Test your IQ with this situation:

There is a mute who wants to buy a toothbrush.

By imitating the action of brushing one's teeth he successfully expresses himself to the shopkeeper and the purchase is done.

Now if there is a blind man who wishes to buy a pair of sunglasses, how should he express himself?

Think and then scroll down for the answer.

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He just has to open his mouth and ask, so simple.

On a side notice, sorry if I post too much... just got internet connection after 3 months of absence

Ahh riddle time.

On your travels you come upon a door. A plaque on the doors reads "one and only one speaks truth".
As you enter the room the door magically disappear behind you. In the room you see 4 creatures; an elf, a human, an orc and a dwarf. Behind them you can see two more doors. One of pure silver and one of gold. They start to speak
Human: "Yesterday Bulwar told me the silver door leads to safety."
Orc: "The Elf always speaks truth."
Elf: "I am not Bulwar."
Dwarf: "Of course not. I am Bulwar."
And as they finish they also vanish, leaving you alone to ponder upon which door to pick.
Which door do you pick and why?

[ 06-21-2005, 07:18 PM: Message edited by: mad=dog ]

heh almost had me... I thought about it for 2 seconds, and then hey, thats a trick question... [img]smile.gif[/img]

Anyway here is a related joke.

A blind man enters a hardware store, with his seeing eye dog, he picks up the dog and starts swinging hime around. The store clerk goes to him and asks, "can I help you?"
"No we are just looking around".

[img]smile.gif[/img]

You take the gold door. If the human is telling the truth, the person who told him to take the silver door was lying. The Orc can't be telling the truth because there is only one honest person; if the orc was telling the truth, the elf would also be telling the truth. The elf can't be telling the truth because that would make the orc's statement true; see above. If the dwarf is telling the truth, that still means that the human lied about their prior conversation and so you should do the opposite of whatever the human says Bulwar said. Ergo... the gold door.

Hey I didn't go through that entire process of thinking but I DID choose the gold door because it was the only logical choice :D

Oh yeah, the blind man. He may ask for sunglasses, however he does nor require them in the slightest.

[ 06-21-2005, 08:15 PM: Message edited by: Kynaeus ]

he might be buying them for someone else [img]tongue.gif[/img]

Precisely. Or to elaborate
If the Orc is telling the truth so is the Elf, which cannot be true. Ergo the Orc lies.
If the Orcs statement is a lie we know that the Elf is lying. Ergo is the Elf is named Bulwar.
If the Elf is Bulwar the Dwarf is lying.
Leaving no alternatives the Humans statement is true. However (and that is the trick) the answer is still the gold door. Bulwar is a lying Elf remember [img]smile.gif[/img] . Illuminas "short-cut" is perfectly valid - you don't need to know who Bulwar is.

A little twist on an age-old riddle.

What is the logic to the following series of numbers?

(4) (14) (11,14) (31,14) (13,21,14) (31,13,12,14) (23,41,12,14)

This one should be simple. All you have to do is think about it for second, or if you played a certain game you may also have figured it out.

[ 06-21-2005, 09:16 PM: Message edited by: Firestormalpha ]

22,13,24,31

For a moment I thought it was 12,13,14,21,12,11,14, but the the sixth term proved that wrong.

A bit like the orc one:

YOu are lost in the wood and there are two paths, one leads to safety and oen further into the wood. There are two twins, only one of which speaks the truth, what question should you ask in order to know which is the right path?

which direction would the other guy tell me to go in, and then go the other direction.

Yes that is the one mine is a variation of. Rather than simply give the answer I'll give the analytical reasoning you need to apply to reach the answer. You know you will either get a lie or truth from either source. You know that each source is consistent to this end. So your only option is to guarantee that you will get both a lie and a true statement.

Another one that tricks the mind into jumping to a conclusion prematurely:

A guest in a price show is shown three closed lockets. He is informed that only one of the lockets contain the price and is asked to make a pick. After he has chosen the host opens one of the other lockets and reveals it to be empty. He then asks the guest if he wants to change his choice.
Which of the following is true
1) It is smartest to stick with his initial choice
2) It is smartest to switch his choice
3) It doesn't matter. The chance is equal

it is smartest for him to stick with his initial choice, because he already knows that the one the host has is empty, thus he does not want it.

