Probability of double matchings

Question

Yesterday, I solved this problem question from probability theory: Two similar decks of $N$ distinct cards are matched against a similar target deck. Find the probability of exactly $m \leq N$ matches.

I proceeded in the following manner. Let $A_i$ denote the event that $i^{\text{th}}$ card is matched (from both the decks) against the target deck. Therefore, let $$P(\text{E = exactly $m$ match occurs})(N,m)$$ (don't mind the bad notation please) then

$$P(N,m) = S_m - \binom{m+1}{m}S_{m+1} + \binom{m+2}{m}S_{m+2} + \ldots + \binom{N}{m} S_{N}$$

where $S_1 = \sum_{1\leq i \leq N} P(A_i)$, $S_2 = \sum_{1\leq i \lt j \leq N} P(A_i \cap A_j) \ldots$

Clearly, we have $$S_{m+k} = \binom{N}{m+k} \frac{(N-m-k)!^2}{N!^2}$$ Therefore, \begin{align*} P(N,m) &= \sum_{k=0}^{N-m} (-1)^k \binom{m+k}{m} \binom{N}{m+k} \frac{(N-m-k)!^2}{N!^2} \\ &= \frac{1}{m!} \frac{1}{N!} \sum_{k=0}^{N-m} (-1)^k \frac{(N-m-k)!}{k!} \end{align*}

After obtaining the above expression, I thought if there exists some nice closed formula for the series. So I plugged it on W|A but it doesn't returns one (in terms of elementary functions).

Next, I started wondering, how does this probability function behaves as $N \rightarrow \infty$. Because this limit might be actual translation of some real world phenomena (although that is something to be considered about, later on).

So, I first tried to check for $m=0$

\begin{align*} \lim_{N \rightarrow \infty} P(N,0) &= \lim_{N \rightarrow \infty} \frac{1}{N!} \sum_{k=0}^{N} (-1)^k \frac{(N-k)!}{k!} \\ &= \lim_{N \rightarrow \infty} \left(1 - \frac{1}{N} + \frac{1}{2!}\frac{1}{N(N-1)} - \ldots \right) \end{align*}

It doesn't strikes me on how to solve this limit, as I cannot evaluate the limit pointwise since it is an infinite sum. So I thought of setting up a recurrence (which may help?). This is what I found:

$$P(N+2,0) = P(N+1,0) + \frac{P(N,0)}{(N+1)(N+2)} + \frac{(-1)^{N}}{(N+2)!^2}$$

But again, I still couldn't figure out much. I even expressed this as an integral (just because sometimes, it does help) and then tried to do some manipulations, but still no clue

$$P(N,0) = (N+1) \int_0^1 \sum_{k=0}^N\frac{ t^k(1-t)^{N-k}}{k!^2} \mathrm{d}t$$

So these are the questions, I am trying to find a solution to:

1. Is there any nice closed form for the expression?

2. How does the probability function, which I derived, behaves when $N \rightarrow \infty$ for a fixed $m$?

3. What happens as $N \rightarrow \infty$ and $m \rightarrow \infty$?

Any help would be appreciated.

Edit 1: I figured out that $P(N,0) \rightarrow 1$ as $n \rightarrow \infty$ with some computations but I guess, it still requires a rigorous proof.

Answer 1

Let $X_k = \mathbf{1}_{A_k}$, $k=1,\dots,N$, be the variable indicating a (double) match in the $k$th card so that $X = X_1+\dots + X_N$ is the number of matches. By linearity, $$ \mathbb{E}[X] = \sum_{k=1}^N \mathbb{E}[X_k] = N\cdot \frac{1}{N^2} = \frac{1}{N}. $$ By the Markov inequality, $\mathbb{P}(X\ge 1) \le \mathbb{E}[X] \to 0$, $N\to\infty$. Therefore, $P(N,0)\to 1$, $N\to\infty$.

