# Probability question about “good” tickets.

The problem:

Out of 20 exam tickets, 16 tickets are "good". Tickets are carefully mixed, and students take turns pulling one ticket. Who has the better chance to draw a "good" ticket the first or the second student in the queue?

My attempt:

Obviously the probability of the first student getting a "good" ticket is $16/20 = 4/5 = 0.8$

Now here's where I'm confused. I considered two cases. If the first student got a "good" ticket then the chances of the second student are $15/19 = 0.789$, so lower than the first student.

However if the first student didn't get a "good" ticket, the chances of the second student are $16/19 =0.842$, so slightly better chances than the first one.

So is the answer "depends on whether the first student gets a "good" ticket"?

• The question is to "predict" the propability of the 2nd student getting a good card not knowing what the first person draws. – Cornman Sep 21 '17 at 20:32
• Hint. Is there any of the 20 tickets that the second student has a different chance of getting from any other? – Henning Makholm Sep 21 '17 at 20:34
• So I have to count the probability of the second student having better chances? That would be equal to $4/20 * 16/19 = 0.168$. But I'm kinda lost after that. – Nick202 Sep 21 '17 at 20:34
• Draw a "probability tree". :) – Cornman Sep 21 '17 at 20:36

You're almost there. You just need to apply conditional probability rules.

Essentially, the unconditional probability that the second student gets a good ticket is equal to a weighted average of the conditional probabilities $\frac{15}{19}$ and $\frac{16}{19}$, where the weights are the probabilities that the first student did or did not get a good ticket, respectively. If you carry out that computation, you get

$$\frac{4}{5} \times \frac{15}{19} + \frac{1}{5} \times \frac{16}{19} = \frac{76}{95} = \frac{4}{5}$$

So the two students have equal probabilities of getting a good ticket.

The same result can be obtained by observing that if you lay the tickets out "face down," so to speak, the only thing distinguishing the first and second tickets is the arbitrary order you laid them out in. Since the first two tickets are already destined for the first two students, they must have the same probability of being good, namely $\frac45$. In fact, the same logic applies to all the students.

• Mild edit after a small math-o. – Brian Tung Sep 21 '17 at 20:37
• I see. I calculated ($4/20 * 16/19$) to find out the probability of the first student getting a bad ticked and second one getting a good ticket. Then I guess I wanted to calculate the opposite(first one getting good and second one getting bad) to see which one would have higher probability. I found out that they are equal. Would that be considered a correct solution? – Nick202 Sep 21 '17 at 20:44
• If I understand you correctly, that would work, but it seems pretty indirect to me. The most straightforward solution to me is the one by symmetry. – Brian Tung Sep 21 '17 at 21:20

If you don't look at the first student's ticket, you should do a weighted average of the two possibilities and you will discover that the second student's chance of getting a good ticket is also $0.8$. The fact that one was already drawn does not matter at all. How can the chance be different from if they each drew, then swapped tickets before looking at them?

Let $A_1$ and $A_2$ denote the Bernoulli random variables of student 1 and student 2 picking a good ticket, respectively. So for example, the probababilty that student one picks a good ticket is $P(A_1 = G)$.

First, lets calculate the probability $P(A_1 = G)$. This is a simple case of the hypergeometric distribution, since the good cards form a "sub-population" within the total "population" of cards:

$$P(A_1 = G) = \frac{{16\choose{1}}{4\choose{0}}}{20\choose{1}} = \frac{16}{20} = .8$$

And thus the probability that the first student chooses a bad card is: $P(A_1 = B) = .2$.

Now for second student: Use the law of total probability.

$$P(A_2 = G) = P(A_2 = G | A_1 = G)P(A_1 = G) + P(A_2 = G | A_1 = B)P(A_1 = B)$$

The conditional distribution of $A_2$ given $A_2$ is also hypergeometric.

$$= \frac{{15\choose{1}}{4\choose{0}}}{19\choose{1}}*.8 + \frac{{16\choose{1}}{3\choose{0}}}{19\choose{1}}*.2$$ $$= \frac{15}{19}*.8 + \frac{16}{19}*.2 = .8$$

A-ha! the probability is the same. Go figures.