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You bid for a coin. You're confident that the price of the coin is between 0 and 100, if your bid is greater than the price, you win and sell it to your friend at the price of 1.5 times price. what's your bid to max your profit?

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16 réponse(s)

6

I loved this question and want to renew this debate. What do you guys think about my two approaches to solve it: 1) If we can only play this game once AND our goal is to maximize profit (as the question states). I agree with above that expected value of a coin is 50. Given that we bid 51 to win auction and pocket 24. Problem is we only win if coin is (0:50) which gives us new expected value of 25, and so we lose. We can deduct this way all the way to zero bid. 2) Nothing beats little Monte Carlo experiment. I created a matrix of 100X1000000. Where 100 is the number of possible bids given certain price. 1M is the number of random uniformly distributed prices 0-100. Calculated expected gain at each bid level 0 to 100. I wish I could post a MATLAB graph here. It looks as downward facing 1/2 of parabola with max value of 0 and min of -25. Results: best gain of 0 achieved at 0 bid, worst average gain of -25 is at 100 bid. Comments appreciated!

Alex B le

2

i had the longest argument with a friend on this. you cannot get a positive expected value no matter what you bet. if you bid $50, then you can discount the EVs if the value of the coin is 51-100 since that'll be 0 (you don't win the auction). if you bid$50 and the coin's worth $50, you sell for$75make $25. but if the coin's worth$0 you lose $50. keep comparing the extremities and you will see in almost all cases you will be losing more than you make...that's the best i can explain it. i had to use a spreadsheet to prove this to my friend. in order to get an EV of 0, you'd need to change the multiplier to 2. which makes sense. if X is your bid, your profit is (X/2) *1.5 - X. willie le 1 I don't think the argument for avg of 50 => bid 75 is accurate. You can't simply use the expected value of theprice to find its area of the profit curve (by integration, the area under the profit curve gives the expected profit. When you can do however, is find the mean of the expected profit, but you will end up with the same answer that the expected profit is negative everywhere except for 0. SP le 1 The answer is 37.5 t le 1 I am 95% sure it is 0. The people saying 100 are not correctly counting other numbers. If you bid 100, then it could be 0, meaning you sell it for 0. Or 1, and you sell it for 1.5(losing 98.5), et cetera until 100. You make a profit if >66 gets drawn, but it doesn't outweigh the loss from what I can see from doing trials on 1,2,3,10 Chase Schwalbach le 2 I think the answer is 75. The expected value of the price of the coin is 50, then the expected value by selling it is 50*1.5, which is 75. Anony le 0 Last comment makes no sense. If expected profit is negative, then it is not the max of itself and zero. Who would take a bet on that game? Huh? le 1 So if B>P we make a profit of 3P/2 -B, and a profit of 0 otherwise. And P follows a continuous uniform distribution over 0,100 (f(P)=1/100) So expected value of profit = integral from 0 to min(100,B) of f(P)*3P/2 - B dP, which, assuming B<=100, = -B^2/400. This is maximized by B=0. I've seen a different version of this problem where P is in a discrete 1 to 100 distribution. Here expected value of the profit is \sum_{P=1}^{ceil(B)-1} (3P/2 - B)/100 =(3*ceil(B)*(ceil(B)-1)/2 - (ceil(B)-1)*B)/100 which I believe is maximized by B=1.01. dv le 0 In continuous case - the result is 0 (see 'Alex B' and 'dv' posts). In discrete case - the result is 2. The reason is the following: the more the bid - the more the expected loss is. And there is a special case of 'bid = 1' - where 'loss = 1'. Loss in case 'bid=2' is 'loss = 1/200'. So, if we are forced to play - it is better to choose 'bid = 2' and loose less. Russ le 0 I originally got 37.5 as follows (Expected Profit) = (Profit from outcome) * (Probability to win the bid) = 0 * (1-bid/100) + (1.5*price - bid) * (bid/100) = (1.5*price*bid-bid^2)/100. Take partial derivative with respect to bid => d(Profit)/d(bid) = (3*price - 4*bid)/200. Find where this derivative =0 gives 3*price=4*bid. Substitute the expected value of the price in (E(price)=50) => bid = (3/4)*E(price) = 37.5. HOWEVER, when I plug this back into a Monte Carlo, this appears to be a local minimum with an avg profit of -$3.65 per game. Monte Carlo shows this loss gets worse as you increase the bid and gets better (closer to zero) as you decrease the bid. bidding even $0.01 results in a small average loss, so you can maximize your profit by bidding either 0 or >100 to ensure you never win, giving you an expected profit of$0. This also makes sense logically: your profit is limited to 0.5*price (assuming you bid infinitesimally more than the price) while your loss could be as great as (bid - 1.5*price), or, as a % of the price: (bid/price - 1.5) for all bid > price. If the price is small, this results in the potential for a MASSIVE loss (in %-of-price-terms), which the max profit of 0.5*price can't ever make up for. Therefore, it makes sense that you shouldn't want to win the bidding!

James le

0

The price is NOT necessarily uniformly distributed between 0 and 100, therefore 0 might not be the right answer

Utilisateur anonyme le

0

Monte Carlo says that the answer should be near zero.

Daniel le

0

The expected value of the coin is 50 (Mean of the parameter 0 - 100). If you can sell you coin 1.5 times the value of the coin the Maximum value you would bet is 1.5*50 = 75. This way your expected profit is 0.

Petar le

0

@chase : You are correct in saying that the expected value of the profit is negative when bidding 100, in fact it is negative for any value of the bid. But since we are bidding only once, we look at the bid at which the probability of making a positive profit is maximum, that bid is 100 with 1/3 probability.

0

I got 100

Utilisateur anonyme le

0

I got the answer 0, but the interviewer said it was wrong

Utilisateur anonyme le

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