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October 22nd, 2014, 04:24 AM   #1
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Card draw probability problem

Hi everyone,

I have been stuck in this question for a while now and I really appreciate any help

Problem:

A game is being played by n people, A1, A2, ...., An, sitting around a table. Each person has a card with their own name on it and all the cards are placed in a box in the middle of the table. Each person in turn, starting with A1, draws a card at random from the box. If the person draws their own card, they win the game and the game ends. Otherwise, the card is returned to the box and the next person draws a card at random. The game continues untill someone wins.

Let W be the probability that A1 wins the game.

Let
$p\:=\:\frac{1}{n}$ and $q\:=\:1-\frac{1}{n}$

(i) Show that

$W$ $=$ $p$$+$$q$$^n$.$W$

(ii) Let m be a fixed positive integer and let Wm be the probability that A1 wins in no more than m attempts. Also if this expression holds true:

$e^{-\frac{n}{n-1}}<\left(1-\frac{1}{n}\right)^n<e^{-1}$

Then show that, if n is large, Wm/W is approximately equal to
$1-e^{-m}$.
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October 22nd, 2014, 04:35 AM   #2
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For part 1, you should consider what happens on everybody's first turn. i.e. once round the table.

For W we require either that A1 wins or that nobody wins. In the latter case, we are back to the initial position, because each round is independent of the others. The two possibilities for each round are mutually independent, so we can add their probabilities. This description is the meaning of the equation you are asked for.
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October 22nd, 2014, 04:46 AM   #3
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For part 2, start by solving the equation of part 1 for $W$ and creating an expression for $W_n$. You should see the middle expression of the inequality emerge quite simply.

For the final answer, notice what happens to the expressions on the ends of the inequality as $n \to \infty$.
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October 22nd, 2014, 08:37 AM   #4
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A further hint would be that the expression for $W_n$ will be a geometric series (with $n$ terms).
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