bushindo
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Posts posted by bushindo
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Excellent work so far. Sorry for not responding earlier. I was on a vacation and didn't have access to a computerI found this interesting 102-digit number: 490000...000077 (“49”, followed by 98 zeros, followed by “77”.) If you multiply that number by 111...111 (one hundred 1-s), the resulting 201-digit number:
54444...4447555...55547 (“5”, followed by 99 “4-s”, followed by “7”, followed by 98 “5-s”, followed by “47”) meets the conditions set forth in the OP, except I cannot guaranty it is the smallest number that does. (I did not find any simple combination of "1-s" and "0-s", which multiplied by all "7-s" would yeild "4-s", "5-s", and "7-s", while avoiding "8-s".)On a side note. While playing with these numbers I found a simple yet curious divisibility rule, of which I was not aware before. I should try and construct a problem based on it.
The number above indeed satisfies the conditions giving in the OP, but there is a smaller number. The number I have in mind is a 201-digit number, but the 201-th (leftmost) digit is a 4.
I'm interested in seeing a puzzle based on the divisibility rule you found. For what it is worth, I also based this puzzle on a divisibility rule as well.
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Found a roundabout way to display formatted text inside a BD post. Same thing as above, but visible.
You are approaching this from the correct angle
Breaking the 100 7s into 7 * 100 1s is the right first step. As to the analysis above, I recommend trying some small case like 11 or 111. It shouldn't be hard to find the solution in those cases.
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I can't find a neat analytic solution for the problem of finding the expected number of steps until the ant leaves the circle, though it could be done by simulation or numerical approximation. This puzzle, however, asks a different question...Mathemagician, on 30 Jan 2013 - 21:37, said:
A long time ago, I posted a problem about an ant in a circle with a diameter of 3 meters. The ant would go 1 meter in a random direction, rest, and go 1 meter in another random direction and repeat this process until it got out of the circle. What is the expected number of 1-meter steps the ant takes before escaping. It was never fully solved and I just recently got back to it and thought of this question: assume f(x) is the expected distance from the center after x steps. Is the answer to the original question merely when does f(x) become greater than 3?
Assume f(x) is the expected distance from the center after x steps. I think that the answer in general is that the expected number of steps to escape the circle is not the same as the x for which f(x) >= 3. The differences between the simulations and bonanova's method already hint at this.
Let's consider a different simpler example to see how these two quantities are different. Consider an ant on the x-axis at the origin (0,0). Let's say that every step, the ant either stays still with probability (999/1000) or moves to the right by 1000 unit with probability (1/1000). We want to know the expected number of steps until the ant passes the marker 3 units away at location (3,0).
The expected distance away from the center for after 1 step is (999/1000 * 0 + 1000 * 1/1000 ) = 1. Also, in general, the expected distance after x steps is x, (i.e., f(x) = x ).
If expected number of steps to pass the marker is the same as the x for which f(x) >= 3, the ant should pass the point (3,0) within 3 step. But we know that is not true. In fact, it will take the ants 1000 steps in average to pass the point (3,0).
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I'm assuming it has to have atleast one 4, 5 and 7?
That is correct. I'll edit the OP to clarify that point. Thanks.
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This is an extension of Prime's excellent puzzle
What is the smallest 101-plus-digit number consisting of only 4's, 5's, and 7's that is divisible by '7777.....777' (one hundred 7's)?
The solution *must contain* at least one 4, one 5, and one 7.
