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March 21st, 2016, 05:34 PM  #1 
Newbie Joined: Mar 2016 From: Japan Posts: 1 Thanks: 0  The probability of the card 7 being on the desk
Hello, dear all. I am preparing for a math exam and solving some sample questions, but cannot figure out this one: Suppose you have ten cards, each with a number from 1 to 10 as follows: 1 2 3 4 5 6 7 8 9 10. Now, you shuffle these cards and place them into your pocket in a random order. With these ten cards in your pocket, we will play the following game: 1. You take out one card blindly from your pocket and place it on a desk. 2. You take another card blindly from your pocket again. If the card that you just took out has a larger number than the card on the desk, put it on the desk, too. Otherwise, you throw away the card (that you just took out of your pocket) into a trash. 3. You take another card from your pocket, and if it is larger than any cards on the desk, you put it on the desk. Otherwise, you throw it away in the trash. You repeat this step 3 until you do not have any more card in your pocket. Then, what is the probability that the card 7 is on the desk, and not in the trash. Thank for your time and kindness! 
March 21st, 2016, 05:56 PM  #2  
Senior Member Joined: Jul 2014 From: भारत Posts: 1,178 Thanks: 230  Quote:
Probability of card number 9 to be on the desk is 0.9, because only one card is larger than card number 9. Working similarly, here are the probabilities of all cards to be on the desk: Card number 10: 1 Card number 9: 0.9 Card number 8: 0.8 Card number 7: 0.7 Card number 6: 0.6 Card number 5: 0.5 Card number 4: 0.4 Card number 3: 0.3 Card number 2: 0.2 Card number 1: 0.1 (because it is the smallest in number)  
March 22nd, 2016, 04:19 AM  #3 
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 141 Thanks: 25 
Card 10 has an obvious 100% chance to be on the table. Card 9 is on the table if it is drawn before Card 10, an obvious 50% chance. Card 8 is on the table if it is drawn before both 9 and 10. There are 3! ways to order these three cards, and 2! ways where 8 is first. So the chances are 1/3. Card C is on the table if it is drawn before all higher cards. There are (11C)! ways to order these (11C) cards, and (10C)! ways where C is first. So the chances are 1/(11C). The answer, for C=7, is 1/4. +++++ The Principle of Indifference says that if a each member in a set of outcomes occurs under equivalent circumstances, that each should have the same probability. It's what we use to say that, for a coin flip, Heads and Tails each have a probability of 1/2, that each number on a sixsided die has a 1/6 probability, and that each of your cards has a 1/10 probability of being drawn at any position. The danger in applying this principle, is that it is tempting to apply it to equivalent descriptions instead of equivalent circumstances. For example, one might be tempted to say the chances, when you flip two coins, of getting 0, 1, or 2 Heads are each 1/3. The descriptions seem equivalent, but the circumstances are not. The correct set is {HH,HT,TH,TT}. Yet even the best mathematicians sometimes confuse the difference between description and circumstance. Last edited by JeffJo; March 22nd, 2016 at 04:38 AM. 

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