probability w/dice

Printable View

Show 60 post(s) from this thread on one page

Nov 30th, 2003, 10:19 PM
DiGiTaIErRoR
Quote:

Originally posted by Spajeoly
yeah, my bad, i was on 1 hour of sleep when I did that.

See, that's the trick you need to learn, admit when you're wrong. Nothing can have a 125% of happening.

I've even commented my code. :lol:

Run this code:
VB Code:
```
Dim x As Long, lRnd As Long, lAvg As Long
Dim c As Long, y As Long
 
Randomize
For y = 1 To 1000000
  'Four rolls
  For x = 1 To 4
    'compute random number 1 to 6
    lRnd = Int(Rnd * 6) + 1
    'if roll is 5 or 6, add it to the counter, c
    If lRnd = 5 Or lRnd = 6 Then c = c + 1
  Next x
  'add counters together, so we can get an average
  lAvg = lAvg + c
  'reset counter
  c = 0
Next y
 
MsgBox lAvg / 1000000 ' display average 5 or 6 per four rolls
```
You'll get a messagebox with ~1.33.

Wow, seems that when you get 4 random numbers, 1 to 6, that 5 or 6 will appear 133% of the time.

Not surprising. Since the single probability is 1/3. And you roll it four times. Thus for every three rolls(1/3) it'll happen once, so every four rolls it'll happen more than once, wow, 1/3*4=4/3 too easy.

Ok. I don't want to repeat myself too much.

:lol:
Dec 1st, 2003, 01:32 AM
WorkHorse

When looking at total posibilities, keep in mind that you may have already succeeded in the previous roll.

On 1st roll, you have have 3 possibles (3^1 = 3). You have 1/3 chances of success = 1/3 * 3 = 1. Probability is chances of success divided by possibles = 1/3.

On 2nd roll, you have have 9 possibles (3^2 = 9). You have 1/3 chances of already having succeeded on 1st roll = 1/3 * 9 = 3. You also have 1/3 chance of success on remaining possibles (9-3 = 6) = 1/3 * 6 = 2. Total chances of success are 3+2=5. Probability is chances of success divided by possibles = 5/9.

On 3rd roll, you have have 27 possibles (3^3 = 27). You have 1/3 chances of already having succeeded on 1st roll = 1/3 * 27 = 9. You also have 1/3 chance of success on remaining possibles (27-9 = 18) = 1/3 * 18 = 6. You also have 1/3 chance of success on remaining possibles (9 used on 1st + 6 used on 2nd = 15 used. 27-15 = 12) = 1/3 * 12 = 4. Total chances of success are 9+6+4 = 19. Probability is chances of success divided by possibles = 19/27.

On 4th roll, you have have 81 possibles (3^4 = 81). You have 1/3 chances of already having succeeded on 1st roll = 1/3 * 81 = 27. You also have 1/3 chance of success on remaining possibles (81-27 = 54) = 1/3 * 54 = 18. You also have 1/3 chance of success on remaining possibles (27 used on 1st + 18 used on 2nd = 45 used. 81-45 = 36) = 1/3 * 36 = 12. You also have 1/3 chance of success on remaining possibles (27 + 18 + 12 = 57 used. 81-45 = 24) = 1/3 * 24 = 8. Total chances of success are 27+18+12+8 = 65. Probability is chances of success divided by possibles = 65/81.

I'm not sure how to express this in an equation, but it would be like this:

Let r’ = rolls
Let t = total possible combinations
Let p = probability of success for one roll
Let c = chance of success

t =3^r
p =1/3

r0 = 0
r1 = ((t-r0)*p) + r0
r2 = ((t-r1)*p) + r1
r3 = ((t-r2)*p) + r2
r4 = ((t-r3)*p) + r3
etc…

c = r’ / t

So for roll 4 you would have

t =81

r0 = 0
r1 = (81-0) * 1/3 = 27 + 0 = 27 (27/81 probability = .3333 or 1/3)
r2 = (81-27) * 1/3 = 18 + 27 = 45 (45/81 probability = .5555 or 5/9)
r3 = (81-45) * 1/3 = 12 + 45= 57 (57/81 probability = .7037 or 19/27)
r4 = (81-57) * 1/3 = 8 + 57 = 65 (65/81 probability = .8025 or 65/81)

So it is like Acidic & Kedaman posted in the begining.
Dec 1st, 2003, 02:19 AM
kedaman

DE: you need to check up the definition of probability.

and here's a final confirmation (enumerating the perms in Haskell) that the probability is... *drum roll*

Code:

Main> fromInt(length [(a,b,c,d)|a<-[1..6],b<-[1..6],c<-[1..6],d<-[1..6],a>4||b>4||c>4||d>4])/6^4 0.802469
Dec 1st, 2003, 03:23 AM
sql_lall

mmmm...

