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I have a pack of cards and use the following method to shuffle them

  • Pick a random card from the deck and replace the first card with it
  • Put the first card back in the deck
  • Move to the second card and repeat the process till the final card

So I select any one of 51 cards to replace the first card, 50 for the second, 49 for the third and so on till the 51st card. Cards once selected and replaced aren't to be put back in the deck

Given the above method

  • Does all cards have equal probability of ending up in any position?
  • If the answer is yes or no, how to calculate and prove it?
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    When you put the first card back in the deck, where in the deck do you put it? Do you mean it switches places with the random card chosen to replace it? And what do you mean, "cards once selected and replaced aren't to be put back in the deck"? If a card is replaced and not put back in the deck, then the deck has fewer than 52 cards, which is not what's usually meant by shuffling.2012-05-17
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    @GerryMyerson Cards once replaced aren't put back. Its a biased shuffle (shuffle may be the wrong term). So for the first card I pick from the next 51 cards, for the second card, I leave the first card and select from the remaining 50 cards and so on. So there would be no selection for the last card since it is automatically select. And just 2 options for the 51 st card.2012-05-17
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    I'm still not with you. You write that you replace the first card, and that you put the first card back in the deck, but now you write that cards once replaced aren't put back. So, is the first card put back in the deck, or isn't it? If it isn't, you only have 51 cards in the deck (and as you continue, you'll have fewer and fewer), and the chance of the first card ending up in any position is zero, because, once replaced, it wasn't put back. So what is going on here?2012-05-17
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    @GerryMyerson For the first card (7C), I pick any card from the remaining 51 cards (6S). Now I am going to put the original card (7C) back in the deck. And for the next card (6D), I am going to select from the remaining 50 cards excluding 6S and 6D and so on. So what is the probability that 6S,6D,7C or any other card would end up in the list and is it same for all the cards in the pack?2012-05-17
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    OK, so where in the deck do you put that 7C when you put it back in the deck? Might you put it on top? Do you always put it on the bottom? Where?2012-05-17
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    I would put the 7C back in place of 6S. If 6S is the 32nd card, then after the first replacement 7C would be the 32nd card and 6S the first card (both would swap places). But 6S could be anywhere in the deck.2012-05-17
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    Ah! So, you're **swapping** the first card with some random card below it, then swapping the second card with some random card below it, etc., etc.2012-05-17
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    Yes and thanks for persisting with me. Can I move this to chat. I have already got an alert!2012-05-17
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    Ignore the alert. Anyway, it's bedtime here. Maybe I'll come back to it tomorrow.2012-05-17

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