## angela210793 4 years ago suppose there are 100 guys... and each of them pick any number between 0 and 100 once everybody has picked a number....you add all of the numbers and multiply that by 2/3 so...if everybody picked 100....the average would be 100 and after multiplying by 2/3, you get 66.67... now, whoever's picked number is closest to the average multiplied by 2/3, wins the game and gets \$100..... what no I should I choose to maximize my chance of winning?

1. Yue27

Just a stab in the dark, but if you multiplied each number by 2/3, and found the average of those. Wouldn't that be the best number to guess?

2. angela210793

Idk....I've no idea at all :/

3. agdgdgdgwngo

Use Bayes' Theorem!

4. angela210793

Bayes????????

5. agdgdgdgwngo

what is the most likely number that 100 guys would pick?

6. Ishaan94

1 ?

7. angela210793

Not getting it..........

8. agdgdgdgwngo

good! so if 100 guys pick one, then the result would be 2/3!

9. agdgdgdgwngo

so if you picked 1, then you would be winning

10. angela210793

come on....i don't think it is solved like this guy1=59 guy2=31 . . . . . . . . . . .guy100=3 Wht now?

11. agdgdgdgwngo

alright, try having all guys pick different numbers: guy 1 pick 1, guy 2 pick 2... guy 100 pick 100

12. agdgdgdgwngo

find$\sum_{i=1}^{100}i$

13. agdgdgdgwngo

this is $\frac{100*(100+1)}{2}=10100/2 = 5050$

14. angela210793

i did smth like |dw:1317896959755:dw|

15. agdgdgdgwngo

yes!

16. KorcanKanoglu

hmm....

17. angela210793

silly me whn i was finding the average i divided it by 2(God knows why i did tht) and i got 1683...O.o Thnx ^^

18. agdgdgdgwngo

If you don't like that answer, set up a simulation. For every trial, make it create a list of 100 random integers. Sum those elements, divide by 100, and multiply that sum by 2./3. Repeat for about 10000-1000000 trials, adding the results of each trial together, and finding the average result.

19. angela210793

|dw:1317897551125:dw|

20. agdgdgdgwngo

yep, I got the same 33.3 answer after 10000 trials