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This is a puzzle of five men guessing how many stars will appear from tossing four coins.

coins


There are five men: Mr. Zero, Mr. One, Mr. Two, Mr. Three, and Mr. Four; playing with four coins. For each coin, one side is a star and the other side is blank (no star on this side).

The men guess how many stars will appear from tossing the coins one by one. There are no duplicate guesses (i.e. all men guess a different number from the others). Also, the number of men that pick their own name as a guessing number is not exactly 1 (may be 0, or more than 1).

  • First coin is tossed, and Mr. Two announces that he lost.
  • After the second coin toss, Mr. Three says "Oh no!" in panic and Mr. Zero surrenders.
  • Third coin is tossed, and Mr. One is still hoping for a win.
  • The last coin is tossed and it shows a side without a star, and now Mr. Four is clearly a winner!

Can you find out what are the men's guesses and which sides appear from all of the tosses?

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  • 18
    $\begingroup$ That's a pretty misleading image there: I counted 13 stars on one side and one on the other. $\endgroup$ – Laurel Jan 22 '18 at 18:15
  • $\begingroup$ I like naming the men "Mr. Brown," "Mr. Green," "Mr. Pink," etc. better :-) $\endgroup$ – Carl Witthoft Jan 22 '18 at 20:07
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    $\begingroup$ Flowery language aside: are we to infer that "surrenders" means he's mathematically lost, as opposed to giving up because the odds are strongly against him? Similar for "oh no" - does this mean he's lost or just that his chances went down to near but not zero? $\endgroup$ – Carl Witthoft Jan 22 '18 at 20:09
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    $\begingroup$ @CarlWitthoft each toss only eliminates one person, so Mr. Zero and Mr. Three cannot both lose at the same toss. The language of the next line, Mr. One still hoping for a win, suggests that Mr. Three is eliminated only at that point. I didn't like the ambiguousness of it, either. $\endgroup$ – Rob Watts Jan 22 '18 at 20:35
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    $\begingroup$ @Carl Witthoft, yes, "surrenders" means he's mathematically lose (i.e. it's impossible for him to win.) And as Rob Watts said, Mr. Three doesn't lose at this toss because of Mr. Zero surrendering. Also, we can deduct that he said "oh no" is to imply that the chance for him to win is smaller compared to if the coin showed the other side. $\endgroup$ – athin Jan 22 '18 at 23:45
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For my answer I'm considering the sides of the coin to either have a star or be blank for ease of writing. I also will write out the number for peoples name, and use the character for guesses to try and keep things legible. (Ex. Mr Three guessed 4). Also when typing out sequences, + means star and - means blank. (Ex ---- means no stars / all blanks)

I agree with Kate Gregory's Answer but there is a flaw in the logic.

in the second flip she said:

Mr Zero surrenders, which he would do if he had guessed (in order of the possible combos) 1, 0 (but he would have done so a turn earlier), 0 or 4, or 4. But for the last one, again, he would have done it a turn earlier. Therefore Mr Zero is 0 or 1, and he can't be 0 so he is 1 and the first two tosses are **. Also Mr Two is a 0.

however that's not true. because its also possible that...

Guessed 3 and the result was -- . It was noted that Mr Zero could have guessed 4 and conceded when +- happened. However this is ignored at the end and it's assumed that they only could have guessed 0 or 1. The assumption of the results being -- is correct but couldn't have been determined by the logic given.

How I got to my answer:

I listed all of the possible flip orders. I've lettered them for ease of reference. There are only 8 because we are told the result of the last flip.

A) ----
B) --+-
C) -+--
D) -++-
E) +---
F) +-+-
G) ++--
H) +++-

After that I assigned guessed to people based on the clues.

From First Flip:

In A through D, Mr Two would had to have guessed 4, and in E through H they would have had to guess 0.

Second Flip:

We have Mr Zero guessing 3 for A/B, 0 for C/D, 4 for E/F, and 1 for G/H. I am ignoring Mr Three for now.

Third Flip:

At this point we are told Mr One is still in it to win it. but with a single flip left, there would only be 2 possible outcomes left. so if Mr Four wins and Mr One is still in, Mr Three must be out. Which means in Alphabetical order Mr Three would have guessed: 2, 0, 3, 1, 3, 1, 4, 2

Final Flip:

Now we set Mr Fours guess to whatever the right answer is and Mr One to one above that.

