probability w/dice

[dolor]

A fair die is tossed four times. What is the probability that it lands with either a 5 or 6 on top at least once?

[Acidic]

I'm not sure about this but:
(N nCr X) * (P^X) * ((1-P)^(N-X))
can be used to find this where
N is the total number of throws, X the number we want and P the probability of it happening each time.
so if N=4, X=1 and P=1/3
then
answer = 0.3951
hmm.. lower than I expected. someone tell me if I'm right.

[Acidic]

Another method that seems to make more sense. The only possible outcome that wouldn't do this is if it NEVER rolled a 5 or a 6. The chances of that is 2/3. The chances of it four times is (2/3)^4
= 0.19753
Now, we want all other outcomes so:
1-0.19753=0.80247

[kedaman]

Originally posted by Acidic
I'm not sure about this but:
(N nCr X) * (P^X) * ((1-P)^(N-X))
can be used to find this where
N is the total number of throws, X the number we want and P the probability of it happening each time.
so if N=4, X=1 and P=1/3
then
answer = 0.3951
hmm.. lower than I expected. someone tell me if I'm right.

the binominal distribution can be used but you have to sum the probabilities for where X is 1,2,3 and 4 (but its easier to do it the other way but it only works in that special case)

[Acidic]

thx for clearing that up for me.

[DiGiTaIErRoR]

Lets make this more simple.

Lets roll 4 dice, what's the chance of getting a 5 or a 6?

4*6 = 24 possibilities
4*2 = 8

8/24 = 1/3

You have a 1/3 chance.

[kedaman]

Originally posted by DiGiTaIErRoR
Lets make this more simple.

Lets roll 4 dice, what's the chance of getting a 5 or a 6?

4*6 = 24 possibilities
4*2 = 8

8/24 = 1/3

You have a 1/3 chance.

I don't follow your logic, 4*2?

[dolor]

Originally posted by kedaman
I don't follow your logic, 4*2?

i think he means that there are 24 total possibilities (4 * 6) and 8 possibilities that it'll be a five or six (4 * 2)

[DiGiTaIErRoR]

Originally posted by dolor
i think he means that there are 24 total possibilities (4 * 6) and 8 possibilities that it'll be a five or six (4 * 2)

Uhm... yea. But I think that's wrong.

My final answer is 4/3.

And I can prove it.

Dim x As Long, lRnd As Long, lAvg As Long
Dim c As Long, y As Long
 
Randomize
For y = 1 To 1000000
  For x = 1 To 4
    lRnd = Int(Rnd * 6) + 1
    If lRnd = 5 Or lRnd = 6 Then c = c + 1
  Next x
  lAvg = lAvg + c
  c = 0
Next y
 
MsgBox lAvg / 1000000

[Spajeoly]

I hope you stopped long enough to consider your logic. Your statement is that the probability of rolling a 5 or a 6 on four dice is equal to the SUM of the probabilities of rolling them on one... which says that not only is it MORE THAN 100% LIKELY to happen, it's even MORE MORE than 100% likely not to happen... and then the universe implodes.

Well, that or you're wrong, of course.. which you are.

Your program didn't experimentally determine anything CLOSE to the probability of getting at least one 5 or 6 in four rolls. What it does is add together the number of times a 5 or a 6 is rolled in 4,000,000 rolls, then divides the total by 1,000,000 - how is THAT useful? Try it again and instead of incrementing your total by "c", increment the total by one if "c" is greater than zero.

The probability of getting a 5 or a 6 on four dice rolls is 65/81 - you can get this mathematically using the formula of the chance of it NOT occurring during each iteration, to the power of the number of attempts... all subtracted from one.. so:

1 - (4/6)^4

Or you can write another program and be a bit more careful in it's coding before using it as proof =)

[kedaman]

DE: even your original logic seemed better in comparison athough theyre both flawed, a 133% probability

i think he means that there are 24 total possibilities (4 * 6) and 8 possibilities that it'll be a five or six (4 * 2)

which permutations exactly? I bet typing out all the 24 and which of 8 contain at least one 5 and 6 can't be that hard.

[TheManWhoCan]

Let X (a DRV) represent the number of times a 5 or 6 is thrown over four throws:

X ~ B(4, 1/3)

P(X > 0) = 1 - P(X = 0) = 1 - (2/3)⁴ = 1 - 0.1975 = 0.8025

[Acidic]

hehe, all this and I was right (on my second method)

[TheManWhoCan]

Indeed, I only provided my post because it seems a lot of the other posts lacked any kind of systematic method.

