Records and covers

Question

Juan is challenging his friend Thomas with the following:

Juan has 5 LP vinyl records, each of them of a different band, and asks Thomas to match them with its respective band name, by showing him only the cover photo, not the names etc.

Thomas will make a first guess and Juan will reveal him the number of correct matchings but without telling him which ones are correct. If not all his matchings are correct, he has the right to do a second guess, by changing as many as he wants.

If he correctly guesses all 5 in these two attempts, then Juan will buy him a nice dinner.

What is the probability for Thomas to win the dinner, in each of the below 3 cases:

1) He can recognize the 3 covers but not the other 2.

2) He can recognize the 2 covers but not the other 3.

3) He knows the matching of 2 covers with 2 of the bands but not which is which. Same also for 2 more covers. For example, he knows that cover1 and cover2 correspond to Band1 and Band2 but not which is which, and also he knows that cover3 and cover4 correspond to Band3 and Band4 but not which is which.

I can easily tell that the requested probability for 1) is 100%: If he knows 123 and he guesses 45, Juan will reply him “3” as the correct number of matchings. Since Thomas knows 123 are correct, he will say 12354 and this will be correct.

For number 3): Thomas must guess 12 or 21, 34 or 43 and he can deduce 5. We have 4 cases but with the 2nd attempt, in essence we have 6:

Suppose the correct choices are 21345:

If he says 21345, Juan replies “5” and we are done.

If he says 12435, Juan replies “1” and we are also done because Thomas knows it is 21345.

If he says 12345, Juan replies “3”. Thomas doesn’t know which are the correct two (he only knows 5). Therefore in the second attempt, he can either say 21345, (correct) or 12435 which is wrong.

Same also for 21435: In the second attempt he can either say 21345 (correct) or 12435 (wrong). So the overall probability is 4/6? Is this correct?

Can you also help me out with the 2nd case? I started examining the cases but it’s very confusing!

Many thanks!

Answer 1

You reasoning for the first case is correct.

In the second case, there are six possible permutations for the three unknown covers, each of which is equally likely to be chosen. We can distinguish three cases:

1. Juan says "5" with probability $\frac{1}{6}$, which means all covers were correctly placed (solved);
2. Juan says "3" with probability $\frac{3}{6}$, which means one of the unknown records was correctly placed. We choose one of the three covers as the correct one and swap the two others, having a probability of $\frac{1}{3}$ that the order is now correct (we chose the right fixed cover);
3. Juan says "2" with probability $\frac{2}{6}$, which means all unknown covers were incorrectly placed. There are two more permutations left to choose from (if we initially chose "345", we can now choose "453" or "534"), each of which is equally likely to be correct.

The probability of ending up with the correct answer thus equals:

$$\frac{1}{6} + \frac{3}{6} \cdot \frac{1}{3} + \frac{2}{6} \cdot \frac{1}{2} = \frac{3}{6} = \frac{1}{2}$$

In the third case, Thomas has four options to choose from, each of which is equally likely to be correct. If he doubts between 1 and 2 and between 3 and 4, he can start by choosing "52341" (i.e., deliberately put the 5 in the wrong place). Then, four answers are possible:

1. "3" - the answer is "12345"
2. "0" - the answer is "21435"
3. "1" - the answer is "12435"
4. "2" - the answer is "21345"

The probability of ending up with the correct answer thus equals $1$.