EDIT: clarified the properties of the required solution
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Seven International Hostages are lined up in a row .Some are standing or sitting ,facing backward or forward.All were blindfolded with different colored duct tapesand ordered not to move. None has any information about the other hostages.The terrorist leader announced to the nationals before him on the speakers that..they shall all be released if they all can guess their blindfold's color correctlywhich are in the order as ff : Red Orange Yellow Green Blue Violet WhiteChi has 2 hostages facing backward on his right, 3 sitting down on his right.Jap has 2 hostages facing forward on his left, 3 standing up on his leftFra has 1 hostage facing backward on his right, 3 standing up on his left.Rus has 3 hostages facing backward on his left, 2 sitting down on his left.Eng has 2 hostages facing forward on his right, 3 sitting down on his right.Ger has 2 hostages facing forward on his left, 1 standing up on his right.Spa has 2 hostages facing backward on his left, 2 standing up on his leftIn 60 seconds 1 hostage should raise both his tied hands then will be pulledbehind the row and will be shot in the head right on the spot in case he failsto shout the correct color of his blindfold .The other nationals will share thesame fate. If he is right, all the others should not fail or else everyone is goingto die. If none of them will guess in a minute all will remain hostages..Who would first raise his hands?
The two possible arrangements are
Jap, Chi, Rus, Spa, Ger, Fra, Eng
Eng, Fra, Ger, Spa, Rus, Chi, Jap
Since the hats are displayed in the following order
Red Orange Yellow Green Blue Violet White
If 'getting pulled behind the row' (see colored part above) means being pulled backwards in the prisoner's orientation, then the prospect of survival for all the prisoners is 1/2. Anyone besides Spa can raise his hand and guess one of his two possible color. For instance, Jap is either Red (beginning of the line) or White (last in line). If his guess is correct, everyone else should be able to guess their color. If his guess is *incorrect*, then everyone else should also be able to guess their color, although it might not do them much good at that point according to the OP.
If 'getting pulled behind the row' means getting pulled behind the row from an observer's perspective, then the prisoners can use that information to improve their survival odds. In that case, Spa should call Green. Once he is pulled 'behind' the row, the other prisoners would know which of the two arrangements they are in.
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I think I finally got it!
D is the best position with approx $1.11 expected payout. The odds and payouts for each position are (rounded):
A: 19/63 $1.06
B: 8/21 $0.95
C: 10/63 $0.95
D: 10/63 $1.11
Great job, k-man. You got it.
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You're right. I messed up. Here are the corrected probabilities...
A: 19/63
B: 32/63
C: 10/63
D: 2/63
Based on these odds, B is still the pest position with the expected payout of almost $1.27.
Comments
A B C D
BB BR BB RR (RR)
RR BR RR BB (BB)
BR BB BB RR (RR)
BR RR RR BB (BB)
So it seems that the sequences that lead to D winning (probability 2/63) are above. But there are a few more sequences that would lead to D winning. For instance, consider the following distributions
BR RR BB BR BR.
There are more missing distributions for D, so you might want to look for the winning cases for A, B, C and find them.
Doh!
I went through everything again and found a couple of mistakes, so now the odds and expected payouts are as follows...
A: 5/21 $0.83
B: 4/9 $1.11
C: 10/63 $ 0.95
D: 10/63 $ 1.11
According to this result positions B and D are equal in the expected payout, but this makes me think that I'm still missing something.
Comments
You're very close. I would recommend checking the cases under B again.
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Prove that you cannot
cover a 10 x 10 chessboard with 25 figures
(Problem from Russian Math Olympiads. 6-th grade, 1964.)
Answer
Let's say that the 10x10 grid is a chessboard. There would then be 50 Black and 50 White cells. Each tetris piece would be one of two types
(1) B W B B (2) W B W W
So, let A be the number of pieces of type (1), and B be the number of pieces of type (2). The following two equations would have to be true if we can fill a 10x10 board
A + B = 25
3*A + B = 50
But obviously, there are no integer solutions for the above, so we can't fill the board with this Tetris shape.
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2
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You're right. I messed up. Here are the corrected probabilities...
A: 19/63
B: 32/63
C: 10/63
D: 2/63
Based on these odds, B is still the pest position with the expected payout of almost $1.27.
Comments
A B C D
BB BR BB RR (RR)
RR BR RR BB (BB)
BR BB BB RR (RR)
BR RR RR BB (BB)
So it seems that the sequences that lead to D winning (probability 2/63) are above. But there are a few more sequences that would lead to D winning. For instance, consider the following distributions
BR RR BB BR BR.
There are more missing distributions for D, so you might want to look for the winning cases for A, B, C and find them.
