5D6 drop the lowest two (math ?)
- Thread starter JiffyPopTart
- Start date
Olgar Shiverstone
Legend
The problem is that the "drop the lowest" of the dice changes the distribution, since the dropped die is itself a random variable, and none of the "typical" discrete distributions -- like hypergeometric -- apply. The convolution of the variables (summing five uniform distributions and then conditionally discarding the two lowest-valued distributions) makes finding an expectation analytically very nasty.
There may very well be a closed form analytical solution, if you want to take the time to evaluate the solution, but frankly, given computing power it's faster just to run a short little program to evaluate all the possibilities.
Note that this is what Nathan's solution above gives us. It's the equivalent (mathematically) of calculating all the possibilities and summing them to get an expectation. Stick Nathan's formula into a basic program and it'll do what I suggested above (though I haven't checked evry step in his math).
There may very well be a closed form analytical solution, if you want to take the time to evaluate the solution, but frankly, given computing power it's faster just to run a short little program to evaluate all the possibilities.
Note that this is what Nathan's solution above gives us. It's the equivalent (mathematically) of calculating all the possibilities and summing them to get an expectation. Stick Nathan's formula into a basic program and it'll do what I suggested above (though I haven't checked evry step in his math).
kkoie
First Post
Now if you had asked ME this question 3 years ago when I was taking a statistical annalysis class for my Masters Degree, I could've answered your question quite rapidly, but as it is... I've slept quite a bit since then, and my statistics book is neck deep in a stack of boxes in the garage. But from what I've seen of the other posts, their formula's look rather sound.
Spreadsheets
Thanks for that spreadsheet TS.
I modified it to give me the results for 3D6 + 1D8 drop lowest.
It has the following results:
3(-4): 0.1%
4-5(-3): 0.8%
6-7(-2): 3.4%
8-9(-1): 8.9%
10-11(+0): 16.4%
12-13(+1): 22.5%
14-15(+2): 23.0%
16-17(+3): 16.7%
18-19(+4): 7.2%
20(+5): 0.9%
The mean is 13.17
I think that I'll modify the spreadsheet a few more times to have results of other possibilities. It can be a pain to copy those numbers, but all the formulas are easy. It is much easier to use a spreadsheet than to try to strain my brain to do it with a standard formula.
Thanks for that spreadsheet TS.
I modified it to give me the results for 3D6 + 1D8 drop lowest.
It has the following results:
3(-4): 0.1%
4-5(-3): 0.8%
6-7(-2): 3.4%
8-9(-1): 8.9%
10-11(+0): 16.4%
12-13(+1): 22.5%
14-15(+2): 23.0%
16-17(+3): 16.7%
18-19(+4): 7.2%
20(+5): 0.9%
The mean is 13.17
I think that I'll modify the spreadsheet a few more times to have results of other possibilities. It can be a pain to copy those numbers, but all the formulas are easy. It is much easier to use a spreadsheet than to try to strain my brain to do it with a standard formula.
JiffyPopTart
Bree-Yark
I haven't had time to read the very long and though out analytical answer (and I will as soon as I get a few moments of peace without having to run around after work) but I just wanted to say YOU GUYS/GALS ROCK!
We have been trying to figure this out at work for a couple weeks now, and got nothin.
DS
We have been trying to figure this out at work for a couple weeks now, and got nothin.
DS
Tabarnak Smokeblower
First Post
glad we could help!
TS
TS
LazarusLong42
First Post
Well, I posted this:
a few months ago in one of these threads. This formula also uses summations--no way around it--to determine the number of different rolls, P, available for a certain result, R, when using k types of dice with d(k) sides each, and rolling n(k) of each type.
I didn't also use the drop-lowest, but I'm going to have to take some time to look through Nathan's solution
dcollins: What you describe would be nice, but using summations like these--though, yes, the mathematical equivalent of the brute-force approach--is exactly what you have to do. The difference being that a mathematical description of the brute force approach is a lot easier to program into a computer.
a few months ago in one of these threads. This formula also uses summations--no way around it--to determine the number of different rolls, P, available for a certain result, R, when using k types of dice with d(k) sides each, and rolling n(k) of each type.
I didn't also use the drop-lowest, but I'm going to have to take some time to look through Nathan's solution
dcollins: What you describe would be nice, but using summations like these--though, yes, the mathematical equivalent of the brute-force approach--is exactly what you have to do. The difference being that a mathematical description of the brute force approach is a lot easier to program into a computer.
Tabarnak Smokeblower
First Post
My brain just blew up again.
TS
TS
LazarusLong42
First Post
Well, explosive cranial decompression isn't all that bad. It could've been your entire body
Morgenstern
First Post
Actually, you can get around the massiv summations somewhat... I solved some of this sort of math for L5R (d10) because you use roll & drop for everything in that game (and it has open ended dice, but we don't need to fret over it here).
