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Posers and Puzzles

Posers and Puzzles

  1. 14 Jul '10 13:55
    three people are having a duel. we will call them Poor, Average, and Excellant. Poor only gets one shot out of three, Average gets two shots out of three, but Excellant hits his target every time. it will be solved by firing in rotation until only one is left. Poor is given first shot. what should he do?
  2. Standard member forkedknight
    Defend the Universe
    14 Jul '10 14:11 / 1 edit
    Originally posted by jasperdash
    three people are having a duel. we will call them Poor, Average, and Excellant. Poor only gets one shot out of three, Average gets two shots out of three, but Excellant hits his target every time. it will be solved by firing in rotation until only one is left. Poor is given first shot. what should he do?
    He fires at Excellent. If he were to fire at Average and kill him, he would be shot by Excellent immediately. This way he probably has at least one more shot.

    Each dueler should fire at the biggest threat.
  3. 14 Jul '10 14:27
    that is good, but not the right one. besides, if he didnt hit him, Excellant would surely kill him.
  4. Standard member forkedknight
    Defend the Universe
    14 Jul '10 15:31 / 1 edit
    Originally posted by jasperdash
    that is good, but not the right one. besides, if he didnt hit him, Excellant would surely kill him.
    Excellent might fire at Average. Think about it. If Poor killed Average, now it's Excellent's turn. Who does Excellent fire at (and kill)?

    You really only have to weigh what happens if poor hits, because if poor misses, Average must fire at Excellent or else he almost certainly looses. Let's assume poor misses and it's Averages turn:
    1) A->P and hits (2/3). E->A. A looses
    2) A->P and misses (1/3). E chooses between threats of 1/3 and 2/3. Kills A.
    3) A->E and hits (2/3). Now it's A vs P (Best case)
    4) A->E and misses (1/3). E chooses between threats of 1/3 and 2/3. Kills A.

    P vs A with P to fire is the best case for both A and P after 1 rotation. Both A and P want E dead.


    Average will never fire at Poor. ::= Poor should fire at Excellent.
  5. 14 Jul '10 16:54
    shoot at ground.
    (average will shoot at excellent coz hes better then poor, then if he hits him then poors gotta better chance coz only 1 person compared to 2. but if he misses, excellent will aim for average coz hes worse then poor. average dies, then poor has a better chance to win now then at the start coz 1 person rather then 2)
    p.s. do u read jasper dash?
  6. Standard member PBE6
    Bananarama
    14 Jul '10 17:19
    It's funny, I've heard this one before so I know what the "thinking outside the box" correct solution is supposed to be, but the math isn't matching up. So far I get the same answer as forkedknight, that Poor (P) should fire at Excellent (E) first, instead of firing at Average (A) or the "thinking outside the box" solution. Let me explain:

    CASE 1: P shoots at E first

    In order for E to win this round, P must miss E on his first shot, and then A must miss E on his first shot. (If A decided to shoot at P instead, E would maximize his chances of survival by eliminating A next, so A shooting at P is a bad idea here.) If this happens, E will eliminate A next, and so he must hope that P misses his subsequent shot, after which E will eliminate P and win the round. The chance of all this happening is (2/3)*(1/3)*(1)*(2/3)*(1) = 4/27.

    There are several ways for A to win this round, but they can be simplified into 2 branches:

    (1) P shoots and kills E. After this, A can win by shooting and killing P on his first shot, or by missing, hoping that P misses on his second shot, then shooting and killing A, or by missing, hoping, missing, hoping, killing, or by missing, hoping, missing, hoping, missing, hoping, killing, or by etc... The chance of this happening is calculated using the formula for the sum of an infinite geometric series S = a/(1-r) with a = (1/3)*(2/3) and r = (1/3)*(2/3). Therefore, S = (1/3)*(2/3)/(1 - (1/3)*(2/3)) = (2/9)/(7/9) = 2/7.

    (2) P shoots and misses E. After this, A can win by shooting and killing E, hoping that P misses on his next shot, then shooting and killing P, or by killing E, hoping, missing, hoping, killing P, or by etc... This is another sum, this time with a = (2/3)*(2/3)*(2/3)*(2/3) and r = (2/3)*(1/3). Therefore, S = (2/3)*(2/3)*(2/3)*(2/3)/(1-(1/3)*(2/3)) = (16/81)/(7/9) = 16/63.

