EPTA..B..C
I....I...G
I...G...I
G..I....I
+ Show Spoiler +
I must be missing something because I don't see how it is anything but 50%. I'm terrible at these types of logic problems anyways, I always miss the point.
Math Problem - Page 3
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Chill
Calgary25986 Posts
A..B..C I....I...G I...G...I G..I....I + Show Spoiler + | ||
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dybydx
Canada1764 Posts
(even if he was innocent the court can still sentence him to death anyways) | ||
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Chill
Calgary25986 Posts
there's 3 socks, one is black. Number 3 isn't black. What's the chance Number 1 is black. What am I missing here? | ||
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dybydx
Canada1764 Posts
not quite.. doesnt work that way. the grouping matters. | ||
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Chill
Calgary25986 Posts
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lololol
5198 Posts
Chance that one of the other 2 is not guilty = 100% Chance that Adam is guilty = 1/3, chance that Chris is guilty = 2/3 Guilty = sentenced... whatever wording you prefer, that's just what the answer is. | ||
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atarianimo
United States82 Posts
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dybydx
Canada1764 Posts
 ING DING DING*+ Show Spoiler + We have a winner! Congratulations to lololol! http://en.wikipedia.org/wiki/Three_Prisoners_Problem Next problem! There are 1000 coins, 1 of em is fake and weighs less than others. using a balance scale, how many times do you have to weigh to find out which coin is fake. http://en.wikipedia.org/wiki/Weighing_scale#Balance [a harder version. there are x coins, and your balance has y bowls (your balance can tilt in m directions). how many times does it need now?] | ||
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RzzE
Germany20 Posts
On January 23 2008 03:28 dybydx wrote: ING DING DING*+ Show Spoiler + We have a winner! Congratulations to lololol! http://en.wikipedia.org/wiki/Three_Prisoners_Problem Next problem! There are 1000 coins, 1 of em is fake and weighs less than others. using a balance scale, how many times do you have to weigh to find out which coin is fake. http://en.wikipedia.org/wiki/Weighing_scale#Balance [a harder version. there are x coins, and your balance has y bowls (your balance can tilt in m directions). how many times does it need now?] You divide the pile of coins in half and measure both halves. You discard the heavier pile. You repeat the procedure. If your pile has an odd number of coins you set one coin aside and measure the 2 piles. if they both weigh the same, you have found your fake coin. I counted a maximum of 18 measurements. Edit: sorry I mean 9 measurements on the balance scale. But the method described below of dividing it into 3 piles is more efficient. | ||
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Qatol
United States3165 Posts
I count a maximum of 7 measurements | ||
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RzzE
Germany20 Posts
 ) I'd like to propose the following puzzle:Four people are on one side of a bridge in the dark. The bridge can only support 2 people at a time. There is only one torch among the four of them and the torch needs to be carried along with the pair crossing the bridge. The people take different times to cross the bridge:It takes them 10 mins, 5 mins, 2 mins and 1 min respectively. A pair crossing the bridge walks at the speed of the slowest of the two of them. What is the minimum time for all of them to have crossed the bridge? Edit: Bad wording, sorry. What is the minimum time it takes to have them all end up on the opposite side of the bridge? Answer: + Show Spoiler + 17 mins | ||
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BlackJack
United States10574 Posts
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RzzE
Germany20 Posts
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Macavenger
United States1132 Posts
On January 23 2008 03:28 dybydx wrote: [a harder version. there are x coins, and your balance has y bowls (your balance can tilt in m directions). how many times does it need now?] + Show Spoiler [Answer] + I'm pretty sure this is ceiling(log base (y+1) (x)). In a less technical format, (y+1)^n = x, solve for n, rounding up, gives the smallest possible worst case number of measurements. | ||
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MPXMX
Canada4309 Posts
You have 12 identical-looking coins and a balance that compares the weight on its two arms and only says which side is heavier and which is lighter. One of the coins is fake and differs only in weight, however, when you get the coins, you don't know if the fake one is lighter or heavier. What you need is a foolproof method of finding the fake coin in 3 weighings. | ||
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spammerA
China355 Posts
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BlackJack
United States10574 Posts
On January 23 2008 04:56 MPXMX wrote: This one is hard: You have 12 identical-looking coins and a balance that compares the weight on its two arms and only says which side is heavier and which is lighter. One of the coins is fake and differs only in weight, however, when you get the coins, you don't know if the fake one is lighter or heavier. What you need is a foolproof method of finding the fake coin in 3 weighings. my favorite riddle of all time. spent like a week trying to figure it out until i gave up, then i nailed it in like 10 minutes months later when waiting for an exam lol | ||
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Muirhead
United States556 Posts
+ Show Spoiler + Label the coins 1-12. Then do the following weighings: 1 4 6 10 against 5 7 9 12 2 5 4 11 against 6 8 7 10 3 6 5 12 against 4 9 8 11 Listing out all the cases shows that these weighings uniquely determine the false coin and whether it is lighter or heavier. Problem/Paradox: There are countably infinite many prisoners sitting in a row (i.e. the prisoners are numbered 1,2,3,4,5,... on to infinity) One day a judge comes and says to them: Tomorrow you will each be given either a black or a white hat. You will be able to see the hats of those prisoners with a higher number than yours, but not the hats of those prisoners with a lower number than yours. You will not be able to see your own hat. You will then each simultaneously write down a guess of either black or white on a piece of paper. If you guess your own hat color correctly, you will be set free. Otherwise you will be executed. The prisoners discuss all night what they will do. One suddenly realizes that, with a proper strategy, they can ensure that almost 100% of them will survive. What is that strategy? | ||
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LossLeader
Canada10 Posts
On January 23 2008 04:35 RzzE wrote: If that was correct (or even if it wasn't ) I'd like to propose the following puzzle:Four people are on one side of a bridge in the dark. The bridge can only support 2 people at a time. There is only one torch among the four of them and the torch needs to be carried along with the pair crossing the bridge. The people take different times to cross the bridge:It takes them 10 mins, 5 mins, 2 mins and 1 min respectively. A pair crossing the bridge walks at the speed of the slowest of the two of them. What is the minimum time for all of them to have crossed the bridge? Answer: + Show Spoiler + 17 mins I guess I must still be in 6th grade, but it seems that this is a bit of a trick question. + Show Spoiler + To answer the question as worded the answer to me would be 16 mins. However, the 1 min person will still be on the same side of the bridge, having crossed the bridge with the 10min person, returning which takes 1 min, and then brought the torch back allowing the 5 min (and 2 min) person to cross with the torch. If the 4 people are to remain on the opposite side of the bridge, would not the answer be 19 mins, as the 1 min person has to come back twice taking an additional 2 mins, to accomany the others on their respective trips. Please explain how 17 min is correct | ||
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iSTime
1579 Posts
To give an everyday sort of example, if you were choosing your schedule for college, and you had a list of math classes, I wouldn't ask "which one are you taking," unless I already knew you were only taking one of them. I don't know any native english speaker who would. | ||
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