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Excellent analysis, Prime. I just have 1 minor thing to addI’ll try again.
Just the case of 6 red and 6 blue labels, disregarding the round before the host revealed two of the stamps.
Let’s use the following terminology:
r = a man with two red stamps; b = man with two blue stamps; x = man with a mix.
x' = NOT (x was unable to deduce his stamps.)
* = AND;
+ = OR;
x(condition) = x can deduce his stamps based upon the condition. E.g., x(r’) means x can deduce his stamps after r has failed.
There are only 6 significant cases (disregarding the order) based on the number of mixed colors in the lineup -- from zero to 5: rrrbb, xrrbb, xxrrb, xxxrb, xxxxr, xxxxx. “r” and “b” may be swapped for one another making logically identical cases.
The 6 cases may be resolved as following:
1) r1 r2 r3 b1 b2 ==> b() + r3(r2’(r1’)).
Example 1: r1’ b1 r2 r3 b2, where b1 solves on his turn; OR Ex.2: r1’ r2’ r3 b1 b2, where r3 solves after r1 and r2 failed.
2) x r1 r2 b1 b2==> x(r’ * b’) + r2(r1’) + b2(b1’)
Ex.1: r1’ b1’ x r2 b2. Ex.2: r1’ r2 b1 b2 x. Ex.3: b1’ x’ r1’ b2 r2.
3) x1 x2 r1 r2 b ==> b() + x(r2’(r1’)).
Ex.1: b r1 r2 x1 x2. Ex.2: r1’ r2’ x1 b x2.
4) x1 x2 x3 r b ==> x(r’ * b’).
Example: x1’ r’ b’ x2 x3.
5) x1 x2 x3 x4 r ==> x1(x2’(x1’ * r’)).
Ex.1: x1' r' x2' x3' x4' (x1 solves on his second turn). Ex.2: x1'' x2' x3' x4' r' (x2 solves on his second turn.)
6) x1 x2 x3 x4 x5 ==> x2(x1’(x’(x1’ * x2’))).
Example: x1'' x2' x3' x4' x5', where x2 can deduce that he is a mix on his second turn after x1 fails for the second time.
I derive deduction strings for each case based on perspectives that the men in the lineup may have and deduction strings from those perspectives. Take for example case (3). An x in that line up may suppose he is either part of the case (2), or (3). Without any additional data, r may think he is in either (3) or (4). Whereas b in lineup (3) simply knows he is b.
It appears, the longest deduction string here is one where the men made one full round of non-guessing whereafter in the second round the second man in the line up must deduce his stamps. There are several possibilities for that:
x1'' x2' x3' x4' x5';
x1'' x2' x3' x4' r';
r'' x1' x2' x3' b'.
The first round, where 14 stamps were in play, does not seem to offer any useful information for those longest deduction variations above.
There is 1 more possibility for the distribution that leads to the longest gameplay
r'' x1' x2' x3' x4'.
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Sure.
Pocket stamps are in parenthesis. in all below combinations D wins in the first round. All have equal probability of 1/126.
A B C DBB BB BR RR (RR)BB BB RR RR (BR)BB BR BB RR (RR)BB RR BB RR (BR)RR RR BR BB (BB)RR RR BB BB (BR)RR BR RR BB (BB)RR BB RR BB (BR)BR BB BB RR (RR)BR RR RR BB (BB)
Comments
A B C D
BB BB BR RR (RR) *
BB BB RR RR (BR) *
BB BR BB RR (RR)
BB RR BB RR (BR) *
RR RR BR BB (BB) *
RR RR BB BB (BR) *
RR BR RR BB (BB)
RR BB RR BB (BR) *
BR BB BB RR (RR)
BR RR RR BB (BB)
You may want to check the distribution with the the star next to them. For instance, for the second distribution
BB BB RR RR (BR)
On his turn, B will reason that his hat is either BB or BR. If A saw [bB *BR* RR RR (BB)], he would have called his hat at turn 1. But A didn't, so B knows his hat.
