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You probably know the trick in the movie The Prestige :

[MOVIE SPOILER] A magician has found an impressive magic trick: he goes into a machine, close the door, and then disappears and reappears in the other side of the room. But the machine is not perfect : instead of just teleporting him, it duplicates him. The magician stays where he is, and a copy is created at the other side of the room. Then, the magician in the machine falls discreetly in a water tank under the floor and is drowned. Edit: The probability of the new copy of the magician being drowned is 1/2 (in other words, the new copy has 1/2 chances of being drowned, and 1/2 chances of popping into the room). Also, the water tank never fails and the chances are 1 that the magician dropping in the tank dies.

So the magician doesn't really like doing this trick, because "you never know where you are going to be, on the other side of the room or drowned".

Now, the paradox is the following : Imagine the magician does the trick 100 times. What are his chances of surviving ?

Edit, additional question: What are the chances of the magician of keeping his physical brain and not having a new one ?

Quick analysis: One one hand, there is one magician alive, and 100 drowned magicians, so his chances are 1 out of 100.

On the other hand, each time he does the trick, he has 1/2 chances of staying alive, so his chances are $(1/2)^{100}=1/(2^{100})$ of staying alive.

What is the right response and why ?

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    $\begingroup$ I agress with G.Jay that tricky question is who "the magician" actually is. I think this is less a statistical than a philosophical question ;). $\endgroup$
    – steffen
    Commented Oct 21, 2011 at 15:08
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    $\begingroup$ @steffen In the interests of making something useful out of an admittedly fanciful question, imagine that each time the clone has an "H" permanently stamped on his forehead. We can ask, then, what are the chances that after doing this trick 100 times, the magician still does not bear an "H"? In this case, 100 copies of him have been created and each copy has died. One yet lives. $\endgroup$
    – whuber ♦
    Commented Oct 21, 2011 at 18:35
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    $\begingroup$ @whuber: The question, as described, states that the clone is the one that might survive, while the one that goes into the machine (the original on the first iteration) will die 100% of the time. After the first time this act is performed, the original is dead. I've not heard of this paradox before, so perhaps the question miss-stated it? $\endgroup$
    – Izkata
    Commented Oct 22, 2011 at 3:37
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    $\begingroup$ You should add a spoiler alert in the top. $\endgroup$
    – Frank Meulenaar
    Commented Oct 28, 2011 at 18:07
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    $\begingroup$ Here's an interesting side question: After 100 performances, the magician will have memories of having survived 100 times and died no times. As a Bayesian, how should he assess his chances of surviving the next time? :-). (I asked a seemingly related question at The Sleeping Beauty Paradox.) I could draw striking parallels between this situation and that of the financial and business wizards who are busy running banks and companies into the ground today, arguing that they--like the magician--are merely lucky survivors. But I won't do that. $\endgroup$
    – whuber ♦
    Commented Apr 5, 2013 at 19:02

3 Answers 3

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This mistake was put in evidence in written conversations among Fermat, Pascal, and eminent French mathematicians in 1654 when the former two were considering the "problem of points." A simple example is this:

Two people gamble on the outcome of two flips of a fair coin. Player A wins if either flip is heads; otherwise, Player B wins. What are player B's chances of winning?

The false argument begins by examining the set of possible outcomes, which we can enumerate:

  1. H: The first flip is heads. Player A wins.
  2. TH: Only the second flip is heads. Player A wins.
  3. TT: No flip is heads. Player B wins.

Because Player A has two chances of winning and B has only one chance, the odds in favor of B are (according to this argument) 1:2; that is, B's chances are 1/3. Among those defending this argument were Gilles Personne de Roberval, a founding member of the French Academy of Sciences.

The mistake is plain to us today, because we have been educated by people who learned from this discussion. Fermat argued (correctly, but not very convincingly) that case (1) really has to be considered two cases, as if the game had been played out through both flips no matter what. Invoking a hypothetical sequence of flips that wasn't actually played out makes many people uneasy. Nowadays we might find it more convincing just to work out the probabilities of the individual cases: the chance of (1) is 1/2 and the chances of (2) and (3) are each 1/4, whence the chance that A wins equals 1/2 + 1/4 = 3/4 and the chance that B wins is 1/4. These calculations rely on axioms of probability, which were finally settled early in the 20th century, but were essentially established by the fall of 1654 by Pascal and Fermat and popularized throughout Europe three years later by Christian Huyghens in his brief treatise on probability (the first ever published), De ratiociniis in ludo aleae (calculating in games of chance).

The present question can be modeled as 100 coin flips, with heads representing death and tails representing survival. The argument for "1 in 100" (which really should be 1/101, by the way) has exactly the same flaw.

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One one hand, there is one magician alive, and 100 drowned magicians, so his chances are 1 out of 100.

That reasoning implicitly assumes that every magician is equally likely to be the one surviving at the end of the process. However, only the original would have to endure all 100 trials, and he would have the worst odds. Contrast the original with the last clone that gets created; he only needs to survive once and he has a 1 in 2 chance of being the lone survivor.

Pretend that instead of clones we are dealing with a single-elimination tournament (like the famous NCAA basketball tournament every March). The original has to last 100 rounds whereas the last clone only has to play in the tourney finals. Not all clones are equally likely to last until the end, and the original has the worst chances of $\frac{1}{2^{100}}$.

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The probability he survives is 1 on every trial, and the probability is 1 that he dies on every trial (notwithstanding water tank failure). After duplicating, there is no "him" anymore; there are "hims".

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    $\begingroup$ BTW: if the duplication is imperfect then $P(\mathrm{dies}) = 1$ on every trial (given a trusty tank) and $P(\mathrm{imperfect\ clone\ survives}) = 1$ on every trial (notwithstanding trigger-happy audience members). $\endgroup$
    – user1108
    Commented Oct 21, 2011 at 14:40
  • $\begingroup$ BBTW: if the machine imperfectly duplicates and randomly selects one to teleport (leaving the other to drown) then you would need more info/assumptions about the random selection. $\endgroup$
    – user1108
    Commented Oct 21, 2011 at 14:42
  • $\begingroup$ @Jay : Edited my question about the teleporting $\endgroup$
    – Benjamin Crouzier
    Commented Oct 21, 2011 at 15:07
  • $\begingroup$ Thanks. You addressed the teleportation but you haven't nailed down whether duplication is perfect or not. If duplication is perfect, then my answer stays the same (see @steffen's comment). If duplication is imperfect (what it sounds like you're looking for) then the answer is $1/2^{100}$ as per the last paragraph of whuber's answer, and the other answer is wrong for the reasons whuber and Michael detailed. $\endgroup$
    – user1108
    Commented Oct 21, 2011 at 16:09
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    $\begingroup$ @downvoter - the idea is to write why you downvote so that answers improve over time. $\endgroup$
    – user1108
    Commented Oct 22, 2011 at 23:49

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