Question

Question about probability

Answer 1

As a general rule, simply posting a problem statement with no work shown is the slowest and least effective way to obtain assistance.

Show your work. What's your plan?

Answer 2

uhm, My train of thought right now, is to have 3 rounds.

Answer 3

So, the 8 sided die is rolled first. then the 8sided and 15 sided a rolled together and these numbers are added. then 8 sided and 15 sided are eahc rolled 2 times each.

Answer 4

And then on that 3rd round, you'd add up the sum of those 4 rolls, and that'd be your nuew number. But I'm really lost on the best approach here. this is just what i"m thinking.

Answer 5

anything?

Answer 6

3 rounds won't do unless you adjust for the fact that you can't possibly get 1 or 3.

Answer 7

well in round 1, i'm just rolling the 8 die once and nothing else, so it'd give me everything from 1-8 right? Or how else should I approach?

Answer 8

Right, but in your next round, you cannot possibly have less than 2.

Answer 9

mhmm, but that's to cover all the number 2-23 inclusive

Answer 10

and then once i add up the results of these 3 rounds, i was thinking of subtracting out the duplicates?

Answer 11

@ganeshie8 @freckles

Answer 12

You're also missing the requirement that each value, 1-50, must have an equal chance. That doesn't happen when a single result can be obtained multiple ways, but some other result has only one way, Keep thinking.

Maybe subtract one from each roll and then add one to the total? That might be closer, but still unequal probability.

Answer 13

hmmm, I thought of that too, But I'm still all over the place with this question. :/ But thank you, i'll keep thinking. It's actually been a long time of thinking of this question. a toughy.

Answer 14

You can effectively turn the 8 sided die into a 2 sided die by only looking at if the number that pops up is even or odd. Similarly you could make it a 4-sided die by only looking at what it is in mod 4. Also we can turn the 15 sided die into a 3 or 5 sided die same idea.

Not sure if that helps us though, something fun to think about.

Answer 15

Random thoughts. Without cutting the dice (as rumored by Kainui), it is required that all 50 values must be equally probable. Since there are 8*15 = 120 possible outcomes, each value, 1-50, must appear 120/50= 2.4 times. Since this is not possible, you will have to scale by a factor of 5. Thus, each value must appear 12 of 600 total buckets.


Not sure if that helps us though, something fun to think about.

Answer 16

aww man, come on guys. hahaha I'm dying here.

Answer 17

lol

Answer 18

Notice that all 120 outcomes are equally likely.
Assign any 100 of these outcomes to the integers 1-50.

Just ignore the remaining 20 outcomes.
When any of these 20 outcomes occur, just ignore and roll the dice again.

Answer 19

what do you mean? sorry, kinda lost.

Answer 20

as in doing an encoding type thing? where every pair up to 100 is 1-50?

Answer 21

Yes.

outcome1, outcome2 ---> integer1
outcome3, outcome4 ---> integer2
....
outcome99, outcome100 --> integer50

Answer 22

outcome101, outcome102, .... outcome120 --> ignore

Answer 23

I think that is the simplest method by which you can get a random sample of 1 integer from 50 integers

Answer 24

Just for fun, here's an encoding that is close.

E = Roll of 8-sided die.
F = Roll of 15-sided die.

8*(E-1) + F

Unfortunately, it goes to 71, not 50.

Really, though, listen to ganeshie

Answer 25

(first dice express first digit ,second dice express second and third digit) mod 50
note some values u need to ignore to make it equally likely

Answer 26

If you are allowed to roll a die more than one time, you can roll the 15 sided die 10 times to get 150 equally likely outcomes. Then you can divide these 150 outcomes into 50 equal groups, assigning 3 outcomes to each of 50 integers.

This is same as the method tkhunny has suggested earlier.

Answer 27

Nice. I didn't think of ignoring the 8.

Answer 28

Hey guys, I don't think that would quite work. I was discussing the problem with my TA and the error he pointed out is, there is only one way to get the number 10. However, there are multiple ways to get the other integers, making the distribution sightly skewed right?

Answer 29

@ganeshie8

Answer 30

In the indexing, keep the 8 and the 15 in order, thus
1,9
2,8
3,7
4,6
5,5
6,4
7,3
8,2
Even though there are 8 ways to get 10, this is of no concern. Simply DON'T add them and the outcomes are equally likely and there is no duplication. We should have made this point more clearly.

Note: There is no 9,1 because you can't roll a 9 on an 8-sided die.

Answer 31

oh that is interesting

Answer 32

Wiat, what do you mean keep it in order?

Answer 33

cause we're only adding the 15 sided die 10 times. nothing to do with the 8 right?

Answer 34

You can use both.

Answer 35

but! by doing that, we have 120 numbers and 120 is not evenly divisible by 50.

Answer 36

so we can't evenly map multiple numbesr to 1-50. right?

Answer 37

This is why we take only the first 50 and discard the rest.

Or, to do less rerolling, double up
#1,#51==> 1
#2,#52 ==> 2
...
#50, #100 ==> 50
#101-#120 ==> Roll Again.

Our task is ONLY to make them equally probable.

Answer 38

but the thing is, if this was an algorithm, there would be no end condition

Answer 39

i was thinking rolling both dice twice, and confirm the sum from base 8 to 10 xD
havn't try it though .