Let's look at the first distribution, [ BB BB BR RR (RR) ]. On his turn, C reasons that his hat is either RR or BR. If it is RR, then B would have won on turn 2 due to the reasoning for distribution 2. There on his turn C would know his color.
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Here are the fractions:
A: 19/63
B: 31/63
C: 8/63
D: 5/63
Sorry about that, my bad. Since D has the lowest probability in the above table,
Could you list out the distributions that would lead to D winning?
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...choose position B. Unless I messed up in the calculations somewhere, the odds of winning are roughly
A: 30.16%
B: 49.21%
C: 12.70%
D: 7.94%
Multiplying the payouts by the odds, we get the best expected payout of $1.23 (per game) for position B.
Some comments
Since we are choosing 5 stamps out of 10, the total number of unique unaggregated stamp distributions is 10C5 = 252. We would then expect the probability of winning for each participant to be a fraction with 252 in the denominator.
Looking at the winning percentages above, I don't think they can be converted into such a fractional form.
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Interesting puzzle, Bushino. Can you clarify something for me?
Based on the method of selecting the stamps described in the OP, the probabilities of different combinations are not equal. For example, BB BB BR RR (RR in the pocket) is not as likely as BB BB RR RR (BR in the pocket). That's because after drawing 4 blue stamps out of the bag the chance of drawing another blue among the next 2 stamps is lower that drawing 2 red stamps.
Does the OP intend to account for these unequal probabilities or should it be assumed that all combinations are equally probable?
Definitely, k-man. Clarification are as follows
Let the patterns on the 4 players' head be the 8-dimensional binary vector V (e.g., (B,B,R,R,B,B,R,R) ), the drawing method in the OP is mean to assure that all instances of V are equally likely. As for the example you gave, the chance of drawing BB BB BR RR (RR in the pocket) is the same as drawing BB BB RR RR (BR in the pocket) since if we ignore the first (BB BB) on both, then
P[ BB BB BR RR (RR in the pocket) ] = (1/6) + (5/6)(1/5) = 2/6
P[ BB BB RR RR (BR in the pocket) ] = (5/6)(4/5)(3/4)*(2/3) = 2/6
Note that even though V is equally likely, the aggregated distribution of Blue = BB, Red = RR, and Mixed = BR or RB are not equally likely. For instance, the distribution (Mixed, Mixed, Mixed, Mixed) is a lot more likely than (Blue, Blue, Red, Red).
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Comments
So the question is "in what circumstance would C give up, and in what circumstances would he not?"
First, let's bring back the first answer, which we both agree on. In the case where giving up is not allowed, then A wins and the game goes as follows.
Turn 1- A: I don't know
Turn 2- B: I don't know
Turn 3- C: I don't know
Turn 4- A: I don't know
Turn 5- B: I don't know
Turn 6- C: I don't know
Turn 7- A: My number is 3.
Obviously, C only gets to answer twice. Once at turn 3 and once at turn 6. The rest of this post will show C has no strategy to win, therefore he has an incentive to give up.
Note that since 'Give up' means 'I don't know' + 'I forfeit my right to play further', an 'I don't know' is the same as 'Give up' on turn 5 or 6, since B and C don't get to play further in any instance.
I think we both agree so far that turn 1 and 2 should both be 'I don't know'. So, let's look at turn 3. C has no idea what his hat number is, so he can't win. If C chooses to give up, the the game goes as follows
A: I don't know
B: I don't know
C: I give up
A: I don't know
B: My number is 3.
If C chooses not to give up but to pass on turn 3, then the game goes as follows
Turn 1- A: I don't know
Turn 2- B: I don't know
Turn 3- C: I don't know
Turn 4- A: I don't know
Turn 5- B: I don't know/Give up (identical choices at this turn)
Turn 6- C: I don't know/Give up (identical choices at this turn)
Turn 7- A: My number is 3.
So essentially, C got no winning options if all hats are 3, so technically he would be required by the rules to give up on turn 3. Meaning that the game would proceed as follows
A: I don't know
B: I don't know
C: I give up
A: I don't know
B: My number is 3.