You want to switch your choice.
When you first pick you have a 1/3 chance of picking the correct locket. Once one of the two other lockets is opened you still have that 1/3 chance of picking the correct locket, so you know that the other locket has a 2/3 chance of containing the prize.

Once the host opens one to reveal the empty locket the # of choices becomes 2, rather than one, each locket has the same probability of being the one with the price, in this case 50%, so it really doesn't matter whether you change or not, you've got the same odds either way.

[ 06-22-2005, 01:33 PM: Message edited by: Morgeruat ]

why would you want to switch?

If the one in the hosts hands has nothing and is taken out of the equation you don't have a better chance of picking the better of the two just because one has been removed, it's gone from a one in three, to a one in two, slightly better odds overall but changing does not increase the odds of getting the correct one.

You are making a false assumption, namely that just because the locket is opened that is suddenly is out of the problem.
Lets say there are three lockets A, B, and C, and lets define A as the locket you pick.
There are three equally probable choices.
1) A contains the prize.
2) B contains the prize.
3) C contains the prize.
In case (1) the host will open up locket B or C and you'll lose if you you switch.
In case (2) or (3) the host will open up the other empty locket, and you'll win if you switch.

By not switching you'll win in case (1), which will happen 1/3 of the time.
By switching you'll win in case (2) or in case (3), which will happen 2/3 of the time.

[ 06-22-2005, 01:59 PM: Message edited by: Seraph ]

If you roll a 6 sided die three times and get a 6 each time, what are the odds of getting another 6? one in six. Statistically it will not be very likely as there are 1296 possible combinations, but the straight odds of rolling a 6 on any die roll is still 1 in six. (barring a weighted die of course)

If I put 5 red stones 3 yellow stones and 1 blue stone in a bag, what would be the probability of drawing the blue stone, the answer 1 in 9, if 3 red stones and 1 yellow were removed it would not continue to keep the odds at 1 in 9 to draw the stone because the 4 stones have been removed from teh equation (just like the one locket) the new chance is instead 1 in 5 as the others have been REMOVED FROM THE EQUATION.

Seraph, you already know that the locket in the hosts hand is empty, so you must rule it out of the equation.

You only have a 50, 50 chance of picking the right one, not 66%

Your logic is faulty.

You have to rule out the opened locket, because picking it has a 100% chance of losing, it's not part of the game.

Of the remaining two you have an equal chance of picking the right one, so it is better to stick with what you've got, since you gain nothing by switching.

Seraph is correct.
The choice the host has is limited to two. If the prices is among those two he can only reveal the other - the empty one. Actually you can cut the possible outcomes down to
Initial pick. Possible outcomes [1] the price with 1/3 likelyhood or [2] empty with 2/3 likelyhood.
[1] In case you picked the price on the first choice you will always LOOSE if you change and always WIN if you stay. Remember this outcome has 1/3 likelyhood.
[2] In case you picked an empty closet the host will open the other empty closet leaving only the price. So you will always WIN if you change and LOOSE if you stay. Again we must recall that this outcome has 2/3 likelyhood.
In summary you have 2/3 chance of winning if you change and 1/3 chance if you stay.

The reason why the mind is tricked is that we have information that the contestant does not have. We KNOW that the host is limited to two choices, but the contestant doesn't know that.

EDIT: Sorry there were some critical typing mistakes.

[ 06-22-2005, 02:38 PM: Message edited by: mad=dog ]

Mad=dog, you do know your wrong don't you?

Once the first one has been removed you must recalculate, it's like working out how many days you get off a year and not counting days that overlap, just not the way to get the right awnser.

Are you saying that the contestant doesn't see which locket was opened, or what? this makes no sense...

mad dog is right!

The host knows which box is empty. So when chosing between an empty and a full one he MUST pick the empty one.
i.e there is a 2/3 chance (when you pick an empty one) that he is "forced" to show you which is full by opening the one which is not.

Ahem

So by what logic do you take it that you have a better chance of choosing the right box by changing?

