BrainDen.com - Brain Teasers  dgreening

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Everything posted by dgreening

1. 'Could be over thinking this' Could be over thinking this! for any 2 positive numbers A and B, there are just 3 possible relationships A = B A > B A < B If we assume that the range of numbers is really large then the P [A = B] is very, very low. If we say that P [A=B] = 0, then there are just the 2 choices [A>B or A<B] the relative magnitude of the numbers doesn't matter and it is simply a matter of which number was chosen first. A random process is just as likely to pick "small then large" as it is to pick "large then small" and the P [A>B] = P[A<B] = 0.5 this holds whether the 2 numbers are (1 and 2) or (1 and a trillion)
2. Clarifying question: Do we get to choose the positions of the fixed points??
3. oops too late ... I will just watch from the side lines
4. Nicely done plasmid!
5. The elegant solution would include integrals, but...
6. You are correct! I must had made a leap of faith, when I was solving - apologies! in fact [in your second solution] you can reverse the order of 24 and 42 in the top right and lower right "large" squares to create one more solution.
7. First, I would like to apologize for clobbering the posting above [pls ignore it!] I tried to cover part of my reply with "spoiler" and ended up with a mess ... sorry! Re: Plasmid's "One thing I can say so far, which might or might not be pertinent, is that" You can simplify this even further by removing one of the numbers ... it still results in a unqiue solution.
8. One thing I can say so far, which might or might not be pertinent, is that
9. Normally, I would say that the dice are just a red herring to confuse us - BUT - I think they are of some significance!
10. Attempt at restating the problem for 4 digits, you are allowed to insert any math operators between the digits and assume there is an "= 10" at the end of the sequence. so 9339, could be arranged as (93 - 3) / 9 = 10 8952 could be arranged as 8 + 9 + 5 - 2 = 10 The question is "is there a method, algorithm .... that will allow me to identify all the 4 digit sequences that fit this criteria?" BTW, use of exponentials could create some interesting options.
11. In that case, I think bubbled has the optimum strategy.
12. One more question: Does it matter how many points you win by?? [say you were paid \$1 per point]; or Is winning by 1 point as valuable as winning by 20 points?
13. yes, nicely done!
14. I agree with bonanova
15. The question is [i think purposely] misleading
16. Another perspective I suspect that the impact of the extra toss is actually very small and it could go either way. For 250 [or 251] tosses, you would expect some sort of distribution centered heavily around 50% [or 125 heads]. This is not even close to being a uniform distribution, so most points below 110 or above 140 are zero. [just a guess on the end points]. The point is that there are really only a small number of points that have to be considered. Another approach might be to do some sort of convolution of the 2 distributions [which look virtually identical]. Since Ties go to John, the probabilityt will be less than 50% Rob G may be pretty close [i am not sure].
17. My guess is that it is a pretty big number! [sorry I couldn't pass that up]
18. Good point! I think you are probably correct ... oh well. I still think yours is a really good approach!
19. Very clever aproach m00li!! I like it!
20. can you please explain how your encoding scheme would work? Suuppose you chose the following cards [generated randomly]: 103 89 86 46 54 74 which card would you keep and how would your friend know the identitiy of the one you kept?
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