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there are 5 people in a party and there will be exchanging gifts. Now, they conducted a lottery writing their names . What is the probability that they will get the paper with there name ?

Initial Analysis:

Suppose we fix the order of draw, say $P_1 , P_2, ..., P_5$ letting $P_1$ be the first person to draw. If the question would be what is the probability of $P_1$ to get the paper with his name, then the answer would be $20$%.

How about $P_2$ ?

The probability depends on the outcome of $P_1$. If $P_1$ got the paper except of $P_2$ 's, then the probability is $25$%. If $P_1$ got $P_2$ 's name, then the probability will be $0$ %.

I can't go any further. It seems complicated to me or I am just complicating the problem.

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Which of these are you asking for? (1) The probability that a specific person gets his own name. (2) The probability that at least one person gets his own name. (3) The probability that everyone gets his own name. –  Brian M. Scott Dec 22 '11 at 0:23
    
it is the first. –  Keneth Adrian Dec 22 '11 at 0:27
    
So you want the probability that Maurice will get his own name. Can you see that the probability of getting one's name is the same for everybody? –  André Nicolas Dec 22 '11 at 0:44
    
@AndréNicolas : I am actually thinking that the probability changes as the people draw. –  Keneth Adrian Dec 22 '11 at 1:47
    
Permutations where no element stays in position are called derangements: en.wikipedia.org/wiki/Derangement –  Ross Millikan Dec 22 '11 at 2:43

1 Answer 1

up vote 2 down vote accepted

The probability that the first person gets his own name is, as you say, $1/5$. The same is true for each of the others: it doesn’t matter where a person comes in the order of drawing.

There are $5!=120$ possible orders in which the $5$ slips can be drawn. I claim that exactly one-fifth of those orders have $P_2$’s slip in the second position. To see this, consider the $4!=24$ possible orders for the slips for $P_1,P_3,P_4$, and $P_5$. You can insert $P_2$’s slip into one of these orders in $5$ possible places. For instance, starting with $$P_3,P_1,P_5,P_4$$ you can get any of these:

$$\begin{align*} &\color{red}{P_2},P_3,P_1,P_5,P_4\\ &P_3,\color{red}{P_2},P_1,P_5,P_4\\ &P_3,P_1,\color{red}{P_2},P_5,P_4\\ &P_3,P_1,P_5,\color{red}{P_2},P_4\\ &P_3,P_1,P_5,P_4,\color{red}{P_2} \end{align*}$$

Thus, each of the $24$ possible orders for the slips for $P_1,P_3,P_4$, and $P_5$ gives you one order of all five slips in which $P_2$’s slip is first, one in which it’s second, one in which it’s third, one in which it’s fourth, and one in which it’s last. Consequently, $P_2$’s slip is equally likely to appear in any one of the five possible positions: for any position, there are $24$ orders in which it appears in that position. In particular, it appears in the second position $24$ times out of $120$, or one time in five.

Similarly, $P_3$’s slip is equally likely to appear in each position, so the probability that it’s in the third position (so that $P_3$ gets it) is $1/5$, and $P_4$ and $P_5$ also have probability $1/5$ of drawing their own slips.

Added: The probability that at least one person gets his own slip is $$1-\mathbb{P}(\text{no one gets his own slip})\;.$$ Calculating the probability that no one gets his own slip is the classic problem of derangements. It turns out that $44$ of the $120$ possible orders of drawing are derangements, i.e., orders in which no one gets his own slip, so the probability that you want is $$\frac{120-44}{120}=\frac{76}{120}=\frac{19}{30}=0.6\overline{3}\;.$$

As the number of participants increases, the probability that at least one of them gets his own slip approaches $$1-\frac1e\approx 0.63212\;.$$ As you can see, the approximation is quite good already when $n=5$.

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Thanks for the answer. Ive gotten an idea –  Keneth Adrian Dec 22 '11 at 1:56
    
hanks for the answer. Ive gotten an idea. My bad. I am actually reffering to no 2. –  Keneth Adrian Dec 22 '11 at 2:06
    
@Ken: I’ve added the information for alternative (2) to my answer. –  Brian M. Scott Dec 22 '11 at 2:25
    
@Ken Dee: My guess is that whoever asked you the question was intending you to find the probability that at least one person gets his/her own name. So the person probably wanted you to find the number of non-derangements. But the person who posed the problem may not quite have stated it precisely enough. For $n=5$, you can find the answer with little theory, just careful listing. There are even a few shortcuts. For $n=6$ this already gets unpleasant, and by the time you hit $7$ or $8$, "general" theory becomes a practical necessity. –  André Nicolas Dec 22 '11 at 18:14

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