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## Question

I once participated in a test where the answers had to be chosen from the first 10 letters of the alphabet (A-J).

Not bothered too much about the solution, I merely answered by matching the letters with numbers 1 to 10 in strict alphabetical order. I managed a score of only 2/10!

 I showed the result to a friend of mine who bettered the score to 5/10 with his attempt. The same test was passed on to five more people, each who was entitled to view all previous results. All of them scored 7/10, meaning that no one was spot on. (1)   What is the probability that the next person will get 10/10 on his first try? (2)   If you were to submit a solution as contestant № VIII, which letters would you enter to ensure that you obtain full marks? Here’s the score sheet for attempts I to VII:

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Spoiler

Two more solutions:

I J C E F D  G B A H

I E C F J D G B A H

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I J C F D E G B A H

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Are we guaranteed that no two questions have the same answer? Or are duplications possible? Player VI duplicated one letter, so I’m wondering about the latter possibility.

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 I J C F B D G E A H
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19 hours ago, CaptainEd said:

Are we guaranteed that no two questions have the same answer? Or are duplications possible? Player VI duplicated one letter, so I’m wondering about the latter possibility.

Intentially done so. Player VI could not make up his mind with two positions and duplicated one of the letters! The solution indeed has 10 different letters.

46 minutes ago, Jim56 said:
 I J C F B D G E A H

Edited by rocdocmac
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From player II to player III, only 3 positions changed: 1, 5, 10. Player III only gained 2 correct, though, so two of those positions changed are correct and one is incorrect.

From III to IV, 3 positions change again: 5, 8, 9. No score changes, however, which could be a result of a few situations:

I) 5-A, 8-E, 9-B (1 right, two wrong)

II)5-B, 8-A, 9-E (1 right, two wrong)

III) 5-E, 8-B, 9-A (1 right, two wrong)

IV) 5-E, 8-A, 9-B (all are wrong in both cases)

E, A, and B must all fit in 5, 8, or 9. Otherwise, the score would be out of necessity lower than 7 (it would have to be 6 or lower).

Therefore, players II, V, and VII all have their spot 5 wrong, since those players chose a letter besides E, A, or B for spot 5. Player I has spot 8 wrong and players I and VI have spot 9 wrong by the same logic for spots 8 and 9.

Also, player I has spot 1 and 2 wrong, player VII has spot 7 wrong, and player V has spot 10 wrong, since E, A, and B must be in either 5, 8, or 9.

so, player I has spots 1, 2, 8, and 9 wrong for sure. player II has spot 5 wrong, player V has spots 5 and 10 wrong, player VI has spot 9 wrong, player VII has both 5 and 7 wrong.

From IV to V, 3 positions change again: 5, 8, 10. No score change again.

But wait: we know for a fact that V has 5 and 10 wrong.

From III to V, 3 positions change again: 5, 9, 10. No score change again.

In fact, between any two players with a score of 7, only three positions change.

III-IV 5, 8, 9.

IV-V 5, 8, 10.

V-VI 5, 9, 10.

VI-VII 5, 7, 9.

VII is the only one that changes spot 7 between all players. (we also know that VII has both 5 and 7 wrong, as per above).

Comparing V to VII, 5, 7, and 10 change. We know that V is wrong in 5 and 10, and VII is wrong in 5 and 7, and only 3 are wrong in either V or VII. Two options, then:

I)It's possible that both V and VII have 5, 7, and 10 wrong, making the correct sequence 5-E, 7-H, 10-B.

II) otherwise, some number they share in common is incorrect, and V has 7 correct and VII has 10 correct, meaning 7-G and 10-H.

The next step is to walk through those last two scenarios... but I'm tired, and it's getting late. I'll leave it for someone else, or otherwise when I have time. Am I on the right track, Rocdocmac?

It's been a long, long time since I played mastermind....

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Flamebirde ...  there's an additional score (7/10) by Jim56 to consider!

Spoiler

Meanwhile, you are virtually on the right track, especially with your concluding remarks!

