Guessing Haircuts

The last prisoner looks at the people in front of him, and guesses "bad" if there are an odd number of people with bad haircut or guesses "good" if there are an even number of people in front of him with bad haircut.

Next, the second last person sees if there is odd or even number of people with bad haircut in fromt of him, and from this, he can guess his own hair.

Repeating until everyone guessed, everyone in fromt of the last person will have 100% chance of getting it right.

The last prisoner however, only has a 50% chance of getting it right.

1^5×0.5×100%= 50%

Answer is 50%

BUT WHAT IF THEY HAVE AN AVERAGE HAIRCUT

Its simple, we just need to assume that have 3 with bad haircuts and 3 with good haircuts (probability 50%). Them everyone will know your own haircut. Example: The last one will see 3 bad and 2 good, so he is good, the next will remember that his friend said good and will see 2 or 1 good, if he saw 1 good, he is good, if he saw 2 good, he is bad. The next one will see 1 or 0 good (if his friend - the second one - said "good") and will do your logical choice. Its easy, see what is there, remember and count. Sry if my english is bad, i just speak portuguese. But i thik this is the best strategy, it doesnt matter if five people are sure (without assuming half percent) in the end everyone needs to do the right choice, so choice the Math!

The chances of getting a good or bad hair cut are not affected by the quality of the others. 50% every time. Just like roulette (without the zero). Seeing 1,2, 3, 4 or 5 bad haircuts in front of you tells you nothing about your own. So you are 50:50. Just like scoring red or black is not affected by any number of times the red or black has been previously. The number of bad or good haircuts is infinite.

I still can't understand all the proposed solutions here which involves assuming "even" and "odd" number of good and bad haircuts. Since the question is "what.... and how likely..." in percentage form, so I simply put 50 as the answer. That is bcos the possible answers are only two: they might all correctly guess or they might be wrong. But I'm aware my method could be void if the given details are different.

If the problem is :

Six prisoners standing in a line , 3 are given either a Good haircut and another 3 a Bad haircut, and each prisoner can only see the haircuts of the prisoners in front of them.

Starting from the back of the line, each prisoner must guess her own hair situation, and they will be given prize if and only if they ALL guess correctly. If they can strategize in advance, then what is the optimal strategy and how likely is it to work? Give answer in %.

Then the solution be :

The last prisoner looks at the people in front of him, and guesses "bad" if there are an odd number of people with bad haircut or guesses "good" if there are an even number of people in front of him with bad haircut.

Next, the second last person sees if there is odd or even number of people with bad haircut in front of him, and from this, he can guess his own hair.

Repeating until everyone guessed, the first has a probability 0% and last has 100% . And middle man's so on. Averaging them we can get the chance 50% .

Answer is 50%

It's 50% because you assume the best strategy will give a 100% chance of the following 5 people to get theirs right meaning the first one has a 50% chance of getting his/hers right.

Why cannot the strategy be much simpler? Like, every prisoner, along with guessing their own haircuts, also tells what is the type of haircut of the person just in front of him/her. That way all prisoners from 1 to 5 can know what their haircuts are (since the process starts from the 6th prisoner) with 100% probability. Only the 6th prisoner has 50% probability of getting it right, and hence the answer (50%) follows.

General Speaking,

There is 50 % Chance to guess since we have two choices Good or Bad.