Gee, i hope no-one asks "How many ppl do you need before you are >50% likely to have 2 with the same birthday"...

Why? because the reasoning is EXACTLY the same. Just take the inverse.

P(5 OR 6) = 1 - P(not 5 nor 6) **for 4 rolls**

for a single roll, P(not 5 nor 6) = 4/6 = 2/3
=> for 4 rolls: (not 5 or 6) AND (not 5 or 6) AND (not 5 or 6) AND (not 5 or 6) = (2/3) ^ 4 = 16/81

=> 1-16/81 = surprise surprise : 65/81!!! :eek:

P.S DE: change ur code so that if multiple 5s/6s are rolled in the same group of 4, that c doesn't increase by more than 1.

DE, I ran your code after adapting it to Excel while @ work, it does what you say it does, but as lall here says, change ur code so that if multiple 5s/6s are rolled in the same group of 4, that c doesn't increase by more than 1.
VB Code:
Dim x As Long, lRnd As Long, lAvg As Long
Dim c As Long, y As Long
 
Randomize
For y = 1 To 1000000
  'This is fine
  For x = 1 To 4
    'This is fine
    lRnd = Int(Rnd * 6) + 1
    'if roll is 5 or 6, add it to the counter, c
    If lRnd = 5 Or lRnd = 6 Then c = c + 1
  Next x
  'add only one to the fücking sum if we have any.
  If c > 0 Then lAvg = lAvg + 1
  'This is fine
  c = 0
Next y
 
MsgBox lAvg / 1000000 ' display the CORRECT average 5 or 6 per four rolls
I think we should all give up and let DE sit in his world, at least until he takes his reality meds.

Dec 1st, 2003, 05:28 AM
sw_is_great

I think this is simple

lets solve it like this ....

if we suppose that we throw only one die then

probability for 5 or 6 = 2/6 = 1/3

if we throw two dicethen
probability = (2+2)/(6+6) = 4/12 = 1/3

||ly
for 4 dice it will be 1/3
Dec 1st, 2003, 06:00 AM
DiGiTaIErRoR

Ok, modified code, I get .802....

:o
Dec 1st, 2003, 09:10 AM
kedaman

Quote:

Originally posted by sw_is_great
I think this is simple

lets solve it like this ....

if we suppose that we throw only one die then

probability for 5 or 6 = 2/6 = 1/3

if we throw two dicethen
probability = (2+2)/(6+6) = 4/12 = 1/3

||ly
for 4 dice it will be 1/3

+ means "disjunct two mutually exclusive outcomes" so it would be chosing one of the 4 die and throwing it.
Dec 2nd, 2003, 06:45 PM
dolor

why must you complicate a problem that is already so simplified...
1-(2/3)^4 ...tada! Jeebus.
Dec 2nd, 2003, 07:18 PM
kedaman

Quote:

Originally posted by dolor
why must you complicate a problem that is already so simplified...
1-(2/3)^4 ...tada! Jeebus.

a formula means so much more if there is an explanation for it
Dec 3rd, 2003, 05:02 PM
dolor

Quote:

Originally posted by kedaman
a formula means so much more if there is an explanation for it

yea, but isn't there a point in your life when you stopped trying to explain why the derivative of 2x = 2 ? ..idunno, that was just an example.
Dec 3rd, 2003, 05:13 PM
TheManWhoCan

I think this thread should be removed because it's almost entirely full of ****!
Dec 3rd, 2003, 05:31 PM
Acidic

if you hadn't said that, this thread would have kept going down the page and ended up in the abyss of "next page". Just let it slowly dissappear.
Dec 4th, 2003, 05:42 AM
Spajeoly

Quote:

Originally posted by TheManWhoCan
I think this thread should be removed because it's almost entirely full of ****!

Look who talkin'.

Show 60 post(s) from this thread on one page