In my head, I deal with this as a series of 6x6x6 'outcome' cubes. Those are the dice you are going to keep. for a 5k3 roll, I have 36 of them. you can then start solving for number of possible outcomes of each result, and then use that to take an average (although the # of outcomes is much more useful since it shows you the -shape- of the probibility curve).
With a flat 3k3 you get
3: 1
4: 3
5: 6
6: 10
7: 15
8: 21
9: 25
10: 27
11: 27
12: 25
13: 21
14: 15
15: 10
16: 6
17: 3
18: 1
Which is convieniently symetrical, allowing you to casually determine the average is 10.5. That one is easy - it uses only one outcome cube and you can count through it with little trouble.
With the first extra dropped die (4k3), the range remains the same, but the shape of the curve distorts rapidly. You now have six cubes, each one reprsenting the above plus the 'extra die than could be used to substitute for any of the dice in above cube. If you build all six cubes and and then sum the possibiltes for each individual result (from 3 to 18), you can again see the curve and take an average. The first cube you have an extra '1' to play with and there is no change- no combination is improved by substituting the 1. In the second cube you have an extra '2', which changes 91 of the 216 outcomes of that cube by +1. Having an extra '3' to play with changes 152 of them (91 by +2, 61 by +1). An extra '4' alters 189 (91 by +3, 61 by +2, 37 by +1). A '5' affects 208 outcomes in it's cube (91 by +4, 61 by + 3, 37 by +2, 19+1), and finally a '6' affects 215 outcomes or all but one the 6+6+6 result (91 by +5, 61 by +4, 37 by +3, 19 by +2, 7 by +1). I may build full results tables on this thread if I get bored.
Even so, computing the average is pretty easy with that information (I'll post how next time). In any event, the important thing to notice is that a larger extra roll impacts the average more, but each incremental increase affects far fewer new outcomes... This is significant because when you bring a second dropped die into consideration, it actually will have FAR less impact than the first one. The jump in the average between 3k3 and 4k3 and 5k3 isn't linear. It CAN'T be! As a simple example, if the increase in the average between 3k3 and 4k3 were +3 AND it was linear for each dropped die added, then the average would by higher than your maximum possible result by 6k3 (average 19.5 by that logic). Let me fiddle around with the tables a bit and see if I can't get you a specific answer for 5k3 using d6s. It can't be nearly as hard as doing it for d10s was (you are only discussing 7776 possible combinations instead of 100,000 )...
In my head, I deal with this as a series of 6x6x6 'outcome' cubes. Those are the dice you are going to keep. for a 5k3 roll, I have 36 of them. you can then start solving for number of possible outcomes of each result, and then use that to take an average (although the # of outcomes is much more useful since it shows you the -shape- of the probibility curve).
With a flat 3k3 you get
3: 1
4: 3
5: 6
6: 10
7: 15
8: 21
9: 25
10: 27
11: 27
12: 25
13: 21
14: 15
15: 10
16: 6
17: 3
18: 1
Which is convieniently symetrical, allowing you to casually determine the average is 10.5. That one is easy - it uses only one outcome cube and you can count through it with little trouble.
With the first extra dropped die (4k3), the range remains the same, but the shape of the curve distorts rapidly. You now have six cubes, each one reprsenting the above plus the 'extra die than could be used to substitute for any of the dice in above cube. If you build all six cubes and and then sum the possibiltes for each individual result (from 3 to 18), you can again see the curve and take an average. The first cube you have an extra '1' to play with and there is no change- no combination is improved by substituting the 1. In the second cube you have an extra '2', which changes 91 of the 216 outcomes of that cube by +1. Having an extra '3' to play with changes 152 of them (91 by +2, 61 by +1). An extra '4' alters 189 (91 by +3, 61 by +2, 37 by +1). A '5' affects 208 outcomes in it's cube (91 by +4, 61 by + 3, 37 by +2, 19+1), and finally a '6' affects 215 outcomes or all but one the 6+6+6 result (91 by +5, 61 by +4, 37 by +3, 19 by +2, 7 by +1). I may build full results tables on this thread if I get bored
.Even so, computing the average is pretty easy with that information (I'll post how next time
). In any event, the important thing to notice is that a larger extra roll impacts the average more, but each incremental increase affects far fewer new outcomes... This is significant because when you bring a second dropped die into consideration, it actually will have FAR less impact than the first one. The jump in the average between 3k3 and 4k3 and 5k3 isn't linear. It CAN'T be! As a simple example, if the increase in the average between 3k3 and 4k3 were +3 AND it was linear for each dropped die added, then the average would by higher than your maximum possible result by 6k3 (average 19.5 by that logic). Let me fiddle around with the tables a bit and see if I can't get you a specific answer for 5k3 using d6s. It can't be nearly as hard as doing it for d10s was (you are only discussing 7776 possible combinations instead of 100,000 )...
Last edited:
Similar Threads
