# Probability of being highest OR second highest draw from a distribution

In my situation, I have a distribution F(x) over some compact interval. Say I take $n$ iid draws from the distribution. I want to find the probability that one draw, $x_i$, is the highest of the $n$ draws OR is the second highest of the $n$ draws.

If it were just finding the probability of being the highest, it would just be $F(x_i)^{n-1}$, I believe.

And I believe that the probability of being exactly the second highest is ${n-1 \choose n-2} F(x_i)^{n-2} (1-F(x_i))$.

But when I'm trying to find the joint probability, I feel like I might be missing some overlap if I just add those together.

Thanks!

EDIT: Maybe I phrased my question incorrectly. I'm looking at an auction situation, where every bidder gets an independent draw as his value in the auction. Each bidder makes a bid & then the highest bidder wins the item and pays his bid. So I'm trying to find any bidder's expected profit, which would be his value minus his bid, but only in the case that he's the highest bidder. So I thought it would be (v−b)F(b)n−1, where F(b)n−1 is the "probability of being the highest," as I phrased it. But I was trying to do the case where both highest & second highest win

I will assume that the distribution is continuous, so that the probability of ties is $0$.

The probability that a specific draw, say the seventh, is the highest is $\frac{1}{n}$. This is because all draws are equally likely to be highest. The probability it is second highest is $\frac{1}{n}$. Add.

• Thanks for your answer! I don't have enough reputation to upvote you. I think I phrased my question incorrectly, as I explained in the comment below. – knielsen May 16 '15 at 23:42
• You are welcome. The most reasonable interpretation of the question, based on the phrasing, is that you are looking for the probability a specific $X_i$ is first (or second). – André Nicolas May 16 '15 at 23:53

The reason that André Nicolas's solution is correct, and yours incorrect, is that you are confusing the ordinal ranking of the draws, which he addresses, and is what you asked for, with the (cunulative) distribution function of the value of the order statistics x1 and x2. He addresses the probability that x1 will be highest or 2nd highest. You are providing formulas for the distribution function (hence value) of the winner (1st place) or runner-up (2nd place).

It's like the difference between predicting which day in the year will get the most rain vs. what the greatest amount of rain to fall on any day will be.

• Thanks for your explanation. Maybe I phrased my question incorrectly. I'm looking at an auction situation, where every bidder gets an independent draw as his value in the auction. Each bidder makes a bid & then the highest bidder wins the item and pays his bid. So I'm trying to find any bidder's expected profit, which would be his value minus his bid, but only in the case that he's the highest bidder. So I thought it would be $(v-b)F(b)^{n-1}$, where $F(b)^{n-1}$ is the "probability of being the highest," as I phrased it. But I was trying to do the case where both highest & second highest win – knielsen May 16 '15 at 23:41
• In your situation (i.i.d.), who wins is independent of expected profit given that they win. Unconditional expected profit = Prob(winning) * expected profit given they win. Similarly for 2nd place. You can add these 2 pieces, via Law of total probability (expectation) to get overall expected profits. Per previous answer, prob(win) = prob(2nd place) = 1/n. Your formulas for the distribution function of the order statistics are wrong - see item 5. in math.uah.edu/stat/sample/OrderStatistics.html for correct info. – Mark L. Stone May 17 '15 at 0:11
• Your math (i.i.d.) only describes the auction if bids are i.i.d. across the players. Expected value of distribution function of order statistics gives you the expected winning and 2nd place bids. Put the pieces together. BTW, David (died last year) and Nagaraja "Order Statistics" 3rd edition has everything you always wanted to know about order statistics, and probably a lot you don't. Nice bedtime reading, ha ha. – Mark L. Stone May 17 '15 at 0:16