Probability of X &lt; Y for two independent geometric random variables, intuitivelyGiven two independent, geometric random variables X and Y with the same success probability p ∈ ( 0 , 1 ), what is P ( X &lt; Y )?The answer can be computed formally using total probability and one gets P ( X &lt; Y ) = 1 − p 2 − p

Question

Probability of   X  &amp;lt;  Y for two independent geometric random variables, intuitivelyGiven two independent, geometric random variables X and Y with the same success probability   p  ∈  (  0  ,  1  ), what is       P    (  X  &amp;lt;  Y  )?The answer can be computed formally using total probability and one gets       P    (  X  &amp;lt;  Y  )  =            1      −      p              2      −      p

ellynnauwh · Accepted Answer

Step 1This can be thought of in terms of conditional probability. We have   X  &amp;lt;  Y if and only if, in the first round where either A or B rolls a six, A rolls the six and B does not. Focusing on that particular round i, we know that at least one six has been rolled, but we don&#039;t know by whom, or if both rolled a six. We can therefore evaluate:Step 2  P  [  A     wins  ]  =      E    i    P [A rolls six in round i and B does not | At least one six is rolled in round i]This expected value is easy to compute: The distribution over which i is the final round doesn&#039;t matter, since this conditional probability is 5/11 for all i.

alexmjn · Accepted Answer

Step 1Let   q  =      P    (  X  &amp;lt;  Y  ). We condition on the outcome of the first trial for players A and B.If player A gets a success, then   X  &amp;lt;  Y iff B fails. This event happens with probability   p  (  1  −  p  ).Step 2If A gets a fail, then we need B to fail too. After this we are in the same state as initially. So in this case   X  &amp;lt;  Y with probability   (  1  −  p      )    2    q.So   q  =  p  (  1  −  p  )  +  (  1  −  p      )    2    q    ⟹    q  =            1      −      p              2      −      p        .

Given two independent, geometric random variables X and Y with the same success probability p in (0,1), what is P(X<Y)?

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Answer & Explanation

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