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I found a bunch of new puzzles as a side-effect of being bored at work. I don't think any of these have been posted on TL before
I'll post a puzzle a day, especially those that are frustrating me, to tap into TL's collective mind  Not ordered in any way, today's could be the hardest or the easiest, or probably somewhere in between
Let's begin!
One fair coin is given to you, and another one to your best friend. You&your friend can decide on a strategy beforehand then start to play the following game (once the game starts, there is to be no further communication between you two):
You flip your coin, and write down what you think your friend's coin will say.
Your friend flips their coin, and writes down a guess as to what your coin says.
There's a third party involved -- me. I will get to look at your guesses and what the coins end up being. If both of you guess each other's coins correctly, I give you guys $2. However, if either one of you (or both) is wrong, you guys give me $1.
We repeat this all day.
Let's assume you're here to win money. Are you happy that you're on your team, or would you rather trade places with me? Why?
GL!
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+ Show Spoiler +25% chance we are both right, 75% chance you are right... bad deal... switch places
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Depends on if you give each of us $2 or just $2 split. If its the latter, no deal unless you make is $3. Ask stenole.
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assuming we give 1$ together and gain 2$ together.
0.25 * 2 = 0.5
0.75 * 1 = 0.75
So each round we'll give .75 and only receive 0.5
Unhappy 
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Yep... too easy... give us more of a challenge
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There's another strategy, where the two guessers collaborate.
I can flip a coin, and then announce what that coin is as my 'guess' as to my opponent's coin. Opponent then flips coin and repeats what I say. That way, it's just a 50-50 guess, and you're basically paying out, on average $0.50 per toss.
Shared between two, that's the same as the expected payout that you're getting as the 'third party', without any collaboration.
Edit: Bah, rereading the question, it doesn't work since the opponent doesn't hear the guess.
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If there is no communication allowed after the first coin is in the air, then I would want to switch places with the game-host
as said above by Aim Here, if the second person can see the first one's guess, then the first always guesses the same as what they got, increasing the odds in your favor to make money.
If people are interested, there is a great website with these sorts of things here:
http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml
My favorite:
A pirate ship captures a treasure of 1000 golden coins. The treasure has to be split among the 5 pirates: 1, 2, 3, 4, and 5 in order of rank. The pirates have the following important characteristics: infinitely smart, bloodthirsty, greedy. Starting with pirate 5 they can make a proposal how to split up the treasure. This proposal can either be accepted or the pirate is thrown overboard. A proposal is accepted if and only if a majority of the pirates agrees on it. What proposal should pirate 5 make?
progressive tips and then solution in nested spoiler tags:
+ Show Spoiler +Solve it in reverse + Show Spoiler +Pirate 5 gets the most money + Show Spoiler +If it is down to pirate 2 making a proposal, he knows pirate 1 will refuse, because then pirate 1 gets all the gold, so pirate 2 will accept any proposal pirate 3 makes so that he will survive. Conversly pirate 1 will refuse pirate 3's proposal no matter what it is, because if it doesn't pass, he gets to off both pirate 2 and 3, and get all the gold. + Show Spoiler +Pirate 1 then doesn't want it to ever be pirate 3's proposal, because pirate 3 can propose 1000 for himself and because himself and pirate 2 both will agree, it will pass and pirate 1 will be SOL, so if pirate 4's proposal is 1 gold to pirate 1 and 1 gold to pirate 2, they both get more then if they refuse it and it gets to pirate 3's proposal, so pirate 4's proposal will be 998 for himself, 0 for 3 and 1 for 1 and 2 + Show Spoiler +So, pirate 3 will want pirate 5's proposal to pass if he gets 1 gold, and pirate 1 or 2 will also vote for it if it gives them more then 1 gold. So since we just need 3 votes, pirate 5 proposes... + Show Spoiler +
Pirate 5's proposal, winning pirate 2 and 3's approval, and since he votes for himself he gets majority
Pirate 1: 0
Pirate 2: 2
Pirate 3: 1
Pirate 4: 0
Pirate 5: 997
This also works to give pirate 1 the 2 gold and pirate 2 0 gold
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On February 11 2010 05:34 Shiladie wrote:If there is no communication allowed after the first coin is in the air, then I would want to switch places with the game-host as said above by Aim Here, if the second person can see the first one's guess, then the first always guesses the same as what they got, increasing the odds in your favor to make money. If people are interested, there is a great website with these sorts of things here: http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtmlMy favorite: A pirate ship captures a treasure of 1000 golden coins. The treasure has to be split among the 5 pirates: 1, 2, 3, 4, and 5 in order of rank. The pirates have the following important characteristics: infinitely smart, bloodthirsty, greedy. Starting with pirate 5 they can make a proposal how to split up the treasure. This proposal can either be accepted or the pirate is thrown overboard. A proposal is accepted if and only if a majority of the pirates agrees on it. What proposal should pirate 5 make? progressive tips and then solution in nested spoiler tags: + Show Spoiler +Solve it in reverse + Show Spoiler +Pirate 5 gets the most money + Show Spoiler +If it is down to pirate 2 making a proposal, he knows pirate 1 will refuse, because then pirate 1 gets all the gold, so pirate 2 will accept any proposal pirate 3 makes so that he will survive. Conversly pirate 1 will refuse pirate 3's proposal no matter what it is, because if it doesn't pass, he gets to off both pirate 2 and 3, and get all the gold. + Show Spoiler +Pirate 1 then doesn't want it to ever be pirate 3's proposal, because pirate 3 can propose 1000 for himself and because himself and pirate 2 both will agree, it will pass and pirate 1 will be SOL, so if pirate 4's proposal is 1 gold to pirate 1 and 1 gold to pirate 2, they both get more then if they refuse it and it gets to pirate 3's proposal, so pirate 4's proposal will be 998 for himself, 0 for 3 and 1 for 1 and 2 + Show Spoiler +So, pirate 3 will want pirate 5's proposal to pass if he gets 1 gold, and pirate 1 or 2 will also vote for it if it gives them more then 1 gold. So since we just need 3 votes, pirate 5 proposes... + Show Spoiler +
Pirate 5's proposal, winning pirate 2 and 3's approval, and since he votes for himself he gets majority
Pirate 1: 0
Pirate 2: 2
Pirate 3: 1
Pirate 4: 0
Pirate 5: 997
This also works to give pirate 1 the 2 gold and pirate 2 0 gold
This is a fun one. I couldn't resist clicking the spoilers. I am such a froob.
