Probability theory problem - the order of drawing tickets doesn't matter 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)?
 A: 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.
A: When you say you've proved it for players $1$ and $2$, I assume you did something like the following. For player $1$ it's immediate. For player $2$ the probability is
$$
\frac{M}{N}\frac{M-1}{N-1}+\frac{N-M}{N}\frac{M}{N-1}=\frac{M}{N(N-1)}(N-M+M-1)=\frac{M}{N}.
$$
You can do something similar for player $3$. The probability is
\begin{align}
&\frac{M}{N}\frac{M-1}{N-1}\frac{M-2}{N-2}+\frac{M}{N}\frac{N-M}{N-1}\frac{M-1}{N-2}+\frac{N-M}{N}\frac{M}{N-1}\frac{M-1}{N-2}+\frac{N-M}{N}\frac{N-M-1}{N-1}\frac{M}{N-2}\\
&\quad=\frac{M(M-1)}{N(N-1)(N-2)}(N-M+M-2)+\frac{(N-M)M}{N(N-1)(N-2)}(N-M-1+M-1)\\
&\quad=\frac{M(M-1)}{N(N-1)}+\frac{(N-M)M}{N(N-1)},
\end{align}
which you recognize as the probability for player $2$ (already shown to equal $M/N$).
This suggests that your idea of using induction will work. Define events
\begin{align}
&E_{j,l}\ :\ \text{exactly $j$ winning tickets drawn in first $\ell$ draws}\\
&F_\ell\ :\ \text{person $\ell$ draws a winning ticket}
\end{align}
The total probability law implies
$$
\Pr(F_{\ell+1})=\sum_{j=0}^\ell\Pr(F_{\ell+1}\mid E_{j,\ell})\Pr(E_{j,\ell})
=\sum_{j=0}^\ell\frac{M-j}{N-\ell}\Pr(E_{j,\ell}).
$$
The event $E_{j,\ell}$ can be written as the disjoint union
$$
E_{j,\ell} = (F_\ell\cap  E_{j-1,\ell-1})
\cup (F'_\ell \cap E_{j,\ell-1}),
$$
from which we get
\begin{align}
\Pr(E_{j,\ell})&=\Pr(F_\ell\mid E_{j-1,\ell-1})\Pr(E_{j-1,\ell-1})
+\Pr(F'_\ell\mid E_{j,\ell-1})\Pr(E_{j,\ell-1})\\
&=\frac{M-j+1}{N-\ell+1}\Pr(E_{j-1,\ell-1})+\frac{N-M-(\ell-1-j)}{N-\ell+1}\Pr(E_{j,\ell-1}).
\end{align}
Inserting this into the expression we got from the total probability law gives
\begin{align}
\Pr(F_{\ell+1})=&\sum_{j=1}^\ell\frac{M-j}{N-\ell}\frac{M-j+1}{N-\ell+1}\Pr(E_{j-1,\ell-1})+\sum_{j=0}^{\ell-1}\frac{M-j}{N-\ell}\frac{N-M-(\ell-1-j)}{N-\ell+1}\Pr(E_{j,\ell-1})\\
=&\sum_{j=0}^{\ell-1}\left(\frac{M-j-1}{N-\ell}\frac{M-j}{N-\ell+1}\Pr(E_{j,\ell-1}) + \frac{M-j}{N-\ell}\frac{N-M-\ell+1+j}{N-\ell+1}\Pr(E_{j,\ell-1})\right)\\
=&\sum_{j=0}^{\ell-1}\frac{M-j}{N-\ell+1}\Pr(E_{j,\ell-1})=\Pr(F_\ell).
\end{align}
In the first line we adjusted the limits of summation using $\Pr(E_{-1,\ell-1})=\Pr(E_{\ell,\ell-1})=0$. In the second line we shifted the summation index of the first sum by $1$.
A: In such cases, there are two parts of the computation, when being rigurous is an issue. The first part concerns modelling the situation, finding a mathematical framework that best suits the presented situation. (One has to model the case by using only the a priori information. This is already a point where people may start to question the process of making a bet story a probability space with a certain probability and an event to be evaluated.) The second part happens only inside mathematics, here we have less problems.
So i suppose that the mentioned textbook has already at the beginning an idea what is a "simplest" probability space $(\Omega,\mathcal P(\Omega),\Bbb P)$, with a finite set $\Omega$, each of its subsets is measurable, the probability $\Bbb P$ being the uniform one. (I hate textbook where probability is done without any probability space, but rules are immediately proven, and conditional probabilities are defined by a formula, which can be written so only in discrete cases. If this is the case, there is no rigorous approach.)
Let us model the situation.
One "faithful" way to model so that the "story" is parallely mirrored is the following one. We can a priori label the tickets by using numbers from $1$ to $N$ so that the tickets with number $\le M$ are winning. Peoples $P_1,P_2,\dots,P_n$ come now to the urn and extract a ticket in their order. This leads to a result, which is a permutation of the set (with elements) $1,2,\dots,N$. Model the situation using
$$
\Omega=S_N\ ,
$$
the set of permutations of the $N$ tickets. Uniform probability on it.
For some permutation $\sigma$, our person, the person labelled $K$ extracts ticket $\sigma(K)$.
The wanted event is
$$
A(K) = \{\ \sigma\in S_N\ :\ \sigma(K)\le M\ \}
\ .
$$
Now we have a simple combinatorial problem to compute the numbers of elements of $A$. Since the group of permutations $S_N$ actions on $\Omega$ (which is obtained by applying the forgetful functor from groups to sets on $S_N$) in a way compatible with the probability, use this action to see that $A(K)$ and $A(1)$ have the same probability. Now enumerate elements $\sigma$ of $S_N$ as usual. Pick one ticket for the first entry in $\sigma$. From the remained pick one ticket for the second entry in $\sigma$... This is a kind of a tree structure used to enumerate $S_N$. Obviously, the first step is enough to decide if the element $\sigma$ is in $A(1)$ or not. There are $M$ good possibilities among the $N$ possible ramifications at the first step, so the probability is $M/N$.
Alternatively, enumerate $S_N$ by building the tree and taking for the first ramification the value of a "random $\sigma$" in its $K$.th position.

