Results 1 to 8 of 8

Math Help - very easy question!

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    5

    very easy question!

    hello everyone!

    i have a very easy question for you guys!!!! but very hard for me!!

    can you tell me what is the pourcentage that 7+ a random number between 2 and 12 will be greater than 5 + a random number between 2 and 12?

    if you could tell me how to find this out, if its not too hard, i would really appreciate it!!

    i am working an a role playing game

    Francois
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by frankinthecity View Post
    hello everyone!

    i have a very easy question for you guys!!!! but very hard for me!!

    can you tell me what is the pourcentage that 7+ a random number between 2 and 12 will be greater than 5 + a random number between 2 and 12?

    if you could tell me how to find this out, if its not too hard, i would really appreciate it!!

    i am working an a role playing game
    Here are a few questions for you first. What do you mean by "a random number between 2 and 12"? Do you mean a whole number, and are the values 2 and 12 included or do you mean an integer strictly between 2 and 12 (in other words not including the end points)? Also, when you say a "random" number, do you mean that each value from 2 to 12 is equally probable, or are you perhaps thinking of throwing two dice, in which case the result is much more likely to be 7 than 2 for example?

    If you can state the problem a bit more precisely then we should be able to show you how to do the calculation.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    5
    Hello Opalg!

    thank you for your help, you got it, its 2 six sided dice!

    i understand now that if it is dice, its not exactly a ramdon number between 2 and 12!!! you have a lot more chances to get 7 like you say

    no wonder i am having problem finding the answer, its a lot more complicated that i thought!!!

    Francois
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Okay, so this is more complicated than you first thought! In the table below, the top row shows the possible outcomes from 2 to 12 when you throw two dice. For each outcome, the second row shows the probability of that outcome occurring, and the bottom row shows the probability of the outcome being at least that much.

    \begin{array}{c|ccccccccccc}k&2&3&4&5&6&7&8&9&10&1  1&12\\ \\P(k)& \frac1{36} & \frac2{36} & \frac3{36} & \frac4{36} & \frac5{36} & \frac6{36} & \frac5{36}& \frac4{36}& \frac3{36} & \frac2{36}& \frac1{36}\\ \\ P(\geqslant k)& \frac{36}{36} &  \frac{35}{36} &  \frac{33}{36} & \frac{30}{36} &  \frac{26}{36} & \frac{21}{36} & \frac{15}{36} & \frac{10}{36} & \frac6{36} & \frac3{36} & \frac1{36} \end{array}

    Now, if the first player (the one with a current score of 5) throws the dice and scores k, then the second player (currently with 7) needs to score at least k1 in order for their total to remain greater than that of the other player. If the first player rolls the dice and gets 2 or 3, there is no way that they can go ahead. But if they get 4, then the second player must get at least 3. The probability of that happening is 3/36 (= probability of the first player scoring 4) times 35/36 (= probability of the second player scoring at least 3).

    Adding up the probability for each of these combinations, the total probability of the second player remaining ahead after both players have rolled the dice is

    \frac{\parbox{5in}{$1\times36 + 2\times 36 + 3\times35 + 4\times 33 + 5\times 30 + 6\times26$\\ {\color{white}.}\hfill ${}+ 5\times21 + 4\times 15 + 3\times10+2\times6+ 1\times3$}}{36\times36}\approx 0.66435...

    If you want it as a percentage then the answer is approx. 66.4%.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2010
    Posts
    5
    Opalg, thank you so much

    ok, i understand the first line, i even understand the second line lol

    but the third i am not sure, listen, if its really too hard to explain on a forum, (i know forum have their limits hehe) just give me another exemple, how about 7 and 6? maybe i will be able to figure out by myself..

    what i need is not complicated at all, i need something like this

    (2 six sided dice = 2d6)

    7+2d6 VERSUS 5+2d6 = 66%
    7+2d6 VERSUS 6+2d6 =
    7+2d6 VERSUS 7+2d6 = 50% (that one i found out by myself LOL)
    7+2d6 VERSUS 8+2d6 =
    7+2d6 VERSUS 9+2d6 =
    7+2d6 VERSUS 10+2d6 =

    8+2d6 VERSUS 5+2d6 =
    8+2d6 VERSUS 6+2d6 = 66% (i think we can apply the same formula here??)
    8+2d6 VERSUS 7+2d6 =
    8+2d6 VERSUS 8+2d6 = 50%
    8+2d6 VERSUS 9+2d6 =
    8+2d6 VERSUS 10+2d6 =

    9+2d6 VERSUS 5+2d6 =
    9+2d6 VERSUS 6+2d6 =
    9+2d6 VERSUS 7+2d6 = 66%
    9+2d6 VERSUS 8+2d6 =
    9+2d6 VERSUS 9+2d6 = 50%
    9+2d6 VERSUS 10+2d6 =

    10+2d6 VERSUS 5+2d6 =
    10+2d6 VERSUS 6+2d6 =
    10+2d6 VERSUS 7+2d6 =
    10+2d6 VERSUS 8+2d6 = 66%
    10+2d6 VERSUS 9+2d6 =
    10+2d6 VERSUS 10+2d6 = 50%

    11+2d6 VERSUS 5+2d6 =
    11+2d6 VERSUS 6+2d6 =
    11+2d6 VERSUS 7+2d6 =
    11+2d6 VERSUS 8+2d6 =
    11+2d6 VERSUS 9+2d6 = 66%
    11+2d6 VERSUS 10+2d6 =

    12+2d6 VERSUS 5+2d6 =
    12+2d6 VERSUS 6+2d6 =
    12+2d6 VERSUS 7+2d6 =
    12+2d6 VERSUS 8+2d6 =
    12+2d6 VERSUS 9+2d6 =
    12+2d6 VERSUS 10+2d6 = 66%

    if you give me the 7 and 6 exemple, and i can apply it to 8 and 7, 9 and 8, and so forth, there is only a few more i need to find!

    Francois
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    This "very easy question" gets more and more complicated. The original question was "Can you tell me what is the percentage that 7+ a random number between 2 and 12 will be greater than 5 + a random number between 2 and 12?" The answer to the question depends on what is meant by "greater than". The natural interpretation is that A's score (x) is greater than B's score (y) if x>y. A second interpretation would be to allow also the possibility that x=y, and to say that A's score is greater if x\geqslant y. A third interpretation (which sounds strange but is actually quite natural) is to split the difference between the first two interpretations.

    Suppose that A and B start out with equal scores. They then each throw a couple of dice. There is a probability of approximately 11.2% that they will both throw the same total, and a probability of about 44.4% that B will throw a higher score than A. So the probability that B then has the greater score is 44.4% if you use the first interpretation of "greater". It is 55.6% if you use the second interpretation, and it is 50% if you use the third interpretation and split the difference between 44.4% and 55.6%.

    Quote Originally Posted by frankinthecity View Post
    (2 six sided dice = 2d6)

    7+2d6 VERSUS 5+2d6 = 66%
    7+2d6 VERSUS 6+2d6 =
    7+2d6 VERSUS 7+2d6 = 50% (that one i found out by myself LOL)
    In those calculations, I used the first interpretation to get the 66%, but you used the third interpretation to get the 50%. If I had used the third interpretation then my answer would have been 7+2d6 VERSUS 5+2d6 = 71%.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2010
    Posts
    5
    if a and b or equal then it doesnt count and i start over, 1 number have to be higher then the other so its really the natural interpretation that is good.

    and for the 7+2d6 VS 7+2d6, i didnt do the calculation to find out, i only said 50% because i said to myself "hey 7 and 7 chances or pretty equal must be 50%..." but again, myabe i am wrong??

    so to keep it simple i will just ask

    Can you tell me what is the percentage that 7+2d6 will be greater than 6 +2d6?

    if you could to the same calculation like you did for the one before maybe i will figured it out and be able to calculate the other stats i need by myself

    thank you again for you patience

    Francois
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jan 2010
    Posts
    5
    i was able to figure it out, someone showed me how to work with PARI

    thanks a bunch

    Francois
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] easy GCD question
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: July 8th 2011, 11:59 AM
  2. Ask a easy question
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 1st 2010, 03:16 AM
  3. Easy Question
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 19th 2010, 10:31 AM
  4. Seemingly easy question... too easy.
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: August 23rd 2009, 09:36 PM
  5. easy, easy question
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: July 14th 2009, 07:38 AM

Search Tags


/mathhelpforum @mathhelpforum