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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.
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Consider that you can decide on a strategy beforehand as opposed to randomly guessing
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I would guess no signals could be used since that also is a form of communication...
... I guess the trick is to come up with at strategy that gives you better chances of being right than wrong.
Hmm...
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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
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On February 11 2010 07:48 Belano wrote:
I would guess no signals could be used since that also is a form of communication...
... I guess the trick is to come up with at strategy that gives you better chances of being right than wrong.
Hmm...
in a pure double 50% where you need to be right both times.
If the second person can receive no signal of any form from the first person, after the first person has made their guess (not even seeing their guess) then there is no way to influence the odds.
I could be proven wrong here, but there is no guessing strategy that does anything if it is truely a 50% chance on each flip.
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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?
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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
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You are a contestant on "Who Wants to Be a Millionaire."
You have a 50/50 life line left, and are at the final question. You have no idea what the answer is, and for mathematical reasons, the odd of any choice being right is the same for all four answers (A, B, C and D). You decide to randomly guess choice A. But before you give your final answer you decide to use your life line. The host takes away two incorrect answers, Choices B and C thus you are left with A and D.
Do you stick to your original choice A or switch to D? Or does it not matter?
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l10f
United States3241 Posts
Switch, as long as you chose the wrong solution the first time (3/4 chance), by switching you choose the right solution.
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On February 11 2010 08:52 JeeJee wrote:Show nested quote +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.
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On February 11 2010 05:34 Shiladie wrote:If there is no communication allowed after the first coin is in the air, then I would want to switch places with the game-host as said above by Aim Here, if the second person can see the first one's guess, then the first always guesses the same as what they got, increasing the odds in your favor to make money. If people are interested, there is a great website with these sorts of things here: http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtmlMy favorite: A pirate ship captures a treasure of 1000 golden coins. The treasure has to be split among the 5 pirates: 1, 2, 3, 4, and 5 in order of rank. The pirates have the following important characteristics: infinitely smart, bloodthirsty, greedy. Starting with pirate 5 they can make a proposal how to split up the treasure. This proposal can either be accepted or the pirate is thrown overboard. A proposal is accepted if and only if a majority of the pirates agrees on it. What proposal should pirate 5 make? progressive tips and then solution in nested spoiler tags: + Show Spoiler +Solve it in reverse + Show Spoiler +Pirate 5 gets the most money + Show Spoiler +If it is down to pirate 2 making a proposal, he knows pirate 1 will refuse, because then pirate 1 gets all the gold, so pirate 2 will accept any proposal pirate 3 makes so that he will survive. Conversly pirate 1 will refuse pirate 3's proposal no matter what it is, because if it doesn't pass, he gets to off both pirate 2 and 3, and get all the gold. + Show Spoiler +Pirate 1 then doesn't want it to ever be pirate 3's proposal, because pirate 3 can propose 1000 for himself and because himself and pirate 2 both will agree, it will pass and pirate 1 will be SOL, so if pirate 4's proposal is 1 gold to pirate 1 and 1 gold to pirate 2, they both get more then if they refuse it and it gets to pirate 3's proposal, so pirate 4's proposal will be 998 for himself, 0 for 3 and 1 for 1 and 2 + Show Spoiler +So, pirate 3 will want pirate 5's proposal to pass if he gets 1 gold, and pirate 1 or 2 will also vote for it if it gives them more then 1 gold. So since we just need 3 votes, pirate 5 proposes... + Show Spoiler +
Pirate 5's proposal, winning pirate 2 and 3's approval, and since he votes for himself he gets majority
Pirate 1: 0
Pirate 2: 2
Pirate 3: 1
Pirate 4: 0
Pirate 5: 997
This also works to give pirate 1 the 2 gold and pirate 2 0 gold
I remember this problem in high school, and one guy on the math team said "these are some pretty smart pirates".
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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
