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Dice  how many walls? 
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#1
Dec112, 04:36 AM

P: 138

Hey! :) I have got question about verisimilitude.
I have got classic one dice. The dice can have 220 walls, where is numers from 1, 2, 3, ...20 (twowalls: only 1, 2; sevenwalls: 1,2,3,...,7). You can throw with dice. When you have got largest number on dice (twowalls: 2, sevenwalls: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 2walles= big verisimilitude you throw again but small numbers  you have got 20walles= very small verisimilitude you throw again but very big numbers... So... Have you got any idea and MATHEMATICAL PROOF for it? 


#2
Dec112, 05:57 AM

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.... 


#3
Dec112, 09:07 AM

P: 138

So... If I understand... This is verisimilitude about twowalls. 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 :) 


#4
Dec112, 09:39 AM

HW Helper
P: 6,189

Dice  how many walls?
Ah well, with an nsided 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: $$E = (\frac n 2 + 1) \cdot \frac n {n1}$$ So for instance for a 2sided dice, you get ##E = (\frac 2 2 + 1) \cdot \frac 2 {21} = 4##. In particular this expression becomes bigger for bigger n. So you will have the biggest result for a 20sided dice, which is ##E = (\frac {20} 2 + 1) \cdot \frac {20} {201} = \frac {220} {19} \approx 11.6## 


#5
Dec112, 10:06 AM

P: 181

Nearly, but the second equation should be
[tex]E = \frac{(n1)}2+1+E[/tex] which gives you [tex]E = \frac{n(n+1)}{2(n1)}[/tex] But the conclusion is the same: The larger n gets, the larger E will be. 


#6
Dec112, 10:52 AM

P: 138

Ou, that looks good :) I think I understand you. Thank both of you very much.



#7
Dec412, 11:25 AM

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Thanks
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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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