Math and Logic Puzzles

Empking · Post Post #100 (ISO) » Sun Jul 10, 2011 8:08 am

A,B,C get 33
D gets 0
E gets one?

Kcdaspot · Post Post #101 (ISO) » Sun Jul 10, 2011 8:09 am

so if A doesn't get it right he's over board.

and the pirates won't accept 20 coins a piece?

KageLord · Post Post #102 (ISO) » Sun Jul 10, 2011 8:11 am

Empking wrote:A,B,C get 33
D gets 0
E gets one?

Nope.

Kcdaspot wrote:so if A doesn't get it right he's over board.

and the pirates won't accept 20 coins a piece?

Right, if they reject A's proposal, he goes overboard.

And they very well might accept 20 coins a piece, but remember that every pirate wants the most he can get. So basically, can A do better for himself than 20 coins?

Kcdaspot · Post Post #103 (ISO) » Sun Jul 10, 2011 8:11 am

heere's something.

what if they can't agree and A B C and D get sent overboard and E can't agree with himself?

Kcdaspot · Post Post #104 (ISO) » Sun Jul 10, 2011 8:12 am

A gets 0

BCDE get 25 a peice.

Tragedy · Post Post #105 (ISO) » Sun Jul 10, 2011 8:12 am

Kcdaspot wrote:so if A doesn't get it right he's over board.

and the pirates won't accept 20 coins a piece?

If the amount is equal, it's a fail.

I think it's 45?

10 for A,
9 for B
8 for C
7 for D
6 for E

KageLord · Post Post #106 (ISO) » Sun Jul 10, 2011 8:15 am

Kcdaspot wrote:heere's something.

what if they can't agree and A B C and D get sent overboard and E can't agree with himself?

Remember that everyone still on board gets a vote, including the one that proposed it and a tie means that it is accepted. So if it gets down to D and E, D will vote yes and can get whatever he wants, including 100-0.

Kcdaspot wrote:A gets 0

BCDE get 25 a peice.

No.

Tragedy wrote:
I think it's 45?

10 for A,
9 for B
8 for C
7 for D
6 for E

Well... there are 100 coins total... but that scheme is wrong anyway.

KageLord · Post Post #107 (ISO) » Sun Jul 10, 2011 8:19 am

If you can get this without looking it up, you are very smart indeed.

StrangerCoug · Post Post #108 (ISO) » Sun Jul 10, 2011 8:22 am

20 pieces of gold apiece?

StrangerCoug · Post Post #109 (ISO) » Sun Jul 10, 2011 8:23 am

Damn it, I just realized I outsmarted myself ><

KageLord · Post Post #110 (ISO) » Sun Jul 10, 2011 8:24 am

lol... we can pretend you didn't guess that. << >>

StrangerCoug · Post Post #111 (ISO) » Sun Jul 10, 2011 8:26 am

34 to A, 33 to B, 33 to C, 0 to D, and 0 to E?

KageLord · Post Post #112 (ISO) » Sun Jul 10, 2011 8:26 am

StrangerCoug wrote:34 to A, 33 to B, 33 to C, 0 to D, and 0 to E?

Sadly no.

StrangerCoug · Post Post #113 (ISO) » Sun Jul 10, 2011 8:27 am

That's actually

FOUR

incorrect answers if I'm counting right, so you've beaten us. What was it?

Kcdaspot · Post Post #114 (ISO) » Sun Jul 10, 2011 8:31 am

KL: that was meant to be a joke.

Espeonage · Post Post #115 (ISO) » Sun Jul 10, 2011 8:34 am

I've done this puzzle before. I can't remember how to do it.

there is a formula that can be applied. And there is also another puzzle to do with now many pirates you can have before the first pirate won't survive.

inspiratieloos · Post Post #116 (ISO) » Sun Jul 10, 2011 8:37 am

98, 0, 1, 0, 1

One of my favourites, I actually answered before reading the complete question then checked if it was indeed this one

StrangerCoug · Post Post #117 (ISO) » Sun Jul 10, 2011 8:40 am

StrangerCoug wrote:Each player gets one chance and one chance only to solve the puzzle. If you get it right, you become the next puzzle maker. If three people miss, the puzzle maker wins and posts a new puzzle.

KageLord wrote:
Empking wrote:A,B,C get 33
D gets 0
E gets one?

Nope.

KageLord wrote:
Kcdaspot wrote:A gets 0

BCDE get 25 a peice.

No.

Tragedy wrote:
I think it's 45?

10 for A,
9 for B
8 for C
7 for D
6 for E

Well... there are 100 coins total... but that scheme is wrong anyway.

Three misses, so KageLord couldn't even accept my answer.

inspiratieloos · Post Post #118 (ISO) » Sun Jul 10, 2011 8:43 am

So, Kage gives a new one?
I already knew the answer so it's not like it took any skill.

StrangerCoug · Post Post #119 (ISO) » Sun Jul 10, 2011 8:45 am

Yeah.

KageLord · Post Post #120 (ISO) » Sun Jul 10, 2011 8:45 am

Kcdaspot wrote:KL: that was meant to be a joke.

Ah. Lol hard to tell sarcasm over the computer. >>

Answer:

A - 98
B - 0
C - 1
D - 0
E - 1

Surprising, no? To solve this, it is best to work backwards. If D and E are left alone, the result will be 100 - 0. If C, D, and E are left, C knows that if his plan fails it will be 100 - 0, so E will accept the plan if he is offered anything. Thus it would be 99 - 0 - 1.

If it's B, C, D, and E, B will know the above and knows that D will accept anything since he will get nothing next round. So it is 99 - 0 - 1 - 0. So knowing all of that, A can buy C and E's votes with just 1 coin each since they would get nothing next round. It pays to be older.

Next question (can't think of another good one so here is one you have probably heard many times):

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice and why?

Tragedy · Post Post #121 (ISO) » Sun Jul 10, 2011 8:48 am

Yes, because it's 1/3 chance to get a Car.
Also, Goats produce milk = You get money.

KageLord · Post Post #122 (ISO) » Sun Jul 10, 2011 8:56 am

Tragedy wrote:Yes, because it's 1/3 chance to get a Car.
Also, Goats produce milk = You get money.

Yes, because door no. 2 is a 1/3 chance at a car?

Tragedy · Post Post #123 (ISO) » Sun Jul 10, 2011 9:03 am

Think so.
Change that to 1/2.

StrangerCoug · Post Post #124 (ISO) » Sun Jul 10, 2011 9:06 am

You should switch; door #2 has two chances in three of having the car as opposed to door #1, which has only one chance in three.