[unreality]

You chose to face KQK in your last zarball competition for freedom. Unfortunately, you lost the second game against the Queen! It's okay though, the King loves playing zarball, you'll get another chance- and you did, when the Prince came to the castle for a visit.

A good thing (for you) is that you can beat the Prince every single time you play him, no matter what. Yeah, he's that bad. You have a 1/2 chance to beat his mother, the Queen, and only a 1/4 chance to beat the King.

The King comes to your cell with the challenge: you will play 5 games of zarball against the royal family (which is the King, Queen and Prince). You can play them in any order you so choose, though none of the family members can play more than twice total, and none of them can play twice in a row either.

If you win the first, fourth and fifth game, you will go free.

If you win both the second and third game, you will go free.

If you win both the third and fourth game, you will go free.

(You could get two of those combinations, or even all three, and still go free. You just have to succeed at one of those combinations)

Maybe it's not the best choice probability-wise, but you don't want to put a higher chance in any of the three combinations, to put more eggs in one basket, so to speak. You want your chances to be equal for all three. So how should you have the King arrange the games if you want the highest equal chance for all three combinations for going free? (ie, an example would be to have a 1/16 chance to win games 1,4 and 5, a 1/16 chance to win games 2 and 3, and a 1/16 chance to win games 3 and 4... though I can tell you right now 1/16 isn't the highest possible)

[TwoaDay]

my guess for 1st 4th and 5th is pkqpq

2 and 3- kpqpq

3 and 4- kqpqp

somebody else will probably figure out the actual probabilitys and the right answers but quickly looking at it this is what i would do

[Guest]

Playing in the following order will give you a 1/8 chance of winning all three of the combinations:

JKQKQ.

However, by using JKQJQ, you increase your odds of being release through the 1st and 3rd situations to 1/2 without affecting your odds of the 2nd situation. This is the order I would use.

[unreality]

Me too, but nobody has gotten the correct solution yet. (And twoaday was in three different universes it seemed lol, cuz he gave three dif answers, none of them working).

And to whoever changed the topic title, thanks! (though I meant "Five Games of Zarball" instead of "Four Games of Zarball", Just "Five" is fine too lol )

The answer:

The order you should arrange the games is:

QPKPQ

Yep, contrary to the other one, this time, King is in the dead center, with the princes surrounding him, and the queens on the outside.

Chances of Success:

1,4,5: 1/2 * 1/2 = 1/4

2,3: 1/1 * 1/4 = 1/4

3,4: 1/1 * 1/4 = 1/4

That can be proved to be the highest, because 1/2 for all three is impossible, due to that you would need a P and a Q for all of them, and for 1,4,5 you would need two P and a Q, and with the constraints given, it's impossible to place those all correctly, with all princes and queens.

[TwoaDay]

o i get it

i was just doin one choice for each way to get free

i guess i didnt really understand the question

i get it now that i saw the answer

if you have anymore of these you should post them, there fun

[Guest]

Take the question a different way. Which combination gives you the best overall odds of winning one of the three?

[bonanova]

call the probability of winning the five matches respectively a b c d e.

for equal odds, ade = bc = cd.

thus b=d and ae=c [suggests a and e are Q and c is K] then b and d are P

Q P K P Q and the probabilities are all 0.25

[bonanova]

Take the question a different way. Which combination gives you the best overall odds of winning one of the three?

Play the Prince twice - either of the two-game matches.

[Guest]

QPKPQ

[Guest]

Maybe KPQPQ? and if not that: QPQPK

[unreality]

Storm and Bonanova got it

MysteryKidakah, try writing out the five letters and looking at the probabilities for all three. I'll give you a hint: the winning probability for the right answer is 1/4, or 0.25, for all three combinations

btw bonanova is wrong about jkyle's followup question's answers- nobody can be played twice in a row. Making a 1/1 chance impossible for any of the three combos.

Here's a follow-up question:

QPKPQ may give you 1/4 chance for all three combos, but PKQPQ gives you:

* a 1/2 chance for 145

* a 1/8 chance for 23

* a 1/2 chance for 34

the questions:

a) which is the best choice? QPKPQ or PKQPQ?

b) the second one has two that are twice as good, and one that is half as good... does that mean it's twice as good?

c) add up the chances- the first one gives 3/4, or 6/8, while the second one gives 9/8... so is it twice as good or 1.5x as good?

[Guest]

Storm and Bonanova got it

MysteryKidakah, try writing out the five letters and looking at the probabilities for all three. I'll give you a hint: the winning probability for the right answer is 1/4, or 0.25, for all three combinations

btw bonanova is wrong about jkyle's followup question's answers- nobody can be played twice in a row. Making a 1/1 chance impossible for any of the three combos.

Here's a follow-up question:

QPKPQ may give you 1/4 chance for all three combos, but PKQPQ gives you:

* a 1/2 chance for 145

* a 1/8 chance for 23

* a 1/2 chance for 34

the questions:

a) which is the best choice? QPKPQ or PKQPQ?

b) the second one has two that are twice as good, and one that is half as good... does that mean it's twice as good?

c) add up the chances- the first one gives 3/4, or 6/8, while the second one gives 9/8... so is it twice as good or 1.5x as good?

From the numbers, I would say that the option that gives you two 50% chances is the best. However, I never got that far in statistics and want to know how you figure odds that add up to 9/8 or over 100%. The same with rolling a die 6 times. The odds of getting any number add up to 100%, but I know this isn't true.

[unreality]

Yeah me too... I think it just means HIGHLY LIKELY... but out of 8 dice rolls, how can you get 9 sixes? etc. It's confuding... like when something has a 200% chance of happening... wtf? lol. I never said I knew the answer to my follow up questions hehe. Though I'm with you in that the 1/2, 1/2, 1/8 is better than the 1/4, 1/4, 1/4

[unreality]

btw, about rolling 6 six-sided die and seeing what your chances are of getting a six, look at the probability of NOT getting a 6. Ie, the probably of getting at least one six after 6 rolls is:

p = 1 - ((5/6)^6)

or add up all the numbers in the binomial distribution except the last one in the line

[Guest]

Yeah me too... I think it just means HIGHLY LIKELY... but out of 8 dice rolls, how can you get 9 sixes? etc. It's confuding... like when something has a 200% chance of happening... wtf? lol. I never said I knew the answer to my follow up questions hehe. Though I'm with you in that the 1/2, 1/2, 1/8 is better than the 1/4, 1/4, 1/4

btw, about rolling 6 six-sided die and seeing what your chances are of getting a six, look at the probability of NOT getting a 6. Ie, the probably of getting at least one six after 6 rolls is:

p = 1 - ((5/6)^6)

or add up all the numbers in the binomial distribution except the last one in the line

Actually, the first case is exactly like the second. Your chances of winning when all three are playing are not 1/4 + 1/4 + 1/4, they are 1 - (3/4)^3 = 0.578125. Same thing with the other setup: 1 - (1/2)*(1/2)*(7/8) = 0.78125. Two 1/2 chances and one 1/8 chance are significantly better odds for you.