# Math Help - A probability question

1. ## A probability question

Hi. I hope someone can help me with this question, because so far from the people I have asked there has been great conflict over the simplicity/complexity of the answer.

Person A thinks of a card for Person B to then try and randomly draw out from a normal deck of 52 cards. Person B goes ahead and picks a card, and lo and behold it's the right card! What are the odds that have just been beaten? What are the odds that Person B has managed to pick the very card that Person A had asked for?

The confusion seems to be over whether the odds are simply 1/52 or far greater than 1/52.

Thank you very much, in anticipation.
Lee.

2. Originally Posted by blackjak512
Hi. I hope someone can help me with this question, because so far from the people I have asked there has been great conflict over the simplicity/complexity of the answer.

Person A thinks of a card for Person B to then try and randomly draw out from a normal deck of 52 cards. Person B goes ahead and picks a card, and lo and behold it's the right card! What are the odds that have just been beaten? What are the odds that Person B has managed to pick the very card that Person A had asked for?

The confusion seems to be over whether the odds are simply 1/52 or far greater than 1/52.

Thank you very much, in anticipation.
Lee.
Why would the probability be any different from 1/52?

There are 52 cards to choose from, and assuming the choice is made at
random the probability that it coincides with a previously selected card
(whatever the method of selection) there is only one way of being right, so
the required probability is 1/52.

Now if the person A is not selecting at random, and neither is person B, then
the probability of picking the right card can be considerably higher than 1/52.

(like asking someone to pick a number between 1 and 10, the frequencies
of the posibilities are not all equal, there are favoured numbers 3 and 7 come
to mind)

RonL

3. The way I understand it, if neither person has no magical powers and does not cheat, the probability that Person B selects, out of 52 cards, the card that Person A is thinking is 1/52. The odds of Person B picking that card are 1 to 51... one success for every 51 failures.

4. Well, the argument that's been put forward for the probability being higher than 1/52 is this:

Person A thinks of one card out of 52 (the odds being obviously 1/52), the same as if he had picked a card himself at random from the deck. Then Person B picks randomly the same card. It seems that we're looking at the odds of an event occuring twice consecutively, as in:

1/52 x 1/52 = 1/2704

just as the odds of rolling two sixes are 1/36 (1/6 x 1/6).

Lee.

5. But aren't the two events of rolling two dice independent? If Person A were to roll a die and get a 6 and then Person B were to roll a die and get a 6, doesn't it still stand that the probability of rolling two sixes is 1/36?

6. One would not be calculating the probability of two events occuring in sequence but the probability that B chooses the card A chose given that A chose a card. Also, events A and B are independent.

7. Yes, the two events of rolling two dice are independent, but calculating the probability that one would get a 6 on one and a 6 on the other would be the probability that two events A and B occur in sequence, so $P(A\, and\, B) = P(A)\cdot P(B)=\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}$.

The way the probability of the two choosing the same card would be 1/52 x 1/52 = 1/2704 if we were finding the probability that both person A and person B choose, say, an Ace of hearts or any other previously decided on card by person C, but the way you put it shows that we only want the probability for the two cards being the same, not both being the same to some previously chosen card.

8. Originally Posted by blackjak512
Well, the argument that's been put forward for the probability being higher than 1/52 is this:

Person A thinks of one card out of 52 (the odds being obviously 1/52), the same as if he had picked a card himself at random from the deck. Then Person B picks randomly the same card. It seems that we're looking at the odds of an event occuring twice consecutively, as in:

1/52 x 1/52 = 1/2704

just as the odds of rolling two sixes are 1/36 (1/6 x 1/6).

Lee.
Yes, but its wrong.

In the die rolling case you want the probability that the two die show the same face,
which is: 1/36+1/36+1/36+1/36+1/36+1/36=1/6

Try a experiment with smaller numbers (try just two cards in the deck) and you will see.

RonL