My Math Forum Probability question involving a deck of cards

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 March 22nd, 2015, 05:34 PM #1 Newbie   Joined: Mar 2015 From: MA Posts: 1 Thanks: 0 Probability question involving a deck of cards Hi, I have a probability question that I was hoping someone could help me with. I was playing cards with a friend and I asked them if they could pick the 3 of hearts from the deck. They tried and were incorrect. I put that card aside and said try again so now they had a 1 in 51 chance of getting it right but they were wrong again. We did this many times about 25 cards before he got the 3 of hearts. We then had a disagreement about the probability of never picking the right card so you try 51 times and are wrong every time and the last card was the one you were trying to pick. I thought the probability of that happening was like 1 in a billion but my friend thought the odds of that happening were a lot smaller. How would one figure this math question out and does anyone know the answer. It would start with having around a 98% chance of not picking the one card and would go all the way down to the end where you had a 50% chance of picking the card you were trying to avoid. Whatâ€™s the probability you never pick the chosen card? Thanks for anyone who could give some insight into the question.
 March 22nd, 2015, 05:51 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs The probability on the first draw of picking the wrong card is 51/52, and on the second draw is 50/51 and so on down to the last draw where the probability of getting the wrong card is 1/2. Thus, the probability at the start of the game of picking the wrong card 51 times is 51!/52! = 1/52.
 March 22nd, 2015, 09:31 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra This is easily seen if you make your first 51 choices before looking at any of the cards. You have picked all but 1 of the 52 cards. This is therefore equivalent to picking 1 card that you won't choose.
 March 23rd, 2015, 10:02 AM #4 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Hello, cardman24! The problem reduces to this question: $\;\;\;$What is the probability that the Three of Hearts is the last card? There are 52 choices for the last card. There is only one Three of Hearts. $\text{Answer: }\;\frac{1}{52}$

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