Need some help with complex probability average: three 8-sided dice.

Well, complex for me anyway. Here's a sample problem.

You will roll three 8-sided dice. Each number on the dice represents a number of points you win if you roll that number. Point values are: 1=0pt, 2=0pt, 3=0pt, 4=2pt, 5=4pt, 6=4pt, 7=8pt, 8=10pt. After you roll the three dice, you only win points equal to the highest point value according to the chart (so max of 10 points). What would the average number of points be for rolling the three dice?

Well, I was looking for the expected average, not an actual sample. I think the formulas from this page could be converted, but I'm not quite getting it. http://math.stackexchange.com/quest...-dice-and-only-taking-the-value-of-the-higher

codykhan said:

Well, complex for me anyway. Here's a sample problem.

You will roll three 8-sided dice. Each number on the dice represents a number of points you win if you roll that number. Point values are: 1=0pt, 2=0pt, 3=0pt, 4=2pt, 5=4pt, 6=4pt, 7=8pt, 8=10pt. After you roll the three dice, you only win points equal to the highest point value according to the chart (so max of 10 points). What would the average number of points be for rolling the three dice?

Click to expand...

Well, I was looking for the expected average, not an actual sample. I think the formulas from this page could be converted, but I'm not quite getting it. http://math.stackexchange.com/questi...-of-the-higher

Click to expand...

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

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Apologies. Below is my work, but this answer is clearly too high.


[FONT=MathJax_Math-italic]E[/FONT][FONT=MathJax_Main][[/FONT][FONT=MathJax_Math-italic]X[/FONT][FONT=MathJax_Main]][/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Size2]∑[FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]8[/FONT][/FONT][FONT=MathJax_Main]3[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]512[/FONT][/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[FONT=MathJax_Main]512[/FONT][/FONT][FONT=MathJax_Size2]∑[FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]8[/FONT][/FONT][FONT=MathJax_Main](3[/FONT][FONT=MathJax_Math-italic]x[FONT=MathJax_Main]3[/FONT][/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]4725[FONT=MathJax_Main]512[/FONT][/FONT][FONT=MathJax_Main]≈[/FONT][FONT=MathJax_Main]9.23[/FONT]

codykhan said:

Well, complex for me anyway. Here's a sample problem. You will roll three 8-sided dice. Each number on the dice represents a number of points you win if you roll that number. Point values are: 1=0pt, 2=0pt, 3=0pt, 4=2pt, 5=4pt, 6=4pt, 7=8pt, 8=10pt. After you roll the three dice, you only win points equal to the highest point value according to the chart (so max of 10 points). What would the average number of points be for rolling the three dice?

Click to expand...

This is a very very complex problem depending how it is read. I don't think that the replies have correct translated the question. I understand this statement was most likely not written or edited by a professional.

Lets take it apart. Toss three Octagonal dice. The instructions as posted.
After you roll the three dice, you only win points equal to the highest point value according to the chart.
Note is does not say: the sum of the chart values. Oh no. You only are awarded THE HIGHEST point value. If I were tasked with editing this for publication, I would insist upon clarification while insisting that the above is what the English construction actually means. If that is correct, it is a wonderful question.

Example: Rolling the triple \(\displaystyle (4,7,3)\) you would be awarded \(\displaystyle 8\) points because the highest rolled is a \(\displaystyle 7\) and by the chart the point value is \(\displaystyle 8\).
Now, how many triples have at least one \(\displaystyle 7\) but no \(\displaystyle 8's\). That is the number of ways to have eight points on a single toss.

If \(\displaystyle \mathcal{S}\) is the point value of a single toss the \(\displaystyle \mathscr{P}(\mathcal{S}=0)=\dfrac{3^3}{8^3}\). Why?

codykhan said:

Well, complex for me anyway. Here's a sample problem.

You will roll three 8-sided dice. Each number on the dice represents a number of points you win if you roll that number. Point values are: 1=0pt, 2=0pt, 3=0pt, 4=2pt, 5=4pt, 6=4pt, 7=8pt, 8=10pt. After you roll the three dice, you only win points equal to the highest point value according to the chart (so max of 10 points). What would the average number of points be for rolling the three dice?

