Dice - how many walls?

by Numeriprimi
Tags: dice, number, proof, verisimilitude
 P: 138 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?
 HW Helper P: 6,189 Welcome to PF, Numeriprimi! 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....
P: 138
 Quote by I like Serena Welcome to PF, Numeriprimi! 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....
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 :-)

 HW Helper P: 6,189 Dice - how many walls? 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##
 P: 181 Nearly, but the second equation should be $$E = \frac{(n-1)}2+1+E$$ which gives you $$E = \frac{n(n+1)}{2(n-1)}$$ But the conclusion is the same: The larger n gets, the larger E will be.
 P: 138 Ou, that looks good :-) I think I understand you. Thank both of you very much.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 By the way- there is no such thing as one "dice". "Dice" is the plural of "die".

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