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Each of the integers from 0 through 11 are written on the back of shirts which David bought for his team. These shirts are packed in a box. David really wants to wear the shirt with number 6 but will be the fourth to pick from the box. If the players are to pick one shirt form the box at random, what is the probability that David gets the number of his choice?

My Solution

The first player selecting the shirt will have 11 other shirts available to pick from, therefore,

$$P(\text{Shirt 6 not selected in first attempt}) = \dfrac{11}{12}$$

$$P(\text{Shirt 6 not selected in second attempt} | \text{Shirt 6 not selected in first attempt}) = \dfrac{P(\text{Shirt 6 not selected in first and second attempt})}{P(\text{Shirt 6 not selected in first attempt})}$$

$$P(\text{Shirt 6 not selected in first and second attempt}) = P(\text{Shirt 6 not selected in second attempt} | \text{Shirt 6 not selected in first attempt}) \times P(\text{Shirt 6 not selected in first attempt})$$

$$P(\text{Shirt 6 not selected in first and second attempt}) = \dfrac{10}{11} \times \dfrac{11}{12}$$

$$P(\text{Shirt 6 not selected in third attempt} | \text{Shirt 6 not selected in first and second attempt}) = \dfrac{P(\text{Shirt 6 not selected in first and second and third attempt})}{P(\text{Shirt 6 not selected in first and second attempt})}$$

$$P(\text{Shirt 6 not selected in first and second and third attempt}) = P(\text{Shirt 6 not selected in third attempt} | \text{Shirt 6 not selected in first and second attempt}) \times P(\text{Shirt 6 not selected in first and second attempt})$$

$$P(\text{Shirt 6 not selected in first and second and third attempt}) = \dfrac{9}{10} \times \dfrac{10}{11} \times \dfrac{11}{12}$$

$$P(\text{Shirt 6 selected in fourth attempt} | \text{Shirt 6 not selected in first and second and third attempt}) = \dfrac{P(\text{Shirt 6 selected in fourth attempt and not selected in first and second and third attempt})}{P(\text{Shirt 6 not selected in first and second and third attempt})}$$

$$P(\text{Shirt 6 selected in fourth attempt and not selected in first and second and third attempt}) = P(\text{Shirt 6 selected in fourth attempt} | \text{Shirt 6 not selected in first and second and third attempt}) \times P(\text{Shirt 6 not selected in first and second and third attempt})$$

$$P(\text{Shirt 6 selected in fourth attempt and not selected in first and second and third attempt}) = \dfrac{1}{9} \times \dfrac{9}{10} \times \dfrac{10}{11} \times \dfrac{11}{12} = \dfrac{1}{12}$$

Ali Baig
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    Alternative approach: there is no advantage or disadvantage in not being the first to select. So, the chances of the 4th to select getting his favorite number are the same as if the person picked 1st. – user2661923 Feb 17 '23 at 21:02
  • @user2661923 Sir, is my solution correct? I am a bit confused because the answer provided in the source is 1/11. Thank you so much :) – Ali Baig Feb 17 '23 at 21:04
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    Yes, your analysis is (also) valid and accurate. – user2661923 Feb 17 '23 at 21:05
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    An answer of $1/11$ probably assumed $11$ shirts in total rather than the $12$ which is correct from "0 through 11". – aschepler Feb 17 '23 at 21:19
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    The underlying question is identical to that of the linked duplicate. The punchline is that with no knowledge of what was selected by the first however many people drawing shirts (or the first however many cards drawn from the deck), any particular draw will be just as likely to be one outcome as any other and we could have imagined that our main character was in fact the first to draw. The question of if your approach was correct or not is already answered that it is indeed correct (though inefficient compared to the aforementioned appeal to symmetry). – JMoravitz Feb 17 '23 at 22:16
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    I think the hypothesis may be that the three players picking ahead of David choose randomly, but that David will choose number $6$ if it's available. If I'm right, the post should be edited to make this clear, and then reopened because it's not a duplicate of the suggested dupe. – Robert Shore Feb 18 '23 at 00:26

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