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On February 11 2010 11:46 ghrur wrote:Show nested quote +On February 11 2010 08:52 JeeJee wrote:On February 11 2010 08:41 Nitrogen wrote:
i took the american math contest yesterday and there was a pretty good puzzle type problem on it.
in a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. four amphibians, brian, chris, leroy, and mike live together in this swamp, and they make the following statements.
brian: mike and i are different species
chris: leroy is a frog
leroy: chris is a frog
mike: of the four of us, at least two are toads
how many of the four amphibians are frogs?
i dont know if these are the only unique solutions but they both work brian = frog chris = toad leeroy = frog mike = frog brian = frog chris = frog leeroy = toad mike = frog works also re my puzzle i mean, i haven't solved it (i don't know the official solution) but initially i thought of 25% as well. however, then.. + Show Spoiler +
what if they decide on "we will guess the same as our coin shows". so A flips H, guesses H. if A flips T, he'll guess T. and same for B.
so then they would win in a TT and HH scenario and lose in a HT and TH scenario
which would make the game worth it
yes/no? is there a flaw in this logic?
edit: re below switch monty hall clone Right, I took the AMC too, and um, neither of the solutions are right. If Chris=toad, then leroy=NOT frog, thus leeroy must also be a toad. If Leeroy=toad, then Chris must be NOT frog, and thus Chris must be toad. Now, taking either of these, we consider Mike/Brian. If this were true, then mike would be correct, there would be at least 2 toads, mike is a frog. But... this means that Brian is at a logical contradiction. If he is a toad, he is telling the truth as they would be different species. If he is a frog, then he is lying because the two would be the same, both frogs. Thus, this can't be the correct answer. Iirc, it goes, Mike=toad because if Brian is a toad, mike must be a toad because he's lying, so him and mike can't be different species, thus Mike must be a toad. If Brian is a frog, then mike is a frog because Brian is telling the truth, so mike must not be a frog, thus a toad. So, establishing that Mike=toad, we next come to realize that there must be less than 2 toads because Mike is lying. There is either 1 or 0 toads. Then, we look at chris and leeroy and realize that if chris=toad, then leeroy must not be a frog, and thus a toad. If Chris=frog, then leeroy must be a frog because chris is a frog. Either way, the two go as a pair. Okay, so the two can't be toads together, otherwise, it would be at least 2 toads, the two must be frogs. Then, if Brian is a toad, then Brian+Mike=2 toads, at least two toads, so brian must be a frog as well. Thus, 3 frogs, Mike is the toad.
i dont know, but your beginning sentence already has a mistake in it, so i imagine the rest of the post does too
If Chris=toad, then leroy=NOT frog, thus leeroy must also be a toad.
this is not true
if chris=toad, then chris's statement of "leroy is a frog" is true, hence leroy is a frog
. . .
my solutions work
also, just reading your conclusion, if mike is the only toad, then how can his statement of "there are at least 2 toads" be true? he would have to be lying, hence he cannot be a toad
ah. i know what your mistake is, you confused who is telling the truth. the choices of amphibians are not arbitrary, frogs statements are false, toads' are true.
but if you were to flip the logic like you have, then your solution works too
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On February 11 2010 08:34 Shiladie wrote:Show nested quote +On February 11 2010 06:44 Phrujbaz wrote:
The solution to the pirate's riddle assumes a lot of things.
What is to stop pirate 3 from saying he will rather be offered 1 gold by pirate 4 than to get an unfair deal from pirate 5? He can demand at least half the money from pirate 5, and if pirate 5 really believes pirate 3 will vote it down otherwise, pirate 5 will agree to it. + Show Spoiler +
Because pirate 3 has no bargaining position, as spelled out, he will either get 0 gold from pirate 4, or 1 from pirate 5, so he, being greedy, goes for the option that gets him 1 gold
What do you mean no bargaining position? He can decide the life or death or pirate 5. He can rightly assume that this is a bargaining position, and decide he won't agree for less than 500. If pirate 5 offers him less, he will vote against, and pirate 5 will die, so the chance that pirate 5 agrees to the demand of 500 is very high. This high chance of 500$ is worth much more than the 1$ he would get if he accepted your offer. He will gladly take the risk of getting nothing if there is such a good chance he would get 500$.
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On February 11 2010 11:46 ghrur wrote:Show nested quote +On February 11 2010 08:52 JeeJee wrote:On February 11 2010 08:41 Nitrogen wrote:
i took the american math contest yesterday and there was a pretty good puzzle type problem on it.
in a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. four amphibians, brian, chris, leroy, and mike live together in this swamp, and they make the following statements.
brian: mike and i are different species
chris: leroy is a frog
leroy: chris is a frog
mike: of the four of us, at least two are toads
how many of the four amphibians are frogs?
i dont know if these are the only unique solutions but they both work brian = frog chris = toad leeroy = frog mike = frog brian = frog chris = frog leeroy = toad mike = frog works also re my puzzle i mean, i haven't solved it (i don't know the official solution) but initially i thought of 25% as well. however, then.. + Show Spoiler +
what if they decide on "we will guess the same as our coin shows". so A flips H, guesses H. if A flips T, he'll guess T. and same for B.
