Fragen in Vorstellungsgesprächen in Yung Shue Wan, Hongkong

I think it is 7. Kannp, the top five are not established after 5 because the second in one race might be better than the first in another race. Have five races of five each and keep the top 3 in each races. Then, take each of the winners and race them against eachother. The two bottom and the 4 who lost to them are discarded. The two who lost to third place is discarded. And, the one who got 3rd in the race with the horse who gets second in the 6th race is discarded. Now, there are six horses left. Race all but the horse who won twice and keep the top two, combined with the horse who sat out. Now you are done in 7 races.

you're over thinking this...... run 5 heats with 5 horses in each one. Using a stopwatch, you take the 3 fastest........ but if you don't have a stop watch - anonymous is wrong... marty is right - anonymous you can't take the winner of each race, because you need the top 3. if a 2nd place from one race is faster than the other 4 winners, you don't have it right. 1st time thru 5 horses by 5 heats drop bottom 2 = 15 left 2nd time thru 5 horses by 3 heats drop bottom 2 = 9 left 3rd time thru one race with 5, other with 4. keep top 3 of each race = 6 left now run any 5 horses, keep top 3 and for last one, run remaining 4,drop last. 12 heats total

7

1) 5 heats take the first 3 =15 horses left 2) run 3rd against each other, take the first, same thing for 2nd and for the 1st take the best 3 = 3 heats 5 left 3) final heat 9 total

He gave a very good explanation. I'll decline to explain why prm is incredibly wrong.

7 is correct in assumption that each round ends up with a clear 1-st, 2-nd and 3-rd places. If horses can come simultaneously - than we need 11 rounds (When we have 9 horses left - we set up a round for first 5 and set up a round for the last 4 and a horse that took 3-rd place in the previous round. This leaves us 5 horses to race in the final 11th round).

I got stuck with his explanation but once I understood it, it seems very brilliant. Let me put it in this way: 5 groups, each of 5 horses -> race (heat) take the 1st from each group and race (semi-final) discard the 4th and 5th one and all other horses which lost to them in the heat race keep the 3rd one but discard all horses lost to it in the heat race keep the 2nd one but discard the 3,4,5-th horses who raced against it in the heat race keep the 1st one but discard the 4,5-th horses who raced against it in the heat race Now there are 6 horses, namely the 1st from semi-final and the two just following him in heat the 2nd from semi-final and the one just following him in heat the 3rd from semi-final the 1st from semi-final is champion. ignore it and race all other 5 horses in final so 7 in total definitely Russ considers more cases but I am not interested in making it too complicated

Hey guys, revicing the things here. Can you guys be more explicit on the 7 figure ? Even with the explainations it’s not clear for me. You say run the 5 winners of the 1st round and discard the last 2, but what if the 5 winners of the first round are actually the 5 firsts ?

Forget it I read with my eyes closed, I was lookin for top 5 and it’s top 3

It matters if you have already answered the interviewed about how confident you are to get the correct answer to each question. For example, if you said that you are 99% confident that you get each of five answers correctly, then the probability that you get 3 answers correctly is 10*(99%)^3(1%)^2 ~ 1%. Now you know it matters!

good point by tmbtw. If the question was, you are 90% confident you get a question right, what are the odds you got all 5 right, then you could say (0.90)^5 was the right answer. That's not easy to do in your head, but it comes out to about 59%. The odds of missing all five is more easy, it's (.10)^, which is .00001, or about .001%

output probability 0000 (1) --> 1/16 0001 (4) --> 1/4 0011 (6) --> 3/8 0111 (4) --> 1/4 1111 (1) --> 1/16 \sum \sqr{prob} = 2 / 16 / 16 + 2 / 16 + 9 / 64 = 1/128+1/8+9/64 = 0.27 (+2 dollars) < 1/3 so No.

it is a correct quation. these qustion are commom in mathematic. you just have to roun down

To the two above me: no, only one face has a 12, but the dice is such that this face comes up 40% of the time

Ben's answer was what I came up. But Pierreou's one is an even better strategy.

Pierreou's answer assumes that your opponent does not have the concept of expectation. It is so easy to win if your opponent is ignorant......So I suggest you take the initiative to bet a guess on 7 then the probability of winning is >=6/11 no matter what your opponent's guess is

console.log(Array.from(new Set([5, 7, 2, 3, 9, 10, 2, 6]))); //javascript soln.

To UPS personal manager