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David Folster wrote:
> [...] He said that my
> betting system wasn't valid because each bet was an event independent of
> previous events and that my next bet would therefore always have a 50/50
> chance of going my way. Yet when I asked him what the odds of one color
> coming up, say, 5 times in a row, he said that the chance of that happening
> was about 3% (.5 * .5 * .5 * .5 * .5).
>
> Now in my mind, these two different probabilities are measuring the *same
> event*. So how, on the one hand, can my probability of winning be only 50%,
> and on the other hand, just 3%?
First of all, I believe the foregoing is incorrect for a very simple reason: the
probabilty shown (namely, 1/(2^5) is the probability of black coming up 5X in a
row, which is also equal to the probability of red coming up 5X in a row.
Therefore, the "odds of one color (either black or red) coming up ... 5 times in
a row" is twice the foregoing, the probability for black plus the probability
for red.
Now, getting down to the problem you asked about, simply for the sake of
simplicity, let's use 3 times in a row. There, the probability of one (but not
both) color coming up 3X in a row is 0.5 x 0.5 x 0.5 = 0.125 = 1/8 = 1/(2^3).
In answer to your question, the point is that the chance of the third flip
coming up red is independent of whatever the previous flip was or the flip just
prior to it --- each flip is an independent event, independent of any prior flip
--- and the probability for each flip is 1/2, just as you indicate above. In
other words, if you had had a run of 10, 100, 1000 black flips in a row, the
probability of the next flip coming up black is equal to the probabillity that
the first flip would be black, namely, 1/2. The coin has no memory.
>
>
> I just can't figure this out. Anyone?
>
> Dave
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