    Adding these two branches together, we get P(A) = 2/7 + 16/63 = 34/63.

    The probability that P wins the round is just 1 - P(A) - P(E), or 1 - (34/63) - (4/27) = (189 - 102 - 28)/189 = 59/189

    Therefore, P(P) = 59/189, P(A) = 102/189 and P(E) = 28/189.

    CASE 2: P shoots at A first

    Following an argument similar to the one above, we find that:

    P(E) = (1/3)*1 + (2/3)*(1/3)*(1)*(2/3)*1 = 13/27

    P(P) = (2/3)*(2/3)*(1/3)/(1-(1/3)*(2/3)) = (4/27)/(2/9) = 4/21

    P(A) = 1 - P(E) - P(P) = 1 - 13/27 - 4/21 = (189 - 91 - 36)/189 = 62/189

    Converting these results to compare to Case 1, we have P(P) = 36/189, P(A) = 62/189 and P(E) = 91/189.

    CASE 3: P misses on purpose!

    This is the "thinking outside the box" solution. The reasoning goes that by deliberately missing on his first shot, P shifts the burden of having to kill E onto A, who is a better shot. If A kills E, then P gets the first shot against lesser opponent A. If A misses E, E will kill A still leaving P with the first shot against the strongest opponent E. However, the math works out as follows (unless I made a mistake somewhere):

    P(E) = (1/3)*(1)*(2/3)*(1) = 2/9

    P(P) = (2/3)*(1/3)/(1-(1/3)*(2/3)) = (2/9)/(7/9) = 2/7

    P(A) = 1 - P(E) - P(P) = 1 - 2/9 - 2/7 = (189 - 42 - 54)/189 = 93/189

    Converting these results to compare with Cases 1 and 2, we have P(P) = 54/189, P(A) = 93/189 and P(E) = 42/189.

    Results and discussion

    Here are the results for all three cases tabulated for easy reading:

    Case 1
    P(P) = 59/189, P(A) = 102/189 and P(E) = 28/189

    Case 2
    P(P) = 36/189, P(A) = 62/189 and P(E) = 91/189

    Case 3
    P(P) = 54/189, P(A) = 93/189 and P(E) = 42/189

    As you can see (unless I made a big mistake in my algebra, which is entirely possible), P maximizes his chance of winning by shooting at E first instead of missing deliberately. I think the explanation is that by having P take a shot at E instead of missing deliberately, P and A have 2 chances instead of 1 to kill the strongest opponent before he goes on a killing spree. This seems to slightly outweigh the benefit of having A do the heavy lifting and having P keep the first shot against a stronger opponent. However, someone PLEASE check the math above to see if I made any errors...I always second-guess myself when trying to disprove a classic solution!
  7. Standard member forkedknight
    Defend the Universe
    14 Jul '10 17:34
    I ran a quick sim. If each player fires at the biggest threat, each player wins with probabilities of approximately:
    P: 31.4%
    A: 54%
    E: 14.6%

    If P->A and A->E: bad decision for P
    P: 26.2%
    A: 25.2%
    E: 48.5%

    If P->E, A->E, but E->P: bad decision for E
    P: 23.6%
    A: 69.1%
    E: 7.3%

    Any player not firing at the biggest thread is at a disadvantage.
  8. Standard member PBE6
    Bananarama
    14 Jul '10 17:45
    Originally posted by forkedknight
    I ran a quick sim. If each player fires at the biggest threat, each player wins with probabilities of approximately:
    P: 31.4%
    A: 54%
    E: 14.6%

    If P->A and A->E: bad decision for P
    P: 26.2%
    A: 25.2%
    E: 48.5%

    If P->E, A->E, but E->P: bad decision for E
    P: 23.6%
    A: 69.1%
    E: 7.3%

    Any player not firing at the biggest thread is at a disadvantage.
    The numbers I got for "P shoots at E first" match the numbers you got from your simulation almost exactly, which gives me a bit of confidence in my approach. The numbers I got for "P shoots at A first" are a little different from your P(P) and P(A) numbers, but P(E) is almost exactly the same, which could be good or bad. I didn't try your third scenario.

