I'd estimate about 1 in a million, based on the following
reasoning;
(1) You shouldn't count the first one. In my mind the
question is "what are the odds of player B hitting a sf
immediately after player A (in this case, players B AND
C)". In other words, the first event has already occurred.
Note that his is NOT the same odds as asking "What are the
odds of these 3 particular players getting SF's nearly
simultaneously, when none have yet occurred.
(2) The SF hits roughly once every 39,000 hands, with
correct play. Assuming a pace of 600 hands/hour (= 10
hands/minute). These SF's hit within 4 minutes, or 40 hands
of each other. That's 40 shots at a (roughly) one in 40,000
shot, which is about 1,000 to 1.
If the question were "what are the odds of EITHER neighbour
hitting within 4 minutes, the answer would be 2 chances at
a 1,000 to 1 shot, = 1 in 500.
If the questions is "what are the odds of BOTH neighbours
hitting, you have to multiply them - so the answer is
1:1,000 x 1:1,000 = 1:1,000,000.
Brian
(Note that none of the above is correct if I'm the one in
the middle. In that case the odds change - the people on
BOTH sides of me are GUARANTEED to hit, while I'm
guaranteed to NOT HIT. 
The oddest thing happened on Saturday last. I was playing
Pick'em at Woodbine (original non-smoking section along
the wall) when
the player at my right suddenly made a straight flush.
Within thirty seconds so did the player at my left.
the player to
my left had hardly finishes saying "It should be your
turn too to get
one" when I straight flushed too! All three within 4
minutes of each other.
What kind of odds would those be?
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-- mofromto2 <mofromto@sympatico.ca> wrote: