Statistics again

[huy.duong@ctsu.ox.ac.uk]

>>To see a paradox as tha^`y Tha('ng has promised, we change the
>>condition a little.  The names of two lucky guys then are written down
>>in two folded pieces of paper.  As it happened A picked up one of
>>them, and saw B's name inside.  So now what is A's chance?
>                                   ^^^^^
>If one of the releaseds is B then the possible combinations
>of the releaseds are AB and BC. As such, there is a 0.5 prob
>that the combiation is AB, ie a 0.5 prob that A is released.

Cha`o anh Du~ng va` ca'c nha` toa'n ho.c,

Anh Du~ng no'i la` reasoning cua? to^i "would have been fine",
co' nghia~ la` no' "not fine". To^'i qua to^i accept, nhu+ng
ba^y gio+` cho phe'p to^i "xe't la.i".

It is true that inititally, ie before the names are decided and written
onto paper, A has 2/3 chance of being released.

However, this chance is refined as events progress. For example:

1) If A picks up a piece of paper and see his own name, then the chance
*now* that he sees his wife (assuming that she hasn't re-married etc)
is 1, not 2/3.

2) If A picks up a piece of paper and sees B's (or C's) name then the chance
*now* that he sees his wife is 1/2, not 2/3.

This sounds paradoxical, ie that his initial chance is 2/3 but his
chance after picking up A is 1, and after picking up B or C is 1/2.
But in fact it is consistent, as I now show:

The total prob of A being released, P, is

P = prob picking A * prob A released if picked A
  + prob picking B * prob A released if picked B
  + prob picking C * prob A released if picked C

  =    1/3         *         1
  +    1/3         *        1/2 
  +    1/3         *        1/2     

  = 2/3, as expected.

Huy