Recent content by Minotar

  1. M

    Probability Problem, Quick question...

    No, this will not work. You cannot think of it that way at all. The reason is because the probabilities are NOT equal. Please view the chart. If they are not equal then you cannot possibly say that there is a 50/50 chance of winning.
  2. M

    Probability Problem, Quick question...

    Thanks for the helpful thoughts, Doc! This might just be it?! Just to tell you, the probs will not add up to 1, though. They will only add up to 1 if you include all of the non useful card combinations as well (which we are not interested in). The reason the P(1 pair)=1 is kinda hard to...
  3. M

    Probability Problem, Quick question...

    I still am not sure how to proceed:/ And, no, this is not homework. It is a project that I have been working on for some time.
  4. M

    Probability Problem, Quick question...

    I don't follow what you are saying here???? Well, I know what you did, but I don't know where to go from there! I know there must be a theorem or rule that lets you compute the prob from my table, but I am not sure what it would be... To me, my data looks more like a distribution. I need...
  5. M

    Probability Problem, Quick question...

    _______________Player1_______Player 2 Straight Flush_______0____________0 4 of a kind_______1/741_________1/741 Full House________5/741_________5/741 Flush___________15/247__________0 Straight___________0___________7/741 3 of a kind_______43/741________43/741 2...
  6. M

    Probability Problem, Quick question...

    This seems similar to interchanges idea, but the only problem with what you have listed, is that the probs don't add up to "1". The probs for each hand are actually very low, except hand 1, in which P(Player one gets hand 1)=1 and P(Player two gets hand 1)=1. All other Probs are very low. I...
  7. M

    Probability Problem, Quick question...

    Hmmm...This seems interesting... Just wondering, is there any theorem or reasoning you used to come up with this?
  8. M

    Probability Problem, Quick question...

    That would work if it was a simple, clear-cut case, but it won't work here... With this problem, assume that the probabilities for each hand, for each player are different. In other words, assume you have already calculated the probability of coming up with a certain hand, and it is different...
  9. M

    Probability Problem, Quick question...

    Ok, here is the quick question for any who remember anything from Problem Stats... Two people are playing a game that has 8 possible "hands" (1 through 8), and hand 8 is the best "hand" that can be achieved, and "hand" 1 is the worst... If each of these 8 "hands" have a probability to come up...
  10. M

    Need to buy a cheap PC, FAST!!!!

    Can you link this?
  11. M

    Need to buy a cheap PC, FAST!!!!

    Hey all! I need to buy a cheap semi-capable gaming tower for my wife... The only requirements I have, are 1). NO Celerons!!!! 2). Must contain fairly recent technology i.e. nothing older than 1st gen A64/64bit Semprons or 800fsb P4's... 3). Must have a video card slot (AGP or PCI-e) 4). Must...
  12. M

    Difficult Probability question for any math wizzes out there!

    How about doing it this way... Part a= [6*(1/12)^2]+[6*(1/12)^3]=13/288 Part b= [6*(C(12,1)/C(12,12))^2]+[6*(C(12,1)/C(12,12))^3]=312/1331 To me these probabilities seem more reasonable, but the way you listed sounds reasonable too:( Very difficult problem for sure. Really, the...
  13. M

    Difficult Probability question for any math wizzes out there!

    When I tried it this way, I got very low numbers....like 10^-39 order magnitude, which is a very low probability. For some reason, this doesn't seem right to me? Any thoughts from anyone else? Thanks in advance!
  14. M

    Difficult Probability question for any math wizzes out there!

    Ok, here is a very difficult math problem for any prob/stat wizzes out there... What is the probability of the following event to occur: Suppose you have a random group of 30 people. What is the probability that 2 each of those people will have a birthday in each month January through...
  15. M

    Probability math question?

    Thanks for the help:) I just wasn't sure if I should add the probs or multiply things out. You guys cleared it up for me... Thanks!