If the first 51 cards are in their places, the probability of the card 52 being in its place is 1
Yes, that's one scenario that will see the 52 card in its correct place - but, having shuffled a pack, the chance that the 52 card will be in its place is 1 in 52.
To determine the probability of a specific card being in a specific position one does not need to concern oneself with what might be in the other positions.
Consider a deck of 3 cards - say 2 of hearts, 3 of hearts and 4 of hearts. Let's designate that the correct position for the 2 of hearts is the top of the deck, the correct position for the 3 of hearts is the middle of the deck and the correct position for the 4 of hearts is the bottom of the deck.
Clearly the chance of the 2 of hearts being in its correct place after shuffling is 1 in 3(same as for the other 2 cards). The fact that that the 2 of hearts *has* to be in its correct place if both the 3 & 4 of hearts are in their correct places doesn't warrant any special consideration ... that I can see, anyway.
Cheers,
Rob | [reply] |
The fact that that the 2 of hearts *has* to be in its correct place if both the 3 & 4 of hearts are in their correct places doesn't warrant any special consideration ... that I can see, anyway.
I was with you until that last sentence, which I don't get. There is a difference between sampling with replacement and sampling without replacement. I would argue that checking all of the cards in the deck, instead of just one, is sampling without replacement: Once the position of the first card has been checked, there are only 51 cards in 51 possible positions remaining, and so on.
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The events are not independent.
In the case of cards, yes! I think this is correct: http://www.wolframalpha.com/input/?i=product+of+1%2F%2853-n%29+for+n%3D1+to+52
But BrowserUk said:
Its not quite right because each card can only end up in one position; whereas with memory corruption; a value may (and usually will) be repeated.
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[d/l] |