2 Really Hard Questions

[yonki]

Originally posted by Chevy+24 April 2004 - 22:39--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (Chevy @ 24 April 2004 - 22:39)</td></tr><tr><td id='QUOTE'> <!--QuoteBegin-yonki@24 April 2004 - 21:37
I got the final answer for number 2,and dont tell me its wrong,because it isnt&nbsp;

5+3+1=9

5 races again.pick 3 from each.now make 3 races:1sts vs 1sts;2nds vs 2nds and 3rds vs 3rds.from here you get the top 5,and besides you know who is the fastest guy.he doesnt even have to run the last race.

what if the 3 fastest happened to be in the same race at the beginning? [/b][/quote]
thats the worst case and i already thought of that.thats why you need the one last race

[Z-95]

For the second question: one race with all 25 people running. The first 3 to finish will be the 3 fastest.

[muchspl2]

Originally posted by hobbes+24 April 2004 - 16:33--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (hobbes @ 24 April 2004 - 16:33)</td></tr><tr><td id='QUOTE'> <!--QuoteBegin-Chevy@24 April 2004 - 22:23
It has to be worked backwards (rewording of previous post)

If there are 2 left, P2 will definately get all the money leaving p1 with nothing.

Because of this, if there are 3 left, P1 will vote even if he has just one coin (leaving p2 with nothing).

Because of this, if there are 4 left, p2 will vote even if he has just coin (leaving p1 and p3 with nothing).

Therefore, with 5 left, all p5 needs to do is give P1 and p3 a coin each as if you die, they will definately get nothing anyway (it will be settled in the scenario above)




@J&#39;Pol, that is the same as my first post but in the next one I got it down to 11 races by a "top 3 stay in and the last 2 go out and repeat" method.

Chevy, if 3 votes "no", 5 will die.

In the next round, 3 and 4 will split the pot 50/50.

So, I think that it is 98, 0, 0, 1, 1.

Because if 5 dies 4&3 will take all the coins.

Getting 1 is all 1&2 can hope for. [/b][/quote]
this is correct

1 2 3 4 5
5. ? ? ? ? ?
4. 0 1 0 99 -
3. 1 0 99 - -
2. 0 100 - - -
1. 100

Once you see the pattern it becomes very clear. You have to realize that when a pirate&#39;s plan does not succeed then that means you are in the same situation with one less pirate.

1. pirate 1 needs 0 other people to vote for him. so he votes for himself and takes all the money.
2. pirate 2 needs 0 other people to vote for him. so he votes for himself and takes all the money. pirate 1 gets 0.
3. pirate 3 needs 1 other person to vote for him. he gives 1 coin to pirate 1 for his vote - if we are reduced to 2 pirates, pirate 1 gets 0 so pirate 1 knows 1 is better than none. pirate 3 takes 99. pirate 2 gets 0.
4. pirate 4 needs 1 other person to vote for him. he gives 1 coin to pirate 2 - if we reduce to 3 pirates, pirate 2 gets 0 so pirate 2 knows 1 is better than none. pirate 4 takes 99. pirate 3 gets 0. pirate 1 gets 0.
5. pirate 5 needs 2 other people to vote for him. its clear now that the 2 people he needs to convince are the 2 who get shafted in the 4 pirate scenario - pirate 3 and pirate 1. so he can give them each 1 coin (which is better than 0 - what they would get otherwise) and keep 98 for himself.

1 2 3 4 5
1 0 1 0 98

[hobbes]

2nd question : 15 races

I used 4 people per race, breaking it down into 3 groups of 8, the 25th runner ran only in the final race.

[Chevy]

Originally posted by Chevy+24 April 2004 - 22:23--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (Chevy &#064; 24 April 2004 - 22:23)</td></tr><tr><td id='QUOTE'>&nbsp;

Therefore, with 5 left, all p5 needs to do is give P1 and p3 a coin each as if you die, they will definately get nothing anyway (it will be settled in the scenario above)
[/b]

<!--QuoteBegin-muchspl2@24 April 2004 - 21:46

5 needs 2 other people to vote for him. its clear now that the 2 people he needs to convince are the 2 who get shafted in the 4 pirate scenario - pirate 3 and pirate 1. so he can give them each 1 coin (which is better than 0 - what they would get otherwise) and keep 98 for himself.
[/quote]
didn&#39;t you say my answer was wrong before?

[Chevy]

Originally posted by hobbes@24 April 2004 - 21:52
2nd question : 15 races

I used 4 people per race, breaking it down into 3 groups of 8, the 25th runner ran only in the final race.

If you give a runner a bye in the semi it can be done in 12, if it&#39;s a "top 3 stay in the other 2 get knocked out and repeat" it can be done in 11.


(is no-one reading other answers?)

[muchspl2]

it less than 11 races

[yonki]

why is 9 wrong?

[hobbes]

I think is 9 as well.

[muchspl2]

because its not right
I&#39;d rather wait before giving away the anwser