Answer 2

I don't see how to get more nice form of the expression, but I'll try to answer questions 2 and 3. So we have this: $$P(N,m)=\frac{1}{m!} \frac{1}{N!} \sum_{k=0}^{N-m} (-1)^k \frac{(N-m-k)!}{k!}$$ Note that the absolute value of the term of the sum decreases with growth of $k$. But if we have sum of the form $$A_n=a_0-a_1+a_2-a_3+\dots+(-1)^na_n$$ with $$a_0>a_1>\dots>a_n>0$$ then this sum clearly lies between $a_0-a_1$ and $a_0$. To prove this, just note that $A_{i+2}$ lies between $A_i$ and $A_{i+1}$ and use induction to show that $A_i$ lies between $A_0=a_0$ and $A_1=a_0-a_1$ then.

For $m=0$ we have $A_1=1-1/N\le P(N,m)\le1=A_0$, which means that the limit is $1$. All limits for $m>0$ then are zeros, because we are working with probabilities of disjoint events. However, it is easy to see this explicitly, because in these cases $A_0\le 1/N$. The same argument applies if both $N\to\infty$ and $m\to\infty$: it doesn't really matter, because as soon as $m>0$ we obtain $P(N,m)\le 1/N$, thus $P(N,m)\to 0$

Answer 3

As you correctly noted, the limit

$$\lim_{ N \rightarrow \infty} P(N,m )= \lim_{ N \rightarrow \infty} \frac{1}{m!} \frac{1}{N!} \sum_{k=0}^{N-m} (-1)^k \frac{(N-m-k)!}{k!}$$

for $m=0 \,$ reduces to

$$ \lim_{N \rightarrow \infty} \frac{1}{N!} \sum_{k=0}^{N} (-1)^k \frac{(N-k)!}{k!} =\lim_{N \rightarrow \infty} \left( 1 - \frac{1}{N} + \frac{1}{2!}\frac{1}{N(N-1)}- \frac{1}{3!}\frac{1}{N(N-1)(N-2)} ... \right)$$

Note that this sum of $N+1\,$ terms tends to $1$ as $N \rightarrow \infty \,\,$. This can be shown in several ways. For example, if we group all terms starting from that in the third position in $\lfloor (N-1)/2 \rfloor \,\,$ consecutive pairs (we can neglect the last term if $N -1\,$ is odd) , these give a (positive) sum $S $ of the form

$$ S= \frac {3!(N-2) -2!) (N-3)!}{2!3! N!} + \frac {5!(N-5) -4!) (N-5)!}{4!5! N!} ... $$

which satisfies

$$S= \sum_{\substack {k=3 \\ k \text {odd}}}^{N} \frac {\left( k!(N-k+1) -(k-1)! \right) (N-k)!}{(k-1)!k! N!} \\ < \sum_{\substack {k=3 \\ k \text {odd}}}^{N} \frac{(N-k+1)(N-k)!}{(k-1)! N!} \\ = \sum_{\substack {k=3 \\ k \text {odd}}}^{N} \frac{(N-k+1)!}{(k-1)! N!} \\ < \frac{ \lfloor (N-2 )/2 \rfloor }{2!N(N-1)} $$

where the last step takes into account that the terms of the sum in second to last expression progressively decreases (and then all terms after the first one are smaller than it). The last expression clearly tends to zero for $N \rightarrow \infty \,\,$, implying that $S $ converges to $0$ as well. So the initial limit reduces to

$$ \lim_{N \rightarrow \infty} \left ( 1 - \frac{1}{N} \right)=1 $$

Now let us consider the case $m>0\,$. For $m=1\,\,$, the limit of the sum reduces to

$$= \lim_{N \rightarrow \infty} P(N,1) = \\ \lim_{N \rightarrow \infty} \frac{1}{N!} \sum_{k=0}^{N-1} (-1)^k \frac{(N-1-k)!}{k!} \\ = \lim_{N \rightarrow \infty} \left(\frac {(N-1)!}{N}- \frac{(N-2)!}{N!} + \frac{(N-3)!}{2! N!} - \frac{(N-4)!}{3! N!}\ldots \right)$$