So the guesses would have to be (in name order)

A) 3,1,4,2,0
B) 3,2,4,0,1
C) 0,2,4,3,1
D) 0,3,4,1,2
E) 4,2,0,3,1
F) 4,3,0,1,2
G) 1,3,0,4,2
H) 1,4,0,2,3

however there are still 2 clues I've ignored up until this point.

We cannot have a set of guesses where only a single person guessed their name/number which means that A, D, and E are all out. leaving B, C, F, G, and H.

Then there is the last clue I've ignored.

Mr Three panicked when the second coin was flipped.

so lets look at our remaining options:

In B Three guessed 0 and I think it would have been weird for them to panic when 2 blanks come up in a row as that's exactly what needed to happen for them to even have a chance. In C Three guessed 3 and the result was a second star. Which while it's not required that they get a star, it's still better than a blank. So I would find it weird for them to freak out. In F they guessed 1 and the second flip had come up blank after a star which again they needed to happen for them to win. I know people who would still do this, which is why I didn't rule this out when first discussing flip 2. In G we see the exact opposite of B, two stars in a row when they guessed 4, which is exactly what needs to happed.
Lastly there is H which had 2 Stars come up in a row. In this case Three guessed 2 and they actually had some leeway in this one. either result could eventually have won for them, but it's less likely if the result was a star. So for me this is the only one that makes sense for them to panic for.

so with that my guess is:

H) +++- , and the guesses in Name order are 1, 4, 0, 2, and 3 respectively with no one guessing their own name/number.

EDIT: Adjusted some formatted that I realized was off and fixed a spot where i excluded C for the wrong reason as was pointed out in the comments

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  • $\begingroup$ This matches what I came up with and the logic for it. I also agree about the flaw in the logic of Kate Gregory's answer for the second flip. $\endgroup$ – Rob Watts Jan 22 '18 at 21:00
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    $\begingroup$ This is a good strategy and explanation. Small nitpick, "We cannot have a set of guesses where only a single person guessed their name." Why does that eliminate C? I show two people, 0 & 3, guessing their name. But C gets eliminated in the next step, so it doesn't matter. $\endgroup$ – BWhite Jan 22 '18 at 21:38
  • $\begingroup$ Apart from what @BWhite said, your explanation is flawless! Thanks for the answer! $\endgroup$ – athin Jan 23 '18 at 0:29
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The first toss is a star (*) or not (-). Mr Two immediately loses.

He therefore guessed either 0, and it's *, or 4, and it's -. He did not guess 1, 2, or 3.

The second flip can be either, so we have **, *-. -*, or --.

Mr Zero surrenders, which he would do if he had guessed (in order of the possible combos) 1, 0 (but he would have done so a turn earlier), 0 or 4, or 4. But for the last one, again, he would have done it a turn earlier. Therefore Mr Zero is 0 or 1, and he can't be 0 so he is 1 and the first two tosses are **. Also Mr Two is a 0.

Also on that flip,

Since Mr Three has said Oh no, he is probably a 2 and is hoping for no more stars. He can't be a 3.

The third and fourth flip

Mr One is hoping to win at **? but does not at **?-. He therefore didn't guess 0 or 1 (they are taken and he would have lost by then) or 2 (he would win if he had 2 after 3 tosses and still had two after the 4th toss.) He is 3 or 4.

and

Mr Four actually won, and can't be a 4 because of his name, so by the same logic as Mr One, he must be a 3. That makes Mr One a 4 and means the third flip was *.

To sum:

Zero: 1. One: 4. Two: 0. Three: 2. Four: 3. Each number is used exactly once, nobody guessed their name. The flips are * * * -

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  • $\begingroup$ It seems like you're ruling people out from guessing the same as their name. The rules state "Also, the number of men that pick their own name as a guessing number is not exactly 1 (may be 0, or more than 1)." $\endgroup$ – ale10ander Jan 22 '18 at 23:04
  • $\begingroup$ That said, I still agree with your answer. $\endgroup$ – ale10ander Jan 22 '18 at 23:15
  • $\begingroup$ it's possible there are some other combinations that work @ale10ander $\endgroup$ – Kate Gregory Jan 22 '18 at 23:17
  • $\begingroup$ I'm pretty sure that this question only has one possible combination. And as @Alex Holder said on his answer, there is a flaw on your second flip. Pointing out: Mr. Zero may guess 4 on second combo or guess 3 on fourth combo, which makes him surrender right on this flip. But kudos for the answer, thanks! :D $\endgroup$ – athin Jan 22 '18 at 23:55
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First coin is tossed, and Mr. Two announces that he lose.

