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There are $100$ pieces of paper in a box, one of which has a black dot on it. If $100$ people go up one by one and pick a paper from the box, which one has the lowest probability of getting the black dot, and which one has the highest probability of getting the black dot?

Tekheny Ghemor
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    Each person has a $\frac 1{100}$ probability. – lulu Jun 02 '16 at 13:43
  • @lulu I disagree, for the last person picking, there is only one paper left. Is that still a 1/100 probability? – Tekheny Ghemor Jun 02 '16 at 13:46
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    Of course. The special paper has an equal chance of being anywhere in the hat. – lulu Jun 02 '16 at 13:47
  • But is there not a lower chance for the last person picking, since 99 people already could have picked it? – Tekheny Ghemor Jun 02 '16 at 13:49
  • If you want to: prove it inductively. Let $p_i$ be the probability for person $i$. $p_1=\frac 1{100}$, yes? So, then, if I believe it for the first $n$, we write $p_{n+1}=\frac n{100}\times 0+ \frac {100-n}{100}\times \frac 1{100-n}=\frac 1{100}$. – lulu Jun 02 '16 at 13:50
  • @TekhenyGhemor To get some understanding: do the experiment with $2$ persons and calculate the probabilities. You will find out that they are equal. – drhab Jun 02 '16 at 13:50
  • @TekhenyGhemor Use conditional probability if it makes it clearer. Say you want to know the probability for the $n^{th}$ person, this is conditioned on the fact that the previous $n-1$ people didn't pick it. Then see the other comments and answers which show how to find the unconditioned probability. – snulty Jun 02 '16 at 13:53
  • Another view: Suppose $n$ papers and $n$ people are queued such that the $i$-th person take the $i$-th paper. Now the people are set to queue in an arbitrary order (any order you like), and the papers are randomly suffled so that each permutation is equally likely. Then does your arbitrary order selection affect the probability of a particular person obtaining a particular paper? – BGM Jun 02 '16 at 15:11

2 Answers2

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An effort to enrich your intuition.

Give every piece of paper a colored dot. Use $100$ distinct colors (green, blue, black etc.).

Suppose that you are person $38$.

What is the probability that you get the one with color blue (or black if you like)?

drhab
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The probability $p_k$ that the $k$-th person, $1\le k\le 100$ has the black dot is equal to $$p_k=\frac{99}{100}\cdot\frac{98}{99}\cdot\;\cdots\;\cdot\frac{100-(k-1)}{100-(k-1)+1}\cdot\frac{1}{100-k+1}=\frac1{100}$$ so it is independent of $k$ and equal for every one.

Jimmy R.
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