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14 replies to this topic

### #1 unreality

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Posted 17 February 2008 - 12:26 AM

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)
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Posted 17 February 2008 - 01:20 AM

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
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### #3 Jkyle1980

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Posted 17 February 2008 - 03:34 AM

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.
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### #4 unreality

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Posted 17 February 2008 - 03:48 PM

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 )

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Posted 17 February 2008 - 06:07 PM

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
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### #6 Jkyle1980

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Posted 17 February 2008 - 07:14 PM

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

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Posted 17 February 2008 - 08:46 PM

Spoiler for double q and p

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Vidi vici veni.

### #8 bonanova

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Posted 18 February 2008 - 04:43 AM

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.
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Vidi vici veni.

### #9 storm

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Posted 18 February 2008 - 05:27 AM

QPKPQ
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### #10 MysteryKidakaJohnDoe

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Posted 18 February 2008 - 09:48 AM

Maybe KPQPQ? and if not that: QPQPK
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