Having the space, it is of course possible to apply Bayes rule as written in the question, and the argument is also valid. (But the counting argument is the direct approach. Since the ingredients in Bayes rule must be also computed by counting.)

A: Since you already know why the probability for first person to draw winning is $M/N$, it remains to show that the probability of winning for any subsequent person is the same as the first one.
To do that, you can consider each individual event as the total outcome of everyone's ticket; and see that the set of events for 1st person winning have the same number of elements as the set of events for any other person winning.

It is ok to order the set of tickets with $T_1,T_2,T_3,...,T_N$ where $T_i$ is a winning ticket iff $i<=M$. And we will consider the tickets as shuffled in the urn. We will actually only talk about the labels (indices) for these tickets. So we may as well think that the tickets are numerals $1,2,3,...,N$ where winning tickets are less than $M$.
Let's label the persons in the order they draw as well: $P_1,P_2,...,P_K$.
An outcome of everyone's ticket is $\{f(P_i)\}_i$ for $i=1,2,...,K$ where $f$ is a permutation on the set of tickets $\{T_i\}_i$.
Then consider the set of outcomes corresponding to person $P_1$ winning, ie. $f(P_1)<=M$ and the same for person $P_k$ winning, ie. $f(P_k)<=M$. Call them set $1$ vs set $k$.
For every such outcome in set $1$, simply swap the ticket between person $1$ and person $k$, you will have an outcome belonging to set $k$.
This swapping has a natural inverse: swapping the tickets back.
So no 2 distinct outcomes in set $1$ can be mapped to the same outcome in set $k$ under swapping. (For any claim that a different outcome in set $1$ gets swapped to the same outcome in set $k$, you swap back and you get an identical outcome in set $1$.)
The image of swapping over set $1$ fully covers set $k$ because, if any outcome in set $k$ was hypothetically missing, you identify an outcome in set $1$ by swapping back from the hypothetical missing outcome in set $k$; swap forward again; the hypothetical missing outcome is in the image of the swap after all.
Since we identified a bijective function between set $1$ and set $k$, ie. swapping the tickets between person $1$ and person $k$, the two sets have equal number of elements.
Thus, you can argue that the two events have equal probability.
Depending on the context, you may satisfy the problem's author by saying: swapping between permutations is bijective and skip some of the concept argument.

As for your current approach, @Henry's comment and @Will Orrick's answer with induction show that it can work. But I suspect that the problem's author has a simpler solution than induction in mind.