Click to expand...

Ok, here's what I'm thinking.

Ok, so there are 8x8x8 = 512 possible permutations. Not too many to generate on a spreadsheet


Pretty close to Denis' estimation!

@Dennis & @Harry_the_cat
I agree with you both.
BTW: I think the OP has a typo: "6=6pts",

Okay, so I thought I had this figured out last night, but my answer is slightly different. I used the equation provided by pka to calculate the probability of each number being the highest rolled, then fed it through calculation for average from probability distribution and came up with the following results:

3 dice (512 combinations)


I've ran some other number sets with varying number of dice rolled through this process and it looks to be accurate, but I'm no math expert. Can someone check my work or point out flaws?
Thanks for all the responses by the way!

The method given above is correct I'm sure, but I would like to know where I went wrong with my logic. The method I used, in detail:

I used pka's equation 3³/8³ to get 0.053 as the probability of not getting dice sides 4-8 of the dice to show up. (0 as highest point value)

I then ran the equation as 4³/8³ to get 0.125 as the probability of not getting dice sides 5-8 of the dice to show up, then subtracted out the 0.053 as these dice combinations do not have the side 4 in them to get 0.072 as the probability of getting 2 as highest point value.

I then ran the equation as 5³/8³ to get 0.244 as the probability of not getting dice sides 6-8 of the dice to show up, then subtracted out the 0.125 as these dice combinations do not have the side 5 in them to get 0.119 as the probability of getting 4(as die side 5) as highest point value.

I then ran the equation as 6³/8³ to get 0.422 as the probability of not getting dice sides 7-8 of the dice to show up, then subtracted out the 0.244 as these dice combinations do not have the side 6 in them to get 0.178 as the probability of getting 4(as die side 6) as highest point value.

I then ran the equation as 7³/8³ to get 0.700 as the probability of not getting dice side 8 of the dice to show up, then subtracted out the 0.244 as these dice combinations do not have the side 7 in them to get 0.178 as the probability of getting 8 as highest point value.

I then ran the equation as 8³/8³ to get 1.0 as the probability of getting dice sides 1-8 of the dice to show up, then subtracted out the 0.700 as these dice combinations do not have the side 8 in them to get 0.3 as the probability of getting 10 as highest point value.

I then multiplied each of these probabilities by the point value of the appropriate number: 0.053*0=0, 0.072*2=0.144, 0.119*4=0.476, 0.178*4=0.712, 0.278*8=2.224, 0.3*10=3.0.

I then found the sum of all the value probabilities: 0+0.144+0.476+0.712+2.224+3=6.556

Denis said:

Harry, your last column should be:
0
0
0
74
244
364
1016
1690
====
3388

3388 / 512 = 6.6171875

And I emerge the WINNAH with my UBasiced 6.616 :cool:

You musta been tired from practicing crypt construction

You and Pka (for agreeing with you) must go to the corner
for 24.4 minutes. You may bring 3 8-sided dice to practice with...

Click to expand...

Yes (I hate to admit it) you are correct. *Hangs head in shame* Sorry for my silly mistake!

codykhan said:

Okay, so I thought I had this figured out last night, but my answer is slightly different. I used the equation provided by pka to calculate the probability of each number being the highest rolled, then fed it through calculation for average from probability distribution and came up with the following results:

3 dice (512 combinations)

roll/points	probability	average
3(0)	.053	0
4(2)	.072	0.144
5(4)	.119	0.476
6(4)	.178	0.712
7(8)	.278 .248	2.224
8(10)	.300 .330	3
sum	1	6.556

I've ran some other number sets with varying number of dice rolled through this process and it looks to be accurate, but I'm no math expert. Can someone check my work or point out flaws?
Thanks for all the responses by the way!

Click to expand...

See corrections in red.

Okay, I found my error thanks to Harry-the-cat. The probability for rolls not including 8 calculated to 0.6699. I saw it as 0.699 and immediately rounded it to 0.7. Otherwise the method works. Thanks to everyone who provided input!

Yes, thanks Denis. I have made a simple spreadsheet with the method you provided. Now all I have to to is plug in my k, f, v and I have my answer to any combination.