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For the First-Price Sealed Bid, I know that the optimal bid is $$ (n-1)/n * v_i$$

However, I am confused about a step in finding this value. We are told that there are $n$ players each with a personal valuation of $ v_i $.

Everyone plays the same strategy, bidding the optimal listed above. The values are i.i.d. draws from a uniform distribution $U[0, V_{max}\frac{n-1}{n}] $$

I am confused why the probability that a given bidder who bids "b" wins is $$ (b/ [ (n-1)/n] V_{max} )^{n-1} $$

I was wondering if someone could explain how this probability is obtained?

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  • $\begingroup$ You omitted the assumption that private values are i.i.d. draws. $\endgroup$ – mlc Mar 24 '17 at 23:34
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Suppose that each value $v_j$ is an i.i.d. draw from the uniform distribution on $[0,1]$. (The argument is the same if you rescale the support by the constant $V_{\max}$, but I like to keep notation tidy.)

Bidder $i$ wins if his bid $b_i > \max_{j \ne i} b_j$. The bid $b_j$ of a player $j \ne i$ depends on his type $v_j$ according to the bidding function $b_j (v_j) = \frac{n-1}{n} v_j$. Hence, bidder $i$ wins if his bid $$b_i > \max_{j \ne i} b_j (v_j) = \max_{j \ne i} \frac{n-1}{n} v_j = \left( \frac{n-1}{n} \right) \max_{j \ne i} v_j = \left( \frac{n-1}{n} \right) V^{(n-1)}$$ where $V^{(n-1)} = \max_{j \ne i} v_j$ is the maximum of $(n-1)$ i.i.d. uniformly distributed r.v.'s. The cdf for $V^{(n-1)}$ is $F(v) = v^{n-1}$. (If you are unfamiliar with this result, see f.i. Distribution of maximum of $n$ i.i.d. random variables uniform on $(a,b)$?)

Rewrite $b_i > \frac{n-1}{n}V^{(n-1)}$ as $V^{(n-1)} < \frac{n}{n-1}b$. Then $$P \left( V^{(n-1)} < \frac{n}{n-1}b \right) = F \left( \frac{n}{n-1}b \right) = \left( \frac{n}{n-1}b \right)^{n-1}$$

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