Almost there...
Save for two things. First, a clarification of the problem statement: the game continues until each logician has either named his number or given up. So more than one of them can potentially "win". Secondly: would C give up if A had a 2 instead of a 3?
Ah, more twist and turns. Thanks for such a great puzzle.
I erroneously interpreted a 'win' as being the first to name the hat number. My mistake. Defining a win as being able to name a hat number changes the games, since now C is motivated differently. The game now goes as follows
Turn 1- A: I don't know
Turn 2- B: I don't know
Turn 3- C: I give up
Turn 4- A: My number is 3 (C would not have given up if A's number is 2)
Turn 5- B: My number is 3.
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Getting warmer! Consider: in what circumstance would C give up, and in what circumstances would he not? When C gives up, the fact is very informative... to both A and B equally by symmetry...
Comments
So the question is "in what circumstance would C give up, and in what circumstances would he not?"
First, let's bring back the first answer, which we both agree on. In the case where giving up is not allowed, then A wins and the game goes as follows.
Turn 1- A: I don't know
Turn 2- B: I don't know
Turn 3- C: I don't know
Turn 4- A: I don't know
Turn 5- B: I don't know
Turn 6- C: I don't know
Turn 7- A: My number is 3.
Obviously, C only gets to answer twice. Once at turn 3 and once at turn 6. The rest of this post will show C has no strategy to win, therefore he has an incentive to give up.
Note that since 'Give up' means 'I don't know' + 'I forfeit my right to play further', an 'I don't know' is the same as 'Give up' on turn 5 or 6, since B and C don't get to play further in any instance.
I think we both agree so far that turn 1 and 2 should both be 'I don't know'. So, let's look at turn 3. C has no idea what his hat number is, so he can't win. If C chooses to give up, the the game goes as follows
A: I don't know
B: I don't know
C: I give up
A: I don't know
B: My number is 3.
If C chooses not to give up but to pass on turn 3, then the game goes as follows
Turn 1- A: I don't know
Turn 2- B: I don't know
Turn 3- C: I don't know
Turn 4- A: I don't know
Turn 5- B: I don't know/Give up (identical choices at this turn)
Turn 6- C: I don't know/Give up (identical choices at this turn)
Turn 7- A: My number is 3.
So essentially, C got no winning options if all hats are 3, so technically he would be required by the rules to give up on turn 3. Meaning that the game would proceed as follows
A: I don't know
B: I don't know
C: I give up
A: I don't know
B: My number is 3.
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Three logicians had stickers placed on their foreheads so that none could see their own sticker but each could see one another's. They were told that each sticker has a single positive integer written on it (i.e. 1, 2, 3, ...), and that the sum of the integers on all three stickers is either 8 or 9. They were then asked, in turn, to identify the number on their own sticker. Upon being asked, each logician would name their number if they were sure that they knew it, give up if they were sure that they would never know it, or otherwise 'pass' so that the question would be posed to the next person. The question was repeated, again in turn, until each of the three logicians had either named their number or given up. All three stickers had the same number written on them. Who among the three logicians was able to deduce his number?
Answer
Let's label the logicians A, B, and C. The game will go as follows
A: I don't know
B: I don't know
C: I don't know
A: I don't know
B: I don't know
C: I don't know
A: My number is 3.
Reasoning: start from logician A. On his first turn, if he were to see (b, c) such that (b + c) = 8, then he would have raised his hand and declare his number as 1. From the second turn, if B sees (a,c) such that (a + c) = 8 then he'd raise his hand and say 1. If he sees that a = 1, then he'll raise his hand and declare his number b is (b = 7 - c). Continues the reasoning until the 7th turn as above.
You are right that A would answer with his number the third time he is asked, assuming no one has given up by that point. But what about the other two logicians? Can they also answer, or should they give up? ...Or should they have given up sooner than this?
If the two logicians are required to give up if they figure they would never know it, then the game would go as follows
A: I don't know
B: I give up
C: I give up
A: My number is 3.
