P(shuffle)=1/52 and P(friend)=1/52,,, theyre both independent events


so if im shuffling AT THE SAME TIME as my friend says the card, the two events are occurring at the same time (and neither is given), therefore shouldnt the probability of the shuffle leading to say the 5 of clubs AND my friend saying the 5 of clubs equal P(shuffle) multiplied by P(friend) = 1/52 times 1/52?
 
rishrush said:
while someone is shuffling, and someone says a card, what is the probability that the card will be the same as the one that is chosen? ( this will be better explained below)
Could someone please explain if my reasoning is wrong?
because if I hadn't shuffled the cards yet, but my friend says "5 of clubs" , then the card that will be on top will be determined by the shuffle. so the probability of the shuffle ending up with the "5 of clubs is on top" is 1/52.
ANSWER:: 1/52?
Click to expand...
The ace of spades is equally likely to be first as last: ANSWER:: 1/52
 
rishrush said:
Hi

while someone is shuffling, and someone says a card, what is the probability that the card will be the same as the one that is chosen? ( this will be better explained below)

Could someone please explain if my reasoning is wrong?


because if I hadn't shuffled the cards yet, but my friend says "5 of clubs" , then the card that will be on top will be determined by the shuffle. so the probability of the shuffle ending up with the "5 of clubs is on top" is 1/52.


then if i say


ive already shuffled and theres a card on top, lets say its the 5 of clubs, then the probability of my friend choosing the 5 of clubs would be 1/52


so theres an event that P(shuffle)=1/52 and P(friend)=1/52,,, theyre both independent events


so if im shuffling AT THE SAME TIME as my friend says the card, the two events are occurring at the same time (and neither is given), therefore shouldnt the probability of the shuffle leading to say the 5 of clubs AND my friend saying the 5 of clubs equal P(shuffle) multiplied by P(friend) = 1/52 times 1/52?
Click to expand...

Or, to put it another way: When you stop shuffling, there are only 52 different cards which can be on top.

This is, in an off hand way, the same mistake made when some one is asked whether there is a difference between the probabilities of getting two heads by flipping two (fair) coins at a time and getting a heads twice in a row by flipping a (fair) coin twice. There isn't a difference.
 
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