First toss was a star, Mr. Two announced 0 stars so it's over for him.

After the second coin toss, Mr. Three says "Oh no!" in panic and Mr. Zero surrenders.

This toss was a star again, Mr. Three announced 2 and it's over for Mr. Zero who announced 1.

Third coin is tossed, and Mr. One is still hoping for a winning.

This toss was a star too, Mr. One announced 4 stars, Mr. Three is out.

The last coin is tossed and it shows a side without a star, and now Mr. Four is clearly a winner!

Mr. Four announced 3 stars, this toss wasn't one so Mr. One is out.

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  • 2
    $\begingroup$ how did you rule out the case where the first toss was the opposite of what you had, but Mr Two still knows he lost? $\endgroup$ – Kate Gregory Jan 22 '18 at 15:44
  • $\begingroup$ If the first toss is not a star, Mr. One will announce 1 star, but "the number of men that pick his name as his guessing number is not exactly 1" $\endgroup$ – Saeïdryl Jan 22 '18 at 15:50
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    $\begingroup$ That does not follow at all. If the first toss is not a star, the second might be, so why would the person who had guessed one say anything? $\endgroup$ – Kate Gregory Jan 22 '18 at 16:11
  • $\begingroup$ If the first is not and the second is a star, why would Mr. Zero surrenders and why would Mr. Three panicked ? And I was talking about Mr. One, if we begin with a non-star toss, you will end with Mr. One announcing 1, and Mr. One says something at the 3rd toss. $\endgroup$ – Saeïdryl Jan 22 '18 at 16:17
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    $\begingroup$ There is no problem with mister one announcing 1: If another player announce his own number. You are right but your explanation missed the clue: last coin is not a star! $\endgroup$ – Untitpoi Jan 22 '18 at 16:37
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Going through the flips

First

Mr. Two is out. He picked 0 or 4.

Second

Mr. Three panics. This means he couldn't have lost on this flip, but could lose on the next one. This is only possible if he guessed 2 and the second flip was the same as the first.
Mr. Zero is out. Since the two flips had the same outcome, he guessed 1 if Mr. Two guessed 0 and guessed 3 if Mr. Two guessed 4.

Third

This isn't useful yet, so we move on.

Fourth

The last flip is _ and Mr. Four wins. This means he didn't guess 4. Further, Since Mr. Three lost, the number of stars flipped wasn't 2.

From this information, we know that the only possible solution sets are

A) Zero: 1, One: 4, Two: 0, Three: 2, Four: 3, Flips: ***_
B) Zero: 3, One: 0, Two: 4, Three: 2, Four: 1, Flips: __*_
C) Zero: 3, One: 1, Two: 4, Three: 2, Four: 0, Flips: ____

Going back to the third flip eliminates B, as Mr. One would know he can't win, while C is eliminated by the knowledge that exactly one name can't match. Thus, the solution is

Zero: 1, One: 4, Two: 0, Three: 2, Four: 3, Flips: ***_

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  • $\begingroup$ I do really like this solution, it's simple and perfect! Oh my, I don't know which answer should I accept :< (vs @Alex Holder's answer, as he answered earlier but still having a minor error.) $\endgroup$ – athin Jan 23 '18 at 0:32
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I wrote out the outcomes in a tree in notepad:

Line Format: Surname: Guessed number of stars (Current number of stars[Remaining coins])

Outcome Tree for Coin Flips

Explanation

First Flip:

Due to Mr. Two losing on the first flip, he must have either guessed 0 or 4 stars would show. This is because if the first coin shows a star, the only way Mr. Two could lose would be if he guessed no stars would show up; likewise, if the first coin showed no star, he must have guessed 4 stars would show up.

Second Flip:

Mr. Three:

My explanation of Mr. Three's choice is long winded enough that it has been shifted to the bottom of my answer.

Mr. Zero:

Following the tree, if Mr. Two guessed 0, Mr. Zero must have either guessed 1 or 4 stars to lose after the second flip; however, if Mr. Two guessed 4, Mr. Zero must have either guessed 0 or 3.

Third and Fourth Flips:

Mr. One and Mr. Four:

Interestingly, the tree shows us that the result of the third flip must be a star, because otherwise Mr. One and Mr. Four will share the same guess (At this point I will stop listing all the possible values in this text box), and the guess of Mr. One must be larger than the guess of Mr. Four by 1 star.