Can't you see the flaws in your logic, one of the lockets has been taken out of the equation, you already know it's a duff one, if the host does not revel the contents your back to the orginal one in three chance.

It's not that hard.

There are only three possibilities, and they are all equally as likely.
The player picks empty locket one. The game host picks empty locket number two. Switching will win the car.
The player picks empty locket number two. The game host picks empty locket number one. Switching will win the car.
The player picks the prize. The game host picks either of the empty lockets. Switching will lose.
You win 2/3 of the time by switching, it is pretty straight forward.

Draw a probability tree, use the Bayes' Theorm, or (or better yet 'and') search "Monty Hall Problem" on google and Wikipedia if you don't belive me.

No, there are two events to consider, not one as you are.

1st event: possible outcomes: 3
desired outcomes (x): 1

P(x)=1/3

Then a locket is removed:

2nd event: possible outcomes: 2
desired outcomes (x): 1

P(x)=1/2

Therefore it does not matter

My logic is firther confimerd by the fact that the probabilites of all possible outcomes add up to 1, thus i have covered all possible outcome.

[ 06-22-2005, 04:17 PM: Message edited by: Aragorn1 ]

ahh, no.

your counting two lose senerio's as one.

There are two ways to lose, and two ways to win. If you discount the host picking the right locket.

so again, a 1/2 chance of winning.

a couple of questions that are muddled by the question itself and haven't been sufficiently answered.

1: does the host reveal his pick to the guest, or just the audience
2: if it is shown to the guest is it revealed that the hosts choice is empty does it not remove it from the realm of possible choices bringing the number of possible locations for the locket down to 2 from 3 (the one he's chosen or the one he hasn't).
If he shows the emtpy one then there are 2 possibilities, it is in the one he chose, or it is not, and he then as 2 choices, stay or change. which still leaves a 50% chance of either being wrong.

If the host does not show the guest whether or not the one he (the host) chose then his chance is 1 in 3 of any particular choice being correct.

Impossible, there are three lockets, so there can only be three outcomes, there is only one way to win, picking the correct locket, there are, in the first event, two ways to lose, picking either of the wrong lockets.

No you know the one the host has is not the correct one taking it out of the equation, thus it becomes 50/50, choosing to stay is just as risky as choosing to change.

Impossible, there are three lockets, so there can only be three outcomes, there is only one way to win, picking the correct locket, there are, in the first event, two ways to lose, picking either of the wrong lockets. </font>[/QUOTE]what are you yabbering on about?

Ignoring the host picking the right locket you can pick either of the wrong ones and the host picks the other wrong one, or you pick the right one and the host picks either of the wrong ones.

See, two possible outcomes, no?

NO

Four possible outcomes, as we're discarding any outcome in which the host picks the right one.


It's realy, realy easy!

Solution #1

There is a 1/3 chance that you'll hit the prize door, and a 2/3 chance that you'll miss the prize. If you do not switch, 1/3 is your probability to get the prize. However, if you missed (and this with the probability of 2/3) then the prize is behind one of the remaining two doors. Furthermore, of these two, the host will open the empty one, leaving the prize door closed. Therefore, if you miss and then switch, you are certain to get the prize. Summing up, if you do not switch your chance of winning is 1/3 whereas if you do switch your chance of winning is 2/3.
Solution #2

After the host opened one door, two remained closed with equal probabilities of having the prize behind them. Therefore, regardless of whether you switch or not you have a 50-50 chance(i.e, with probabilities 1/2) to hit or miss the prize door.


from your search (the #2 answer on google) Aragorn and Myself are arguing #2, you're arguing #1, both are valid.

[ 06-22-2005, 04:34 PM: Message edited by: Morgeruat ]

TERRY PASCAL

Alex,

I've been reading the Monty Hall problem and would like to post the answer to demonstrate where most people math is incorrect.

Given 2 doors, one with a prize and one without, which should you pick?

I think you'd agree that odds are 50-50 you'll be right. The fact that I may have had 3 doors initially and opened one, or 1000 doors and opened 998, does not change the current problem. There is still only a 50/50 chance of winning.

Because the probability of the problem can be expressed in three different ways.

