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In an Auction , two players are bidding. Their bids will be a unknown fraction of their valuations. The valuations come from a uniform distribution $$[0,1] $$

If Player 2 bids $$ v/2 $$ and Player 1 bids $$b1<1/2$$

What is the probability player 1 wins ? Clearly for player 1 to win,players 2 bid has to be less than player 1 bids. $$P(v/2 < b1)$$ $$P(v < 2b1)$$ I follow the question up to this stage. Now it says since its uniformly distributed the probability player 1 wins is $$2b1$$ Im confused how can you just get 2b1 from the inequality ?

Thanks in advance

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  • $\begingroup$ have I incorrectly types the question ? does it not make sense $\endgroup$ – S F Apr 22 '14 at 20:21
  • $\begingroup$ I guess it should be $2b1$, not $2b$. $\endgroup$ – Michael Greinecker Apr 22 '14 at 20:27
  • $\begingroup$ I guess from the answer that the question actually asked said that $v$ was uniformly distributed on $[0,1]$ and you are to find the probability that player 1 wins given that player 1 bid $b_1$. (The prior distribution of player 1's valuation is irrelevant for that given information.) But that is just a guess. I think you have not accurately stated the problem. $\endgroup$ – David K Apr 22 '14 at 21:24
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We want to know the probability that $b_1>v/2$ when $v$ is uniformly distributed and $b_1<1/2$. This is clearly the same as the probability that $2b_1>v$. Now the uniform distribution on $[0,1]$ is characterized by the property that for each $0\leq x\leq 1$, a realization from the uniform distribution is smaller than $x$ with probability exactly $x$. In particular, since $v$ is uniformly distributed, the probability that $v<2b_1$ is exactly $2b_1$.

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