Same reasoning as above. B would give up at 2nd turn since he figures the (a,c) = (3,3) would not allow him to win at 5th turn anyways. C would also give up because given that B gave up and (a,b) = (3,3), his hat could still be 3 or 2. A then wins.
In fact B is in the most advantageous position. He will never give up, because in any case he can decide his number. He learns at least as much as A or C in each round of questioning.
I see what you're saying
If B knows that C would give up, then he shouldn't give up. The game would then be
A: I don't know
B: I don't know
C: I give up
A: I don't know
B: My number is 3.
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Three logicians had stickers placed on their foreheads so that none could see their own sticker but each could see one another's. They were told that each sticker has a single positive integer written on it (i.e. 1, 2, 3, ...), and that the sum of the integers on all three stickers is either 8 or 9. They were then asked, in turn, to identify the number on their own sticker. Upon being asked, each logician would name their number if they were sure that they knew it, give up if they were sure that they would never know it, or otherwise 'pass' so that the question would be posed to the next person. The question was repeated, again in turn, until each of the three logicians had either named their number or given up. All three stickers had the same number written on them. Who among the three logicians was able to deduce his number?
Answer
Let's label the logicians A, B, and C. The game will go as follows
A: I don't know
B: I don't know
C: I don't know
A: I don't know
B: I don't know
C: I don't know
A: My number is 3.
Reasoning: start from logician A. On his first turn, if he were to see (b, c) such that (b + c) = 8, then he would have raised his hand and declare his number as 1. From the second turn, if B sees (a,c) such that (a + c) = 8 then he'd raise his hand and say 1. If he sees that a = 1, then he'll raise his hand and declare his number b is (b = 7 - c). Continues the reasoning until the 7th turn as above.
You are right that A would answer with his number the third time he is asked, assuming no one has given up by that point. But what about the other two logicians? Can they also answer, or should they give up? ...Or should they have given up sooner than this?
If the two logicians are required to give up if they figure they would never know it, then the game would go as follows
A: I don't know
B: I give up
C: I give up
A: My number is 3.
Same reasoning as above. B would give up at 2nd turn since he figures the (a,c) = (3,3) would not allow him to win at 5th turn anyways. C would also give up because given that B gave up and (a,b) = (3,3), his hat could still be 3 or 2. A then wins.
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Three logicians had stickers placed on their foreheads so that none could see their own sticker but each could see one another's. They were told that each sticker has a single positive integer written on it (i.e. 1, 2, 3, ...), and that the sum of the integers on all three stickers is either 8 or 9. They were then asked, in turn, to identify the number on their own sticker. Upon being asked, each logician would name their number if they were sure that they knew it, give up if they were sure that they would never know it, or otherwise 'pass' so that the question would be posed to the next person. The question was repeated, again in turn, until each of the three logicians had either named their number or given up. All three stickers had the same number written on them. Who among the three logicians was able to deduce his number?
Answer
Let's label the logicians A, B, and C. The game will go as follows
A: I don't know
B: I don't know
C: I don't know
A: I don't know
B: I don't know
C: I don't know
A: My number is 3.
Reasoning: start from logician A. On his first turn, if he were to see (b, c) such that (b + c) = 8, then he would have raised his hand and declare his number as 1. From the second turn, if B sees (a,c) such that (a + c) = 8 then he'd raise his hand and say 1. If he sees that a = 1, then he'll raise his hand and declare his number b is (b = 7 - c). Continues the reasoning until the 7th turn as above.
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Approach
Call the two fuses A and B. The strategy is as follows
1) Light A from both ends and simultaneously light 1 end of B
2) When A burns out after 30 sec, immediately light the other end of B.
3) B will burn out at the 45 sec mark.
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1
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Here is yet another puzzle based on
Suppose that there is a game as follows
* There is a host with 10 stamps, 5 red and 5 blue. There are 4 players- A, B, C, and D.
* In the beginning, the host affixes two stamps to each of the 4 players' head. The choice of stamps for each player is completely random (i.e. the host puts all stamps into an opaque bag and then draws them one by one). The remaining 2 stamps go into the host's pocket. Each player can see the stamps on the remaining 3 players, but can not see his own stamps nor the two in the host's pocket.