    Would it be relatively easy for you to run the simulation again, this time letting P miss deliberately and letting A shoot at E first? I'm curious as to the results.
  9. Standard member forkedknight
    Defend the Universe
    14 Jul '10 17:48
    Originally posted by PBE6
    It's funny, I've heard this one before so I know what the "thinking outside the box" correct solution is supposed to be, but the math isn't matching up. So far I get the same answer as forkedknight, that Poor (P) should fire at Excellent (E) first, instead of firing at Average (A) or the "thinking outside the box" solution. Let me explain:

    [b]CASE 1: P sho ...[text shortened]... ny errors...I always second-guess myself when trying to disprove a classic solution!
    I'm wrong. The "think outside the box" solution appears to be correct, and in my simulation it provides P with almost 40% chance to win.

    I get these numbers (for 100k samples)
    P: 39.9%
    A: 38.1%
    E: 22.1%
  10. Standard member PBE6
    Bananarama
    14 Jul '10 17:50
    Originally posted by forkedknight
    I'm wrong. The "think outside the box" solution appears to be correct, and in my simulation it provides P with almost 40% chance to win.

    I get these numbers (for 100k samples)
    P: 39.9%
    A: 38.1%
    E: 22.1%
    Hrm, I was afraid I had made a mistake somewhere. Time to recheck!
  11. Standard member PBE6
    Bananarama
    14 Jul '10 18:13
    Aha! Found my error. When considering the case where P misses deliberately, I calculated P(E) correctly and the series for P(P) correctly, but I missed one branch where P wins after A has already been killed. This error was carried into the solution because I took the shortcut of calculating P(A) using the incorrect value of P(P). The actual equations are:

    P(E) = (1/3)*(1)*(2/3)*(1) = 2/9

    P(A) = (8/27)/(7/9) = 8/21

    P(P) = (2/9)/(7/9) + (1/3)*(1)*(1/3) <--(this is the part I forgot) = 2/7 + 1/9 = 25/63

    Converting them for comparison to the other cases, we have:

    P(P) = 75/189 = 39.7%
    P(A) = 72/189 = 38.1%
    P(E) = 42/189 = 22.2%

    So indeed, letting A do the heavy lifting and retaining the first shot against the strong opponent does indeed outweigh the advantage of taking 2 shots at the strongest opponent. Neat! Thanks to forkedknight for the simulation and helping me find my error.

    PS One of these days, I'll learn to do algebra correctly the FIRST time around...
  12. 15 Jul '10 00:03 / 1 edit
    Poor shoots into ground
    Excellent quickly throws his gun ten metres away
    Average sees excellent is no threat,turns, shoots at poor, misses
    Excellent dives, rolling across terrain, comes up with gun and shoots both with one shot.
  13. Standard member forkedknight
    Defend the Universe
    15 Jul '10 02:05
    Originally posted by iamatiger
    Poor shoots into ground
    Excellent quickly throws his gun ten metres away
    Average sees excellent is no threat,turns, shoots at poor, misses
    Excellent dives, rolling across terrain, comes up with gun and shoots both with one shot.
    Poor shoots the ground.
    Average drops his gun.
    Excellent thinks he's the only real remaining threat and kills himself.
    Poor is a prick and shoots Average while he's unarmed.
    ...?
    Profit!
  14. 15 Jul '10 10:55
    when p shoot ground, nobodys going to shoot him coz hes the worst. then A will try to kill E. if he hits him thats good becoz its p's turn and he has a 1/3 chance of winning (by shooting a) same thing if a misses coz then e will shoot a coz hes more dangerous.
    either way he has a better chance-1 opponent to 2

    but if he first shoots at e...if he misses, thats good becoz a will shoot at E and then he has a better chance at winning. but if he hits him then a will shoot him but luckily this is not too likely. (thats a 1/3 chance times 2/3 the chance that a hits him...= 2/9)
    but its still a better chance to just shoot at the ground.

    if he first shoots at a. theres a 1/3 chance that he'll lose right away. (shoots A, E shoots him ) and a 2/3 chance that he'll end up with 1 opponent to two.
    but if he shoots at ground theres a 3/3 that he'll be left with 1 to 2
  15. 16 Jul '10 22:07
    Originally posted by Banana King
    shoot at ground.
    (average will shoot at excellent coz hes better then poor, then if he hits him then poors gotta better chance coz only 1 person compared to 2. but if he misses, excellent will aim for average coz hes worse then poor. average dies, then poor has a better chance to win now then at the start coz 1 person rather then 2)
    p.s. do u read jasper dash?
    YES!! you are right banana king!

    and if you mean the book Jasper Dash and the Flame Pits of Delaware, i have read that. if you mean read in general, then yes.