For $m=2\,\,$, the limit of the sum reduces to

$$= \lim_{N \rightarrow \infty} P(N,2) = \\ \lim_{N \rightarrow \infty} \frac {1}{2!} \frac{1}{N!} \sum_{k=0}^{N-2} (-1)^k \frac{(N-2-k)!}{k!} \\ = \lim_{N \rightarrow \infty} \frac {1}{2!} \left(\frac {(N-2)!}{N}- \frac{(N-3)!}{N!} + \frac{(N-4)!}{2! N!} - \frac{(N-5)!}{3! N!}\ldots \right)$$

and so on. In the general case we get

$$= \lim_{N \rightarrow \infty} P(N,m) = \lim_{N \rightarrow \infty} \frac{1}{m!} \frac{1}{N!} \sum_{k=0}^{N-m} (-1)^k \frac{(N-m-k)!}{k!} \\ = \lim_{N \rightarrow \infty} \frac {1!}{m!} \left(\frac {(N-m)!}{N!}- \frac{(N-m-1)!}{N!} + \frac{(N-m-2)!}{2!N!}-\frac{(N-m-3)!}{3!N!} + \ldots \right)$$

To study this limit, we can procede again as above, by grouping the terms, starting from the third, in $\lfloor (N-m-2) /2 \rfloor $ consecutive pairs. We obtain a positive sum $S'$ of the form

$$ S'= \frac {3!(N-m-2) -2!) (N-m-3)!}{2!3! N!} + \frac {5!(N-m-4) -4!) (N-m-5)!}{4!5! N!} ... $$

which satisfies

$$S'= \small{ \sum_{\substack {k=3 \\ k \text {odd}}}^{N-m} \frac {\left( k!(N-m-k+1) -(k-1)! \right) (N-m-k)!}{(k-1)!k! N!} \\ < \sum_{\substack {k=3 \\ k \text {odd}}}^{N-m} \frac{(N-m-k+1)(N-m-k)!}{(k-1)! N!} \\ = \sum_{\substack {k=3 \\ k \text {odd}}}^{N-m} \frac{(N-m-k+1)!}{(k-1)! N!} \\ < \frac{ \lfloor (N-m-2)/2 \rfloor (N-m-2)! }{2!N!} } $$

where again the last step takes into account that, in the sum shown in the second to last step, all terms after the first one (i.e., that corresponding to $k=3$) are smaller than it. As above, the last expression clearly tends to zero for $N \rightarrow \infty \,\,$, and this implies that $S' $ converges to $0$ as well. So the initial limit for the general case reduces to

$$ \lim_{N \rightarrow \infty} \frac {1!}{m!} \left(\frac {(N-m)!}{N!}- \frac{(N-m-1)!}{N!} \right) $$

which, for $m \geq 1 \,\,$, converges to $0$.

Therefore, we can conclude that

$$\lim_{N \rightarrow \infty} P (N,m) = \begin{cases} 1 \,\,\,\, \text{if $m=0$} \\ 0 \,\,\,\, \text{if $1 \leq m \leq N$} \end{cases}$$

This can be interpreted as follows: for $N \rightarrow \infty\,\, $ , the probability of getting any number of matches from our two decks tends to zero, so we have probability $1$ to get no matches. A calculation by WA for the case $m=0\,\,$, performed by setting $N=1000\,\,$, is shown here. Similar calculations for the cases $m=1$ and $m=2$ are shown here and here. The values provided by WA confirm the results obtained above (unfortunately, WA seems to have some problem to calculate the sums for values of $N>1000$).

Lastly, regarding the first question, in my opinion there are no particular reasons to think that a closed form necessarily exists.