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Dice - how many walls?

  1. Dec 1, 2012 #1
    Hey! :-) I have got question about verisimilitude.

    I have got classic one dice. The dice can have 2-20 walls, where is numers from 1, 2, 3, ...20 (two-walls: only 1, 2; seven-walls: 1,2,3,...,7).

    You can throw with dice. When you have got largest number on dice (two-walls: 2, seven-walls:7) you can throw again.

    AND NOW - Which dice (how many walls) is the best for largest mean value totals throws?

    Logical - you have got 2-walles= big verisimilitude you throw again but small numbers
    - you have got 20-walles= very small verisimilitude you throw again but very big
    numbers...

    So... Have you got any idea and MATHEMATICAL PROOF for it?
     
  2. jcsd
  3. Dec 1, 2012 #2

    I like Serena

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    Welcome to PF, Numeriprimi! :smile:

    Suppose we start with a dice with 2 walls.
    And let's call the mean total value E.

    Then the chance is 1/2 that you get a 1, which would be the final total.

    And there is another chance of 1/2 that you get 2, in which case you throw again.
    After this you can expect the same mean total value E to follow.
    So the actual total would become (2+E).

    In other words:
    $$E = \frac 1 2 \times 1 + \frac 1 2 \times (2 + E)$$

    Solve for E and you get your mean total value....
     
  4. Dec 1, 2012 #3
    There is some problems... I'm not sure to I understand you because maths is difficult and more difficult when is in English, which I only a few years learning :-)

    So... If I understand...
    This is verisimilitude about two-walls. Yes, there is quite big verisimilitude to throwing again, this I know. However, what others numbers? I still don't know which dice is best for the bigger mean value and why :-( I maybe badly wrote the example or I didn't understand you... So, can you explain me this problem again? Thank you very much.

    Greetings from Czech Republic to PF :-)
     
  5. Dec 1, 2012 #4

    I like Serena

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    Ah well, with an n-sided dice, your expectation is:
    $$E=\frac 1 n \times 1 + \frac 1 n \times 2 + ... + \frac 1 n \times (n - 1) + \frac 1 n \times (n + E)$$
    I am skipping a few steps in between, getting:
    $$E = \frac n 2 + 1 + \frac E n$$
    Which can be solved to the final formula:
    $$E = (\frac n 2 + 1) \cdot \frac n {n-1}$$

    So for instance for a 2-sided dice, you get ##E = (\frac 2 2 + 1) \cdot \frac 2 {2-1} = 4##.

    In particular this expression becomes bigger for bigger n.
    So you will have the biggest result for a 20-sided dice, which is ##E = (\frac {20} 2 + 1) \cdot \frac {20} {20-1} = \frac {220} {19} \approx 11.6##
     
  6. Dec 1, 2012 #5
    Nearly, but the second equation should be
    [tex]E = \frac{(n-1)}2+1+E[/tex]
    which gives you
    [tex]E = \frac{n(n+1)}{2(n-1)}[/tex]
    But the conclusion is the same: The larger n gets, the larger E will be.
     
  7. Dec 1, 2012 #6
    Ou, that looks good :-) I think I understand you. Thank both of you very much.
     
  8. Dec 4, 2012 #7

    HallsofIvy

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    Staff Emeritus
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    By the way- there is no such thing as one "dice". "Dice" is the plural of "die".
     
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