Edited by rocdocmac
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19 hours ago, flamebirde said:
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From player II to player III, only 3 positions changed: 1, 5, 10. Player III only gained 2 correct, though, so two of those positions changed are correct and one is incorrect.

From III to IV, 3 positions change again: 5, 8, 9. No score changes, however, which could be a result of a few situations:

I) 5-A, 8-E, 9-B (1 right, two wrong)

II)5-B, 8-A, 9-E (1 right, two wrong)

III) 5-E, 8-B, 9-A (1 right, two wrong)

IV) 5-E, 8-A, 9-B (all are wrong in both cases)

E, A, and B must all fit in 5, 8, or 9. Otherwise, the score would be out of necessity lower than 7 (it would have to be 6 or lower).

Therefore, players II, V, and VII all have their spot 5 wrong, since those players chose a letter besides E, A, or B for spot 5. Player I has spot 8 wrong and players I and VI have spot 9 wrong by the same logic for spots 8 and 9.

Also, player I has spot 1 and 2 wrong, player VII has spot 7 wrong, and player V has spot 10 wrong, since E, A, and B must be in either 5, 8, or 9.

so, player I has spots 1, 2, 8, and 9 wrong for sure. player II has spot 5 wrong, player V has spots 5 and 10 wrong, player VI has spot 9 wrong, player VII has both 5 and 7 wrong.

From IV to V, 3 positions change again: 5, 8, 10. No score change again.

But wait: we know for a fact that V has 5 and 10 wrong.

From III to V, 3 positions change again: 5, 9, 10. No score change again.

In fact, between any two players with a score of 7, only three positions change.

III-IV 5, 8, 9.

IV-V 5, 8, 10.

V-VI 5, 9, 10.

VI-VII 5, 7, 9.

VII is the only one that changes spot 7 between all players. (we also know that VII has both 5 and 7 wrong, as per above).

Comparing V to VII, 5, 7, and 10 change. We know that V is wrong in 5 and 10, and VII is wrong in 5 and 7, and only 3 are wrong in either V or VII. Two options, then:

I)It's possible that both V and VII have 5, 7, and 10 wrong, making the correct sequence 5-E, 7-H, 10-B.

II) otherwise, some number they share in common is incorrect, and V has 7 correct and VII has 10 correct, meaning 7-G and 10-H.

The next step is to walk through those last two scenarios... but I'm tired, and it's getting late. I'll leave it for someone else, or otherwise when I have time. Am I on the right track, Rocdocmac?

It's been a long, long time since I played mastermind....

Alright, here we go...

Spoiler

In either case, 7 must be either H or G, and 10 must be either B or H. So, to extend: players I, II, and V all have slot 10 wrong.

to recap:

I has 1, 2, 8, 9, 10 wrong.

II has 5 and 10 wrong.

V has 5 and 10 wrong.

VI has 6 wrong.

VII has both 5 and 7 wrong.

But hold on - if case I is true, and the correct sequence is 5-E, 7-H, 10-B, then this contradicts our initial conclusion above: i.e., that E, A, and B must correspond to slots 5, 8, and 9. B cannot fit in slot 10, or else the given scores cannot be correct!

Therefore, case II must be right. V has 7 correct and VII has 10 correct: 7 must be G, and 10 must be H.

This means that a G anywhere besides 7 is wrong, and a H anywhere besides 10 is wrong. so II must also have 1 wrong.

Next, we analyze III vs VI. What we see is really interesting: III and VI differ only in number 9, and both give a result of 7. Therefore, neither E nor I can fit in spot 9. Players II and III have spot 9 wrong.

to recap:

I has 1, 2, 8, 9, 10 wrong.

II has 1, 5, 9 and 10 wrong.

III has spot 9 wrong.

V has 5 and 10 wrong.

VI has 6 wrong.

VII has both 5 and 7 wrong.

