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In a game of draw straws, all turns are equally good. There are $n$ straws in a box, $m$ of them being golden, the rest are black. They are drawn without replacement (the straws will not be put back after drawing). Suppose that the straws are of the same shape and people do not look when they draw. What is the probability of the $k$-th person in line getting a golden straw?

I did it for $k$=1 and 2. So basically if k=1, probability is just $m/n$. If $k$=2, I used total probability theorem conditioning on drawing golden straws on the first draw and drawing black straws for the first draw, and the probability is still $m/n$. But I'm not sure how to do for the $k$=3 and so on

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    Can you please show some working on your part. Maybe try with an explicit example like 5 straws and 2 of them gold, this should help you grasp how to solve it in the general case. – 2017-02-09
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    I did it for k=1 and 2. So basically if k=1, p is just m/n. if k=2, I used total probability theorem conditioning on drawing golden straws on the first draw and drawing black straws for the first draw, and the p is still m/n. But I'm not sure how to do for the k=3 and so on – 2017-02-09

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So for k=1 we need the first straw to be golden, so yes it's prob m/n. For k =2 (assuming that m>k) there are two possibilities, either the first straw drawn is black or gold and the second has to be gold. So let's add the probs for the allowed cases since either one will give us our desired result:

p1 = p(black then gold) = $\frac{n-m}{n} + \frac{m}{n-1}$

p2 = p(gold then gold)= $\frac{m}{n} + \frac{m-1}{n-1}$

Hence,

p(second person getting gold) = p1 + p2 = $1 + \frac{2m-1}{n-1} = \frac{n + 2m-2}{n-1}$

Notice that we simply worked out the possible cases and then added the corresponding probabilities. For the general case you'll need to work out summation type formulas rather than calculating everything for each case as I did here (also considering the case k>m), but the principle is just the same, if the kth person is drawing gold, then what are the possible things that can happen before that while still allowing for at least one gold straw to be left by the kth draw.