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flamebirde

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Posts posted by flamebirde


  1.  
    1 hour ago, mtngoat said:

    Happened on this site this afternoon, and saw your post. Looks like fun. Here's your answer.

    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...)


  2. 19 hours ago, flamebirde said:
      Hide contents

    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.

     


  3. Spoiler

    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....


  4. Is the owner of the gun automatically the murderer?

    Spoiler

    It was Joe

    Initial assumption: If Bill was shot with Joe's gun then that makes Joe the killer. Therefore, if we assume this statement to be true, then Jack didn't kill Bill, making that statement true as well, making the other two false.

    Jack's statements:

    Jack:
    I didn't do it. True - assumption.
    Al did it. False - by process of elimination.
    Joe and I were at the movies last night when Bill was shot. False - by process of elimination.
    Bill was shot with Joe's gun. True - assumption.

    Now onto Joe's statements:

    Joe:
    I was asleep when Bill was shot. False - he had to be awake to shoot Bill.
    Al lied when he said that Tom killed Bill. True - Joe killed Bill, not Tom.
    Jack is the only one of us who owns a gun. False - Joe owns a gun, the one used to kill Bill.
    Tom and Bill were pals.True by process of elimination.

    And for Tom?

    Tom:
    I've never fired a gun in my life. True - by process of elimination.
    I don't know who did it. True - by process of elimination.
    Joe doesn't own a gun. False - Joe shot Bill with his own gun.
    I never saw Bill until they showed me the body False - Tom and Bill were pals.

    Lastly, Al:

    Al:
    I didn't do it. True - Joe did it.
    Tom did it. False - Joe did it.
    Sure I own a gun. True - by process of elimination.
    Joe and I were playing poker last night when Bill was shot. False - Joe could not have played poker at the same time as shooting Bill.

     


  5. Spoiler

    So the chances that the first two will match is 1/52. That much should be fairly obvious I think.

    If they do, then the chances of the next two matching is 1/51, and if they do then the chances of the next two are 1/50, and so on.

    If a pair doesn't match, then we're looking at an issue -- at least one more pair is guaranteed not to match... but there's some finicky stuff going on there, since you could have a loop of 3 or more unmatched pairs if it chains down a bit. So at minimum if you draw a non-matching card then you have at least 2 unmatched pairs right there and potentially more. Not to mention that you could spawn different "loops" in one run through.

    if a pair doesn't match, then the chances of the next pair matching are... what? That's where the riddle lies next I guess. If we number the first deck 1-52 and the second 1'-52' and the very first pair is 1-2', then we know that the next pair (the one starting with 2-) must also not match. Then there's a 1/51 chance that the loop ends there -- a 1/51 chance that the next pair is 2-1', in which case the probabilities continue as before. But there's also a 50/51 chance that the chain extends, and we draw something like 2-5', which means that now the fifth pair (the one starting with 5-) is guaranteed to be a non-match with a 1/45 chance of terminating.

    I'm sure there's an easier way to think about it, but I'm gonna let it percolate through the back of my head for now.

    Seems similar to that airplane seating problem a while back, I'll see if I can't find and link it here.

    Edit: here it is:

     

     


  6. 19 hours ago, rocdocmac said:

    Both Captain Ed and Thalia have transmitted vital clues in their previous remarks. Eish man, now you ought to get it!

    Spoiler

    ../ -  ....  ..  -.  -.-/ .. / ..-.  ..  --.  ..-  .-.  .  -../ ..  - / ---  ..- !

    What hath God wrought!

     

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