Probability theory problem – the order of drawing tickets doesn’t matter

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I found this problem in a textbook. It is given right after the theory about the Bayes' rule and the total probability rule.

Problem: We have an urn with $N$ lottery tickets of which $M \le N $are winning tickets.
$K$ persons $K \le N$ take turns drawing tickets from the urn in order.
Each person draws one ticket.

Prove that each person (no matter of his order number)
has a probability of $M/N$ for drawing a winning ticket.

  • I can prove this statement for persons $1$ and $2$ using the total probability law, but I cannot quite formalize the proof for the $K$-th person. I have the feeling that the total probability law has to be used here.

  • I have this approach in mind which I am not sure if it's rigorous enough. Here it is:

Obviously the probability of each ticket being a winning ticket is $M/N$.
Let's suppose it's person #$K$'s turn to draw and he draws some ticket $A$.

Now we define these events:

$H_1$: ticket A is a winning ticket
$H_2$: ticket A is not a winning ticket
$B$: person #$K$ has drawn a winning ticket

Using the total probability law we get:
$P(B) = P(H_1) P(B|H_1) + P(H_2) P(B|H_2) = ( M/N ) \cdot 1 + ((N-M)/N) \cdot 0 = M/N$

But this solution is weird to me because I feel like
I am already assuming what I need to prove.
I don't know if this approach is valid, is it?

If it's not, how can this problem be solved more rigorously?

  • And finally, I was also thinking of another approach: some sort of induction by K.
    But it didn't lead me anywhere (at least for now).

So… is the above approach valid and if not, what is the best way to solve this problem rigorously (without using any complex apparatus of course, because this problem is in the very beginning of the textbook, only basic things are known so far)?

Best Answer

It can be solved by applying the rule: $$\text{probability=number of favorable outcomes divided by number of possible outcomes}$$This rule works if the outcomes are equiprobable (which is the case below).

It is IMV the best way to prove this problem rigorously.

For convenience number the tickets with $1,2,\dots,N$ and let the tickets with a number $\leq M$ be the winning tickets.

Fix one of the persons that draws a ticket.

For $i=1,2,\dots,N$ let $E_i$ denote the event that ticket $i$ is drawn by this person.

If $i$ and $j$ are distinct ticket numbers then there is no reason at all to think that the ticket with number $i$ has more (or less) chance than the ticket with number $j$ to become the ticket that is drawn by this person.

In short: $$P(E_i)\text{ does not depend on }i\tag1$$

Next to that is for sure that one of the tickets is drawn by the person so that:$$\sum_{i=1}^NP(E_i)=1\tag2$$

From $(1)$ and $(2)$ we conclude for $i=1,\dots,N$ that: $$P(E_i)=\frac1N$$

Then consequently if $E$ denotes the event that the person draws a winning ticket:$$P(E)=P\left(\bigcup_{i=1}^ME_i\right)=\sum_{i=1}^MP(E_i)=\frac{M}N$$

Edit (concerning your efforts):

"Let's suppose it's person #$K$'s turn to draw and he draws some ticket $A$...."

Under that supposition $H_1$ and $B$ (as defined in your question) denote exactly the same event so that directly:$$P(B)=P(H_1)$$ The application of the total law of probability is not incorrect but is redundant (hence confusing).

"Obviously the probability of each ticket being a winning ticket is $M/N$..."

Indeed, and that tells us that: $$P(H_1)=\frac{M}N$$

So your end result is:$$P(B)=\frac{M}{N}$$ which agrees with the answer that I provided above.

IMV your approach is correct but is up to some level tarnished by the redundant use of the total probability law.

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