so then they would win in a TT and HH scenario and lose in a HT and TH scenario
which would make the game worth it
yes/no? is there a flaw in this logic?
edit: re below switch monty hall clone Right, I took the AMC too, and um, neither of the solutions are right. If Chris=toad, then leroy=NOT frog, thus leeroy must also be a toad. If Leeroy=toad, then Chris must be NOT frog, and thus Chris must be toad. Now, taking either of these, we consider Mike/Brian. If this were true, then mike would be correct, there would be at least 2 toads, mike is a frog. But... this means that Brian is at a logical contradiction. If he is a toad, he is telling the truth as they would be different species. If he is a frog, then he is lying because the two would be the same, both frogs. Thus, this can't be the correct answer. Iirc, it goes, Mike=toad because if Brian is a toad, mike must be a toad because he's lying, so him and mike can't be different species, thus Mike must be a toad. If Brian is a frog, then mike is a frog because Brian is telling the truth, so mike must not be a frog, thus a toad. So, establishing that Mike=toad, we next come to realize that there must be less than 2 toads because Mike is lying. There is either 1 or 0 toads. Then, we look at chris and leeroy and realize that if chris=toad, then leeroy must not be a frog, and thus a toad. If Chris=frog, then leeroy must be a frog because chris is a frog. Either way, the two go as a pair. Okay, so the two can't be toads together, otherwise, it would be at least 2 toads, the two must be frogs. Then, if Brian is a toad, then Brian+Mike=2 toads, at least two toads, so brian must be a frog as well. Thus, 3 frogs, Mike is the toad.
lolol. you got the right answer, but your reasoning is completely wrong. either chris is a toad and leroy is a frog, or chris is a frog and leroy is a toad. we don't know for sure, but we're guaranteed 1 frog and 1 toad. for proof, if we assume that chris is a toad, then we can assume he's telling the truth and that leroy is a frog, which works out because since leroy's lying, chris must be a toad, and vice versa if we assume that chris is a frog.
so now we have to look at brian. if we assume that brian is a toad, then that means that brian is a toad while mike is a frog. however, this creates an impossible situation. mike stated that of the four, there are at least two toads. if we assume that mike is a frog, then that would mean that there had to be 0 or 1 toads. however, in this situation, we have 2 toads: brian, and either chris or leroy.
Therefore, brian has to be a frog, and since brian and mike are the same species, mike is also a frog, and it works out, since either chris or leroy, but not both, has to be a toad. Therefore, JeeJee is correct. 3 frogs. brian, mike, and either chris or leroy
Source: I took AMC as well. this problem took me like 5 minutes to solve.
for those who took AMC, weren't the last half of the problems seemingly impossible? wtf. i really underestimated it.
edit: oh, and as jeejee pointed out, you confused who was telling truth and lies. you got really lucky that both ways, it comes out as 3 frogs. :O
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OH, wow, lol!
I realize now, I had a diff question, with nearly the same wording.
Blah. Mine was toads always lie and frogs always tell truths. >_< My bad, lol.
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hmm it's interesting that both ways come out to having one toad and 3 frogs. that's a sign of a poorly designed question.
However.. before I post day 2's puzzle in a couple of hours, I'd still like for someone to point out the flaw in the logic outlined earlier, where this makes the game worth it (the original puzzle). The logic being, flip the coin, and guess whatever your coin ends up being. i.e. if you flip heads, guess heads for your friend, and your friend does the same. This should give you a win in HH and TT situations, making the game worth it. comments?
(since the consensus by far is that the game is not worth it, i think it's worthwhile to look at it)
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On February 11 2010 08:52 JeeJee wrote:..... also re my puzzle i mean, i haven't solved it (i don't know the official solution) but initially i thought of 25% as well. however, then.. + Show Spoiler +
what if they decide on "we will guess the same as our coin shows". so A flips H, guesses H. if A flips T, he'll guess T. and same for B.
so then they would win in a TT and HH scenario and lose in a HT and TH scenario
which would make the game worth it
yes/no? is there a flaw in this logic?
On February 11 2010 23:16 JeeJee wrote:
.......
However.. before I post day 2's puzzle in a couple of hours, I'd still like for someone to point out the flaw in the logic outlined earlier, where this makes the game worth it (the original puzzle). The logic being, flip the coin, and guess whatever your coin ends up being. i.e. if you flip heads, guess heads for your friend, and your friend does the same. This should give you a win in HH and TT situations, making the game worth it. comments?
(since the consensus by far is that the game is not worth it, i think it's worthwhile to look at it)
after reading your spoiler, it become 50-50 chance, my team win and get $2 ($1 each), and my team lose only pay $1 ($0.5 each)
so the game is worth it 
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There is no flaw in logic with the coin toss.
If you write down your answer after your coin flip has been done, it's understood that you're choosing the right answer (even if it's for your friend) so the .50 probability of your own coin flip is negated.
In other words, whichever coin you flip becomes the RIGHT answer. So your own odds aren't 50% but 100%. The only really variable is your friend's coin flip in which he has a 50/50 chance of matching your answer.
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Definitely a good deal. 50% * 2 + 50% *-1 = 0.5 Positive expected value from your strategy makes the game totally worth it.
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edit: I read the question wrong
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EPT
