Re:Math question

From: mlorrey@datamann.com
Date: Mon Oct 28 2002 - 16:34:43 MST


From: "Mike Lorrey" <mlorrey@datamann.com>
X-Mailer: YaBB

[quote from: Eliezer on 2002-10-28 at 16:08:43]
Suppose that we flip a fair coin 20 times, and that a biased sampling
procedure then randomly selects 5 coinflips from the set of coinflips that
came up heads and reports on those coinflips; if there aren't 5 coinflips
that came up heads, the biased sampling procedure reports all available
heads and enough randomly selected tails to make up the gap.

Suppose the biased sampling procedure reports that the first, sixth,
eleventh, fifteenth, and eighteenth flips came up heads. Is there a
simple general formula for calculating the probability that any given
other coinflip came up tails? Clearly the probability is greater than
50%, but by how much?

(I don't know the answer to this one; it's posed as a genuine math
question, not a math problem.)

The probability of what you ask is 50%. The prob that any given flip is one or
the other is 50% in any situation. The odds that every other flip IN SEQUENCE
is tails is dependent upon how long your sequence is. In the case of getting
five tails, the odds are 3.125%.

This is done by the fact that for each additional instance of getting tails,
your odds halve. i.e.

1 flip: h=50% t=50% (cumulative odds given in parentheses below)
2nd flip: h=50% t=50% (2 heads = 25%, two tails = 25%, 1 head & 1 tail = 50%)
3rd flip: h=50% t=50% (3 heads = 12.5%, two tails = 12.5%, 1 head & 2 tails =
37.5%, 2 heads & 1 tail = 37.5%)
etc...

----
This message was posted by Mike Lorrey to the Extropians 2002 board on ExI BBS.
<http://www.extropy.org/bbs/index.php?board=61;action=display;threadid=53582>


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