P1 - 1/3 chance prize is door #1, 2/3 chance it is not
P2 - 1/3 chance prize is door #2, 2/3 chance it is not
P3 - 1/3 chance prize is door #3, 2/3 chance it is not

Most of the logic I've seen has the following flaw. As soon as the contestant picks door #1, they immediately assume that P2 and P3 are not valid. Well, they are. So they follow only P1. Regardless of which door I pick, all 3 are still valid. It is only when Monty opens a door that things change. By opening door #3 we now have

P3 - 0/0 chance prize is door #3, 3/3 chance it is not.

Now here everyone says, "Hey by P1, then the 2/3 chance all goes to door #2' But then why not, "By P2, then the 2/3 chance all goes to door #1"? Both are incorrect. When Monty picks #3, the whole problem is changed to:

P1 - 1/2 chance prize is door #1, 1/2 chance it is not
P2 - 1/2 chance prize is door #2, 1/2 chance it is not
P3 - 0 chance prize is door #3, 1 chance it is not

Look at it another way.

What's being said is that if I pick #1 and monty shows #3, then 2/3 of the time #2 will win. So If I had picked #2 to start with and monty opens #3, then the odds go 2/3 to #1. Why would they change? What if I don't have to tell monty? What if I could write it on a secret ballot?

Terry Pascal

I get what you guys are saying now!!

It depends on your interpretation of the events, as both use valid probability theory, and both can't be right.

[ 06-22-2005, 04:53 PM: Message edited by: Aragorn1 ]

Morgurat: The contestant doesn't know the host is limited to a choice of two. To him an _apparently_ random locket was just opened to reveal nothing. So if you mistakenly try to place yourself in the contestants place you erronically arrive to the conclusion that the choice is between the remaining lockets is indifferent.

However it is NOT indifferent. The host does NOT make a randomized choice like you do. The host makes a systematic choice based on the following
1) He never opens your locket.
2) He always opens an empty locket.

I made a simulation of this in Java last time someone refused to see the logic. If you fail to recognise the logic then know that it can be backed by an experiment. Switching is smartest.

My point is that you choose 1 of three, one will always be empty, this is always the hosts choice.

now stop and look at it as a fresh problem, you have 2 choices, one is wrong, and one is right. The third option has always been gone from this new problem thanks to the host, leaving only 2 choices, aka 50% of either choice being correct. Staying or switching doesn't matter as there are only 2 outcomes to this problem, right and wrong.

I realise that is where you and mr. Pascal has made an error. You think of it as a new problem with equal chance.
However it is NOT a new problem. You have information about the doors from the previous incidents.

Marilyn Vos Savant addressed this in a column a couple of years ago. And Bob Barker has done this on the Price Is Right for thirty-some years. Your best move is always to switch.

Would you rather have a 1/3 chance of being right, or a 2/3 chance?

If you stick with your original choice, you have a 1/3 chance of being right.

If you switch, you now have a 2/3 chance, even though one of the other two items is gone.

Why? Because originally, the other two items had a 2/3 chance of one of them holding the prize. All the host did was to eliminate one item that wasn't. So you're *still* picking the 2/3 option, even thought there's only one locket there.

Stick to the original problem. Don't put it into a new one.

<font color = lightgreen>Examine the problem as stated.

Suppose you choose an empty locket, which will happen 2/3 of the time. The host also picks an empty locket, which will happen 1/2 of the time (after your choice). These events are independent so you multiply the probabilities: (2/3)(1/2) = 1/3 = p(getting the prize by keeping your choice). P(getting the prize by switching) = 1 - p(getting the prize by not switching) = 1 - 1/3 = 2/3. Thus, it is smarter to switch once the host opens his empty locket.

Suppose you choose the locket with the prize, which will happen 1/3 of the time. The host chooses an empty locket, which happens all the time. Multiply these two events to get (1/3)(1) = 1/3. P(getting the prize by switching) = 1 - p(getting the prize by not switching) = 1 - 1/3 = 2/3. Thus, it is smarter to switch once the host opens his empty locket.

In either case, it is smarter to switch once the host opens his empty locket. Simple. [img]graemlins/petard.gif[/img] </font>