Topic: The questions, riddles and puzzles thread (Read 38366 times)

Kogut · « **Reply #435 on:** April 24, 2011, 07:34:09 am »

Spoiler: Bad idea, at least it is correct for 1 dice (click to show/hide)

100000 tests
0: 0
1: 5.99858
2: 8.7113
3: 10.5729
4: 11.9307
5: 13.029
6: 13.9192
7: 14.7433
8: 15.3995
9: 16.0117
10: 16.5679

Spoiler: etc (click to show/hide)

Spoiler: test.cpp (click to show/hide)

EDIT: tests to 900

Strife26 · « **Reply #436 on:** April 24, 2011, 08:01:20 am »

Quote from: Max White on April 23, 2011, 07:41:33 am

That is totally cheating.

Then again, I remember once using a program to prove that you should always switch doors when shown a goat.

I totally did the same thing when I first looked at the Monty Hall problem.

ed boy · « **Reply #437 on:** April 24, 2011, 08:03:20 am »

This is one question where inductive logic is a great help. This question is all about conditional expectation.

Think about the case with one dice. Consider the first roll. With the first roll, there is a 1/6 chance of it being eliminated, and a 5/6 chance of it not being eliminated. If it is eliminated on the first roll, then it will have taken one roll to eliminate all your dice. If it not eliminated on the first roll, then the expected number of rolls is one plus the expected number of rolls to eliminate it on the second throw. However, the expected number of throws to eliminate it stays the same on all the throws, so you have.

E(1)=1 * 1/6 + (E(1) + 1) * 5/6
E(1)=6

Now, when you roll two dice, you can either end up with two dice, one dice, or none, etc.

ed boy · « **Reply #438 on:** May 08, 2011, 03:01:25 pm »

There are ten wizards, sitting in a circle. Each is wearing a coloured hat, and can see all nine hats but his own. There are 10 possible hat colours, and any combination of hats is possible, including repeats; so there are 10^10 possible permutations. Each wizard will be called upon to speak one word, which must be the name of one of the colours. They all speak simultaneously, so no information can be transferred between them. If a wizard speaks his own hat colour, he lives; else he dies.

Beforehand, they had a chance to confer, and agree upon a general strategy. Can you think of a strategy they could use that will guarantee that at least one wizard escapes?

Darvi · « **Reply #439 on:** May 08, 2011, 03:03:43 pm »

Always speak the color of the dude to your right. Or left, whatever.

Oh wait, that doesn't work.

Cthulhu · « **Reply #440 on:** May 08, 2011, 03:14:35 pm »

There's no rule that prevents them from just telling each guy what color his friend's hat is beforehand, unless they're not wearing hats.

Aside from that, they all say the same color. At least one of them will make it.

The Merchant Of Menace · « **Reply #441 on:** May 08, 2011, 04:00:44 pm »

Quote from: Cthulhu on May 08, 2011, 03:14:35 pm

Aside from that, they all say the same color. At least one of them will make it.

Repeats are permitted, so this doesn't guarantee the survival of anyone

ed boy · « **Reply #442 on:** May 08, 2011, 04:06:57 pm »

Quote from: Cthulhu on May 08, 2011, 03:14:35 pm

There's no rule that prevents them from just telling each guy what color his friend's hat is beforehand, unless they're not wearing hats.

Aside from that, they all say the same color. At least one of them will make it.

They don't get the hats until after they confer.

They don't know what the possible colours are beforehand. If they all say the same colour, they might pick a colour that is not being used for the hats.

Sowelu · « **Reply #443 on:** May 08, 2011, 04:25:46 pm »

You want to guarantee that at least one wizard escapes?

Everyone agree on a leader. Everyone says the leader's hat's color, except for the leader, who says a color that he DOES NOT see. (There's ten possible colors, so there will always be some color he doesn't see.)

If at least one non-leader wizard has a different colored hat from the leader, that wizard will say the leader's hat's color and survive.

If everyone has the same colored hat, the leader will say a different color and survive.

I'm assuming that they know what colors are -possible- at least.

ed boy · « **Reply #444 on:** May 08, 2011, 04:31:44 pm »

That doesn't guarantee it. That only works when at least one person shares the hat color with the leader. If they all have different color hats, or if everybody apart from the leader has the same color, then that will not work.

I re-examined my source for the question, and it doesn't make it very clear whether or not the wizards know the possible hat colors beforehand. If you have a solution that requires knowing the possilbe colors, then I'll accept that (though a solution that does not rely on knowing the possilbe colors would be better).

ed boy · « **Reply #445 on:** June 09, 2011, 11:50:10 am »

I've got two more.

1: How can you take half of twelve away and get seven?

2: Consider the following dots:

o	o	o
o	o	o
o	o	o

Can you draw four straight lines, with each line (except the first) starting from the end of the previous one, that go through all the dots?

Darvi · « **Reply #446 on:** June 09, 2011, 11:52:27 am »

Easy. Lemme upload a pic first.

Or I just type it.

__
l\/
l/\

Sorta like this.

Spoiler: Despite my crappy drawing, this looks better. (click to show/hide)

Heliman · « **Reply #447 on:** June 09, 2011, 11:59:28 am »

Quote from: ed boy on June 09, 2011, 11:50:10 am

I've got two more.

1: How can you take half of twelve away and get seven?

If you take half of twelve away from thirteen you get seven.(13-6=7)

Strife26 · « **Reply #448 on:** June 09, 2011, 12:03:56 pm »

XII becomes VII when you take half horizontally.

Darvi · « **Reply #449 on:** June 09, 2011, 12:07:02 pm »

^This.

Technically Heliman's answer is also correct but I guess it's not what was meant by the question.

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Author Topic: The questions, riddles and puzzles thread (Read 38366 times)

Kogut

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Strife26

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ed boy

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ed boy

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Darvi

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Cthulhu

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The Merchant Of Menace

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ed boy

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Sowelu

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ed boy

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ed boy

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Darvi

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Heliman

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Strife26

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Darvi

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