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# Dice again!

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Six fair 6-sided dice are rolled. What is the probability of getting a "Long Straight", or all six faces showing?

Any help would be great. Thank you.

Nov 3, 2018

#1
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$$\text{A long straight can appear }6! \text{ different ways}\\ \text{There are }6^6 \text{ total possible different dice rolls (dice are distinguishable)} \\ P[\text{long straight}]=\dfrac{6!}{6^6} = \dfrac{720}{45656} = \dfrac{5}{324}$$

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Nov 3, 2018
#2
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Rom: Thank you for this. How much more complicated would the problem become if we rolled 10 dice instead of 6 and expecting all faces to show?

Nov 3, 2018
#3
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what do you mean?

out of the 10 dice you can find (1,2,3,4,5,6)?

you won't get (1,2,3,4,5,6,7,8,9,10) unless they are 10 sided dice

Rom  Nov 4, 2018
#4
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No, I mean just the 6 faces, 1, 2, 3, 4, 5, 6 to show when 10 dice are rolled.

Nov 4, 2018
#5
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Hi Guest,

What do you think the answer might be?

Melody  Nov 4, 2018
#6
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It does become fairly complicated with the roll of 10 dice!. However, there is this "general formula" that gives the correct answer to any number of dice => 6. The reason for subtracting and adding is because of "overcounting" and "undercounting" as is discussed extensively here:
https://web2.0calc.com/questions/probability_882
So, the actual formula looks like this:
1 - (6 nCr 1 *(5/6)^n - (6 nCr 2*(4/6)^n) + (6 nCr 3*(3/6)^n) - (6 nCr 4*(2/6)^n) + (6 nCr 5*(1/6)^n)), where n=10 in this case. When plugged into the above formula, the result is =0.2718121285, or 27.18% probability.

Nov 4, 2018
#7
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Thanks Ginger

Melody  Nov 4, 2018
#8
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Hi Melody,

This guest post isn’t mine.  This is Mr. BB(2). I suppose I could imitate him using a “monkey see, monkey do” method. To make it more natural, I’d need to eat a few pot-laced brownies and chase them with banana daiquiris infused with 1800 imperial minims of over-proof rum. But I usually just stay sober and troll the BB’s typical dumbness. Here, JB trolls this BB by using my proxy: https://web2.0calc.com/questions/dice-help#r2. JB did an excellent job of throwing this BB off the Troll’s bridge.

This post is much better than his usual inept, sloppy fare. He actually replies with a coherent and usable formula giving its reference, instead of a general equation with an incoherent narrative of BS describing how he derived his answer.  Mr. BB has referenced this post before, here: https://web2.0calc.com/questions/the-minimum-number#r3 Apparently he has a great affinity for Naus’ generating function leading to a modified Sterling number.

Nauseated’s presentation is a wonder, for sure!

GA

PS You're welcome

GingerAle  Nov 6, 2018
#9
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Ok thanks Ginger, and thanks guest too :)

Melody  Nov 8, 2018