* Starting from A to D (and then looping back to A and so on), the host asks if each player definitively knows his color (RR, BB, or RB). If the player does not know, the host goes on to the next player. First player to know his color wins. No guessing is allowed.
Suppose that the host likes you, so he secretly offers you a side bet before the game. You have to pay the host 1
dollar before the game starts, and then you can choose whether to be A, B, C, or D. The payout by position if you win is as follows
A: 3.5 dollars
B: 2.5 dollars
C: 6 dollars
D: 7 dollars
Which position should you choose for the greatest expected winnings? -
It may be fruitful to assume a distribution and then see how it can be determined.
It seems profitable to ask: Are any cases undecidable?
The most ambiguous case is likely xy xy xy xy xy (xy).
Everyone would see 5x and 5y and say (responses 6-10) I don't know. (idk)
Everyone knows they could be xy, therefore.
Suppose a player wonders if he is xx. Can he decide? No. At least not after responses 6-10.
If a player is xx (and sees the others are xy) he knows the others would see still see only 6 x's.
They could not therefore conclude (during 6-10) that they are yy.
After responses 6-10, no player can decide whether his colors are similar or not.
Go to the next round: Responses 11-15.
If one player is xx, (and responses 6-10 are all idk) the others can decide something.
They can decide they are not also xx: the non-xx players know they are either xy or yy.
But that's not enough to know which. Responses (11-15) will all be idk.
Go to the next round: Responses 16-20.
Every player knows the distribution might be xy xy xy xy xy (which it is)
but to be certain, he must know that he himself is xy:
he must rule out the possibility that his colors are similar.
Even If another player saw his (possible) xx, the other player could not
(during 11-15) declare his colors: he could be xy or he could be yy.
So (16-20) (which must still be all idk) must prove to at least one player that his colors are dissimilar.
For now, I don't see how. So provisionally, this case seems undecidable.
Of course, Bushindo would not do this to us. So I will think some more.
See the part in red. My calculations show that the players would definitely end the game during that phase for the starting case of all RB stamps.
I'm confused, as often. I thought we were facing one particular run of this game, and we are seeking the one case and reveal for which A does NOT announce his stamps at step 6. The phrase "starting case of all RB stamps" makes me think we have to make a general solution for all runs of the game.
I see your point. The confusion is entirely my fault. I revised the OP to remove this confusion.
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It may be fruitful to assume a distribution and then see how it can be determined.
It seems profitable to ask: Are any cases undecidable?
The most ambiguous case is likely xy xy xy xy xy (xy).
Everyone would see 5x and 5y and say (responses 6-10) I don't know. (idk)
Everyone knows they could be xy, therefore.
Suppose a player wonders if he is xx. Can he decide? No. At least not after responses 6-10.
If a player is xx (and sees the others are xy) he knows the others would see still see only 6 x's.
They could not therefore conclude (during 6-10) that they are yy.
After responses 6-10, no player can decide whether his colors are similar or not.
Go to the next round: Responses 11-15.
If one player is xx, (and responses 6-10 are all idk) the others can decide something.
They can decide they are not also xx: the non-xx players know they are either xy or yy.
But that's not enough to know which. Responses (11-15) will all be idk.
Go to the next round: Responses 16-20.
Every player knows the distribution might be xy xy xy xy xy (which it is)
but to be certain, he must know that he himself is xy:
he must rule out the possibility that his colors are similar.
Even If another player saw his (possible) xx, the other player could not
(during 11-15) declare his colors: he could be xy or he could be yy.
So (16-20) (which must still be all idk) must prove to at least one player that his colors are dissimilar.
For now, I don't see how. So provisionally, this case seems undecidable.
Of course, Bushindo would not do this to us. So I will think some more.
See the part in red. My calculations show that the players would definitely end the game during that phase for the starting case of all RB stamps.
A society that prizes girls
in New Logic/Math Puzzles
Posted
I think there is a discrepancy with bonanova's answer because...