Mr. Three:

By examining the results of the previous 4 coin flips we can choose the value of Mr. Three's guess by simply choosing the guess which hasn't yet been taken by another man.

On the panic of Mr. Three (thanks to eyeballfrog):

The range of possible stars from the remaining coin flips is can be represented as [Current number of stars, (Current number of stars + Remaining coins)]. One would expect that the only cause Mr. Three would have for panic would be if his guessed approached one of the bounds of this range. The two possible ranges (in order of the tree) before the second coin flip are: [1, 4], and [0, 3], while the four possible ranges after the second coin flip are: [2, 4], [1, 3], [1, 3], and [0, 2]. Given the four possible values of Mr. Three's guess, and comparing them to the change in ranges caused by the second coin flip, the only valid guess which would cause Mr. Three to become panicked is a guess of 2 where the range changed from [1, 4] to [2, 4]. All of the other valid paths would have given Mr. Three no cause for panic.


Result

Mr. Two guessed 0, Mr. Zero guessed 1, Mr. One guessed 4, Mr. Four guessed 3, and Mr. Three guessed 2. The coin flips (in order) are star, star, star, no star.

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    $\begingroup$ Your logic about the panic is flawed. Mr. Three does panic in scenario 1 because he could lose on the next flip, whereas he couldn't before. This makes it the unique solution. $\endgroup$ – eyeballfrog Jan 22 '18 at 23:54
  • $\begingroup$ This is actually a good answer, I don't know why it doesn't get upvote hmm.. anyway, as @eyeballfrog said, there is a flaw on the panic logic, maybe you can check my comment on the question to make it clear. And ya, here a +1 from me ^^ $\endgroup$ – athin Jan 23 '18 at 0:15
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Ok...

Mr. Zero said 0 (matches his name)
Mr. One said 1 (matches his name)
Mr. Two said 4
Mr. Three said 3 (matches his name) Mr. Four said 2

Explanation

toss #1 = no star, Mr. Two loses because he said 4
toss #2 = star, Mr. Zero surrenders because he said 0; Mr. Three panics because he must get stars on the next 2 tosses (good luck Mr. Three)
toss #3 = no star, still only 1 star tossed so far so Mr. One is still hoping for a win, but...Mr. Three is out because he said 3 stars and that is impossible now (doesn't say but he's a goner)
toss #4 = star, and with the 2nd star, Mr. One is out and Mr. Four has won because he said 2 stars

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  • $\begingroup$ Welcome to Puzzling Stack Exchange! Please use >! to hide your intended solution and analysis behind 'spoilers'. $\endgroup$ – Glorfindel Jan 23 '18 at 18:17
  • $\begingroup$ We're told that toss #4 is not a star $\endgroup$ – Rob Watts Jan 23 '18 at 20:55
  • $\begingroup$ waaaah...i missed that...should i delete mine then? new here.... $\endgroup$ – ryno Jan 23 '18 at 21:13
  • $\begingroup$ Ryno - an answer you know doesn't solve the puzzle as stated is technically "not an answer" and can be downvoted and/or deleted as such. It's best to clean up after yourself; no sense keeping an answer you know is not useful. $\endgroup$ – Rubio Jan 24 '18 at 20:30
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In my solution let 1 represent stars and 0 represent blanks. 1100 being the final result satisfies all the requirements.

Mr 2:

Guessed 0000 and loses. Guessing 1111 or 0000 as it is the only possible way to lose on the first throw.

Mr 0:

Guessed 1000 and loses with the tosses now being 11??. Guessing 0111, 1111, 1000, or 0000, is the only possible way to be out on the second throw with Mr 2 already ruling out 0000.

Mr 3:

Guessed 1111 and says "Oh no!" as a distraction in this order of coin toss results.

Mr 1:

Guessed 1101 where toss = 110?. Mr 3 loses here.

Mr 4:

Guessed 1100 and wins. Toss = 1100. Mr 1 loses here.

It looks like there are multiple solutions. For instance,

according to ryno's solution, the toss order could be 0110. Mr 2=1111, Mr 0=0000, Mr 3=0111 and needs 2 more 1s, Mr 1=0100 and is out on the 3rd throw, Mr 4=0110 and wins. This satisfies the Oh no! requirement of Mr 3.

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  • $\begingroup$ The guess is about the count, not the sequence. $\endgroup$ – Weijun Zhou Jan 23 '18 at 19:12

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