So... if E cannot fit in spot 9, and also cannot fit in spot 5, then E must fit in spot 8, forcing 5 to be A and 9 to be B. So, 5-A, 7-G, 8-E, 9-B, and 10-H must be correct. Putting this together with the initial four letters I J C F, and via process of elimination, I think this is the correct sequence:

I J C F A D G E B H

Except... it can't be, because III only two letters off and only has a score of 7. I've made a mistake somewhere, and I'm not sure where: I think I can't assume that the first four letters are I J C F. If we look back at II and III, then we see that they differ in 1, 5, and 10, for a net change of 2. Since we know for a fact that II has gotten 1, 5, and 10 all wrong, then that means III must have two of those three positions correct. We also know that 10 must be H, and 5 must be A, making the I in position 1 incorrect.

But this leads to a whole slew of problems with Player V's results... I don't trust the idea of I in position 1 being wrong, not yet at least. I think the italicized section above is where my reasoning goes wrong, but I'm not sure. I think I'll leave it here for now.

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@Flamebirde ...

Spoiler

Not that it will really change anything drastically in your reasoning, but have another look at your six recap statements. One of them is incorrect (possibly just a typo)!

The first player started with two correct answers and 8 wrong. Believing that all of the incorrect answers have been identified, eight changes were made by Player II, but the score was only upgraded by 3 (new) truly correct answers.

Choosing 2 correct answers from 10 for Player I (A&B, A&C, ..., I&J) already gives us 45 combinations to play around with for starters. Now one has to permute the other eight so-called "correct" answers in each of those 45 categories and then it becomes difficult and a lengthy exercise to unravel the true solution. In fact, approximately 1.8 million possible rearrangements altogether!

Believe me, there is a firm solution that satisfies all of the seven test scores.

Thank you for your mighty effort so far!

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1 hour ago, mtngoat said:

I J C F D E G B A H

Spoiler

Ahhh... sucks to be beaten to the punch after investing so much time! Congratulations on the answer. How'd you arrive at it? (I guess my very first conclusion was wrong, and it threw everything else off...)

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flamebirde, sorry to have spoiled your fun. I'm new to this site, and didn't notice I had to hit the eye in order to hide my answer. Please forgive a newbie.

Spoiler

flamebirde, a few things right off:

III & VI differ only on #9 (hence rocdocmac's having VI repeat one answer.) Since they both got 7 right, both missed #9.

II & III agree on 7 questions.  5 of these must be right. ( If only 4 were right, III must have gotten the remaining 3 right, leaving II with only 4.)

III, then, must have gotten 2 of the remaining 3 right.

In the sense of fair play, I used only the original clues; however, as a shortcut,  looking at your previous answer, you differed from III only in that you switched #8 & #9, dropping your score to 6. Therefore, III got one of those right & one wrong. Since we've already seen that they got #9 wrong, their #8 is correct.

On looking at your work, I noted:

On 8/7/2019 at 12:17 AM, flamebirde said:

E, A, and B must all fit in 5, 8, or 9. Otherwise, the score would be out of necessity lower than 7 (it would have to be 6 or lower).

This is where you went wrong, and not exactly  sure how you arrived at that one. Happy puzzling!

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@mtngoat ...

Spoiler

Well done! This is one of three combinations that are acceptable! Not the one I originally wanted, but yes, as the OP currently stands, it works just as well.

@flamebirde

Spoiler

There are two other solutions and maybe you can find one or both of them!

@Brainden

Spoiler

The challenge ...

Suppose that the 10 answers all had to do with colors, e.g. amber, blue, crimson, dandelion, ebony, fuchsia, green, orange, indigo and juniper. If you were tipped by the examiner that the answer to no. 6 is definitely D, with no other letter considered for this position,  what would the solution then look like?

Edited by rocdocmac
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I came up with this answer. pls see attached file.

Edited by SamSam11
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15 hours ago, mtngoat said:
Hide contents

Two more solutions:

I J C E F D  G B A H

I E C F J D G B A H

That's it!

Spoiler

Also correct, but mtngoat was first with all three solutions!

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