How many rings can you guess? -Riddle

[Andreas C]

Inspired by micromass' statistics challenge, I hope you enjoy it!

Recently your financial situation has been less than optimal, and you are left with no other choice but to beg for money in the square in front of the king's palace. However, begging is strictly forbidden in that particular square, and, unfortunately, you are caught. The usual punishment is 20 lashes, but the king gives you a grace. He gives you the option to be let go unharmed, or solve a puzzle, with a reward, but also a punishment if you don't manage to solve it. Intrigued, you chose to accept it.

The King's Puzzle: The king will hide 5 identical diamond and 5 identical amethyst rings in 10 boxes. Both types are precious, but diamond rings worth more than amethyst ones. Your task is to guess which box contains which type of ring. Once you make a guess, you don't see if you guessed right or wrong, you have to move on to the next box. If you guess either 0, 5 or 10 rings right, you will be sold as a slave (which is unfortunate). If, however, you guess exactly 8 rings right, since the king's birthday is on the 8th of August, you will be gifted all of the rings that you guessed right. If you guess any other number of rings right, you won't be awarded anything, but you won't be punished either.

Is there something that you can do to maximize your chances of getting mostly diamond rings (I forgot to mention that you're so greedy, amethyst rings just aren't good enough), or even any rings at all, but also minimize the chances of you being sold as a slave? Justify your answer.

Optional challenge 1: Say now that the king retains this 0-5-10 = punishment, 8 = reward rule, but changes the number of rings? For example, what if he gives you instead 10 diamond and 10 amethyst rings to guess? Can you generalize the rule?

Optional challenge 2: Are there numbers of rings to guess would you chose not to waste your precious time (which you could spend wandering around with no purpose) with, if the punishment-reward rule remains unchanged? Is there a number of rings that would maximize your chances of getting precisely 8 rings right?

Happy solving! If you enjoyed this puzzle, I will try to come up with a "sequel", and also try not to jump the shark with it, like so many movies these days

 

[mfb]

Do we have to guess all at once, or do we see the ring in the first guessed box before we make the next guess?

 

[Andreas C]

Do we have to guess all at once, or do we see the ring in the first guessed box before we make the next guess?

No, you have to guess all at once. I should specify it in the original post, thanks for pointing this out! It would still make an interesting riddle though.

 

[mfb]

Okay, then I can get 0% risk of getting sold as slave, and 1/12 probability to get 5 diamond and 3 amethyst rings. Not sure if the probability is ideal, but I think it is, and the result is ideal for sure because you cannot get more than that.

Spoiler

Guess 7 boxes as "diamonds" and 3 as "amethyst".

You get at least two diamond guesses wrong and at least two right, so 0 and 10 correct guesses are out. There is also no way to get exactly 5 guesses right, because every wrong amethyst guess also gives a wrong diamond guess, which means you are always right for an even number of guesses. You are never sold as slave.

You get exactly 8 guesses right if the three amethyst guesses are right, for a probability of 5/10*4/9*3/8=1/12.

Edit: How does optional challenge 1 work? Does the king pick 10 out of the (e.g.) 20 rings at random? Optional challenge 2 also depends on that question.

 

[Andreas C]

Okay, then I can get 0% risk of getting sold as slave, and 1/12 probability to get 5 diamond and 3 amethyst rings. Not sure if the probability is ideal, but I think it is, and the result is ideal for sure because you cannot get more than that.

Spoiler

Guess 7 boxes as "diamonds" and 3 as "amethyst".

You get at least two diamond guesses wrong and at least two right, so 0 and 10 correct guesses are out. There is also no way to get exactly 5 guesses right, because every wrong amethyst guess also gives a wrong diamond guess, which means you are always right for an even number of guesses. You are never sold as slave.

You get exactly 8 guesses right if the three amethyst guesses are right, for a probability of 5/10*4/9*3/8=1/12.

Spot on! There is a way to increase your chances though (not going to say what), but you risk being sold as a slave. Now only the optional challenges remain unsolved. Optional challenge 2 should be easy enough, Optional challenge 1 is a little bit trickier.

 

[mfb]

You can increase your chance to win to 25/252 or nearly 10%, but with a 1/126 risk to get sold as slave, sure.

Spoiler

Guess diamond 5 times and amethyst 5 times

For the optional challenges, see my edit in the previous post: how are they defined?

 

[Andreas C]

You can increase your chance to win to 25/252 or nearly 10%, but with a 1/126 risk to get sold as slave, sure.

Spoiler

Guess diamond 5 times and amethyst 5 times

For the optional challenges, see my edit in the previous post: how are they defined?

Optional challenge 1: Instead of showing you 5 boxes containing rings of each type, he shows you 10 of each type (a total of 20). You have to make a guess for each one of them, just like in the case of 5 boxes of each. The second part of optional challenge 2 should be very easy to solve when defined like that, the first is just a little bit trickier, but generalizing the rule in optional challenge 1 should be just a bit harder, depending on your knowledge.

 

[mfb]

Ah, more boxes then. Well, you can evaluate the hypergeometric distribution for all those cases, and see which one gives the best combination of "avoid 0 or 10 but aim for 8". Things quickly get dangerous as the difference between 8 and 10 becomes less significant.

 

[Andreas C]

Ah, more boxes then. Well, you can evaluate the hypergeometric distribution for all those cases, and see which one gives the best combination of "avoid 0 or 10 but aim for 8". Things quickly get dangerous as the difference between 8 and 10 becomes less significant.

Indeed they do. In fact, after a certain number (not too hard to figure out), it's impossible to have any chances of getting 8 rings right without also having chances of getting 10 rings right. And then there's an optimal number (even easier to figure out) where you have the maximum chances of getting them right.

 

[Andreas C]

Ok, it looks like so far, the main question has been answered, the optional challenges remain unanswered. Anyway, I have come up with the next puzzle (hint: it involves dices), and will probably be posting it tomorrow (I haven't yet solved it in its entirety, because some of the questions are a bit... too hard for me...)!