(New page: ==Part(a)== Show that P(B) > P(C) > P(T) > P(A): - P(H) = p , 0 < p < 1 <math> P(B) = p + p(1-p)^4 + p(1-p)^8 + \dots + p(1-p)^{4(n-1)} </math> Recall geometric series: <math> \...)
 
(Part(a))
 
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<math> P(B) = p + p(1-p)^4 + p(1-p)^8 + \dots + p(1-p)^{4(n-1)} </math>
 
<math> P(B) = p + p(1-p)^4 + p(1-p)^8 + \dots + p(1-p)^{4(n-1)} </math>
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Recall geometric series:
 
Recall geometric series:
  
<math> \sum_{i=0}^\inf x^i = 1\{1-x}, for |x| < 1 </math>
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<math> \sum_{\imath=0}^{\infty} x^{\imath}= \frac{1}{1-x}</math>  for |x| < 1
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<math> P(B) = p\sum_{\imath=0}^{\infty} (1-p)^{4\imath} = \frac{p}{1-(1-p)} </math>
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Repeat this for Carol, Ted, and Alice to show that the order of your toss affects your probability of winning.

Latest revision as of 16:09, 9 September 2008

Part(a)

     Show that P(B) > P(C) > P(T) > P(A):

- P(H) = p , 0 < p < 1

$ P(B) = p + p(1-p)^4 + p(1-p)^8 + \dots + p(1-p)^{4(n-1)} $


Recall geometric series:


$ \sum_{\imath=0}^{\infty} x^{\imath}= \frac{1}{1-x} $ for |x| < 1


$ P(B) = p\sum_{\imath=0}^{\infty} (1-p)^{4\imath} = \frac{p}{1-(1-p)} $


Repeat this for Carol, Ted, and Alice to show that the order of your toss affects your probability of winning.

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Ryne Rayburn