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Machine generates one random integer in range $[0;40)$ on every spin.

You should choose 5 numbers in that range.

Then the machine will spit out 5 numbers (numbers are independent of each other).

what is the probability that you will get exactly two numbers correct?

My logic:

You should get two of them right. chance of that is: $r = left( 1 over 40 right)^ 2 $

You should get 3 of wrong. Chance of that is: $w = left( 39 over 40 right)^3 $

As order doesn't matter answer should be: $$ans = rw over 2!3!$$
Simulator tells me I'm wrong. Where is my logic flawed?

P.s. Machine can spit out duplicates

This is a simple binomial problem.

Note: I am assuming that both you and the machine choose with replacement. That is, either (or both) of you might have duplicates.

For a single choice you make, the probability that the machine also makes it (as one of the $5$ it chooses) is $psi = 1 - left( frac 3940right)^5$.

As your choices are independent, the answer is then $$binom 52times psi^2times (1-psi)^3approx .0967$$

Let's say, you have chosen 5 numbers. Then the odds of the machine generating one of your chosen numbers is $frac540$. This happens twice and it doesn't happen thrice. So this should be $left(frac540right)^2timesleft(frac3540right)^3$. Now this could have happened in $frac5!2! times 3!$ (or $5 choose 2$).

So, is it $left(left(frac540right)^2 times left(frac3540right)^3 times 5 choose 2 right)$?

However, the assumption is that you choose distinct numbers in that range. In case, your strategy is to maximize the "correct" score. If you are picking at random without a strategy, then lulu's answer is the one you are looking for.

Here is the argument if neither you nor the machine can repeat.

Since all the numbers are equally likely let's assume you chose $1,2,3,4,5$.

There are $T = binom405$ total ways the machine can choose its five numbers. Let's count how that choice might overlap with yours.

There are

$$
binom355 text choices with 0 text overlap
$$

$$
binom51binom354 text choices with 1 text overlap
$$
$$
binom52binom353 text choices with 2 text overlap
$$
$$
binom53binom352 text choices with 3 text overlap
$$
$$
binom54binom351 text choices with 4 text overlap
$$
$$
1 text choice with 5 text overlap
$$

To find the probability of exactly $2$ matches, compute that number as above and divide by $T$, To find the probability of at least $2$ matches, add the probabilites of $0$ and $1$ match and subtract from $1$.

There are a few different errors in your calculation.

• If the numbers you choose are $ n_1, n_2, dots, n_5 $ (I'm presuming these have to be all different), and the numbers generated by the machine are $ m_1, m_2, dots, m_5 $, then $ left(frac140right)^2 $ is merely the probability that any specified ordered pair of numbers generated by the machine are equal to some given pair of your chosen numbers (e.g. $ m_1=n_1 $ and $ m_2=n_2 $, for instance). To allow for the fact that the order doesn't matter, you need to multiply, not divide, by $ 2! $—that is, $ mathrmProbleft(leftm_1=n_1, &, m_2=n_2rightlorleftm_2=n_1, &, m_2=n_1rightright)=2left(frac140right)^2 $. But you still haven't counted the cases where some other pair of the numbers chosen by the machine ($ m_4 $ and $ m_5 $, for example) are equal to your $ n_1 $ and $ n_2 $, let alone any other pair of your chosen numbers.


• As clarified in one of your comments, it is possible for more than $ 2 $ of the numbers generated by the machine to match some of those in your hand even when you have only exactly two right. If you choose numbers $ 1,2,3,4,5 $, for instance, and the machine generates the numbers $ 6,7,1,8,2,1 $, then exactly $ 2 $ of your numbers—namely, $ 1 $ and $ 2 $—are among those generated by the machine, but three of those generated by the machine—namely the $ 2 $ and the two $ 1$s—are among those you chose. Also, once the machine has generated two of the numbers you have chosen, it cannot generate any of the remaining $ 3 $ you have chosen without increasing the number you've got correct to more than $ 2 $. Thus, for both of these reasons, $ left(frac3940right)^3 $ isn't the probability of your getting $ 3 $ wrong.

The easiest way I can see of calculating the probability you're looking for is to treat the accumulating number of matches between the numbers generated by the machine and those you have chosen as an inhomogeneous Markov chain. Let $ C $ be the set of numbers you have chosen (which I assume to be all distinct), $ m_1, m_2, dots, m_5 $ the numbers generated by the machine, and $ X_n = leftvert leftm_iright_i=1^ncap Crightvert, X_0=0 $. Then $ leftX_nright_i=1^5 $ is a Markov chain whose states and transitions are represented in the diagram below:

The numbers on the links are the transition probabilities, while the numbers inside the circles are the probabilities of the states being reached. If my arithmetic is correct, the probability that exactly two of your chosen numbers appear amongst those generated by the machine is $ frac81,0911,280,000approx 0.063 $.

If the machine is choosing with replacement, but the person is choosing five distinct numbers, and if success means two distinct numbers the person chose are picked by the machine, then the problem is much more difficult:

Probability machine chooses five distinct numbers:
$$dfrac40!35!40^5$$

Probability machine chooses four distinct numbers (choose the four distinct numbers the machine selected, choose the one number that is repeated twice, permute the multiset to find all possible orders these numbers could be chosen, divide by the total number of ways to choose five numbers):
$$dfracdbinom404dbinom41dfrac5!1!1!1!2!40^5$$

Probability machine chooses three distinct numbers (choose the three distinct numbers the machine selected, either one number is repeated three times or two numbers are repeated twice each):
$$dfracdbinom403dbinom31left(dfrac5!3!1!1!+dfrac5!2!2!1!right)40^5$$

Probability machine chooses two distinct numbers (choose the two distinct numbers the machine selected. You can have one number four times and the other once or one number three times and the other twice.):
$$dfracdbinom402dbinom21left(dfrac5!4!1!+dfrac5!3!2!right)40^5$$

So, the probability of you matching exactly two numbers:
$$beginalign* & dfrac40!35!40^5cdot dfrac5cdot 4cdot 35cdot 34cdot 3340cdot 39cdot 38cdot 37cdot 36\ + & dfracdbinom404dbinom41dfrac5!2!40^5cdot dfrac4cdot 3cdot 36cdot 35cdot 3440cdot 39cdot 38cdot 37cdot 36 \ + & dfracdbinom403dbinom31left(dfrac5!3!+dfrac5!(2!)^2right)40^5cdot dfrac3cdot 2cdot 37cdot 36cdot 3540cdot 39cdot 38cdot 37cdot 36 \ + & dfracdbinom402dbinom21left(dfrac5!4!+dbinom53right)40^5cdot dfrac2cdot 1cdot 38cdot 37cdot 3640cdot 39cdot 38cdot 37cdot 36 approx 0.0091259765625endalign*$$

Note: It is possible for the machine to choose one distinct number, but there is a zero probability that you wind up with two matching numbers, so I ignored that case. To show that my numbers work out though, you can test that the following holds:

$$dfrac40!35!+dbinom404dbinom41dfrac5!2!1!1!1!+dbinom403dbinom31left(dfrac5!3!1!1!+dfrac5!2!2!1!right)+dbinom402dbinom21left(dfrac5!4!1!+dfrac5!3!2!right)+dbinom401 = 40^5$$

(I verified this is true myself using Wolframalpha).