[Spajeoly]

Originally posted by Acidic
hehe, all this and I was right (on my second method)

Yes you was, my friend

[ZaidGS]

make a tree !!!
the usual and most easy and accurate thing.
we try 4 times, and if 5 or 6 is there then we dont have to finsih,
cuz the point is done !!!!
(starred(**) values should be added)

-- 5 or 6 : 1/3 (**)
-- other: 2/3
-- -- 5 or 6 : 2/3 * 1/3 (**)
here 2/3 is because in a tree we multiply the previous branch
probability
-- -- other: 2/3 * 2/3
-- -- -- 5 or 6: 2/3 * 2/3 * 1/3 (**)
-- -- -- other: 2/3 * 2/3 * 2/3
-- -- -- -- 5 or 6: 2/3 * 2/3 * 2/3 * 1/3 (**)
-- -- -- -- other: 2/3 * 2/3 * 2/3 * 2/3 (this value makes a trick)

but here is a trick, dont add the starred values.
the last value is the only chance for that condition not occuring
so probability is
1 - (2/3)^4
which is 80.2%

[TheManWhoCan]

And after all that you rounded it wrong!

[dolor]

Originally posted by TheManWhoCan
And after all that you rounded it wrong!

ooo! burn! ...now, i just think that's a little too picky but hey, it happens.

[ZaidGS]

i did NOT round it wrong,
i just gave it to 3 significant figures,
and for that accuracy i guess im 100% right !!!
to be more accurate its:
80.2469%

[sw_is_great]

Total possible outcomes = 6+6+6+6 = 24

Favouravble outcomes = 2+2+2+2 = 8

probability = 8/24 = 1/3

[dolor]

Originally posted by ZaidGS
i did NOT round it wrong,
i just gave it to 3 significant figures,
and for that accuracy i guess im 100% right !!!
to be more accurate its:
80.2469%

...fiesty, eh? c'mon, let's step it down a notch

[kedaman]

yeah, after all we are here to help or solve problems for fun, not argue over small things (or getting picky )

[ZaidGS]

the are really and ,
but if sm1 has to be picky he should pick it right!!!!
not be so picky when he's mistaken !!!!

[DiGiTaIErRoR]

I suppose you could say it's 100% probable.

The likelihood of a 5 or 6 being rolled over 4 rolls is 4/3.

Probability: A number expressing the likelihood that a specific event will occur, expressed as the ratio of the number of actual occurrences to the number of possible occurrences.

The probability of it happening on one roll is 1/3. So for every 3 rolls one actual will occur. Every four rolls more than one certain to appear.

To make a simpler analogy,

Take a coin. What's the probabilty that if you flip the coin one hundred times it'll be heads atleast once? Now, just about everyone can tell you that on average it'll occur 50 times in one hundred flips. That's 1/2*100=50.

So, 1/3*4=4/3.

Follow yet?

My code from my previous post demonstrates my logic perfectly.

Using the prior definition, there are only 6 possible occurances. However, with the four rolls, there becomes 8 possible actual occurences. 8/6 = 4/3.

A coin has 2 possible occurances, over 100 flips, that gives 100 possible actual occurances of either side. 100/2 = 50.

[TheManWhoCan]

Right, so if the probability is 4/3 as you suggest, how come the number of actual occurences is greater than the number of possible occurences...

Your logic for the entire last post is flawed. With the coins thing, the probability you'll get heads at least once:

P(H > 0) = 1 - P(0) = 1 - (1/2)¹⁰⁰ = 1.0000 to 4 d.p

[sw_is_great]

probability is 4/3

how can it be more than 1 (100%) ?????

[kedaman]

it can probably be if you define probability like DE does

[Spajeoly]

He must make his own dice with all 5's & 6's, not to mention, one that will somehow display 2 sides up @ the same time.

1/3 * 4 = 4/3? Wouldn't it really equal 4/12?

[ZaidGS]

no offence
the right answer is 80.2%
no matter what you say,
its math not something to be discused freely
a right answer is a right answer !!!!
and its 80.2%
there is nothing in such a problem to get
a probability over 100% thats non-sense

[kedaman]

Originally posted by ZaidGS
no offence
the right answer is 80.2%
no matter what you say,
its math not something to be discused freely
a right answer is a right answer !!!!
and its 80.2%
there is nothing in such a problem to get
a probability over 100% thats non-sense

even some math is concidered controversial, but in this case we can come to agreement on that the answer is precisely 65/81

[Acidic]

agreed, as everyonw should have seen from my post

[dolor]

nonsense! it's all nonsense. just look at all the different answers we've got. i mean, more than 100%?!?

[kedaman]

count

Main> [(a,b,c,d)|a<-[1..6],b<-[1..6],c<-[1..6],d<-[1..6],a>4||b>4||c>4||d>4]
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1,2,1),(5,1,2,2),(5,1,2,3),(5,1,2,4),(5,1,2,5),(5,1,2,6),(5,1,3,1),(5,1,3,2),(5,1,3,3),(5,1,3,4),(5,1,3,5),(5,1,3,6),(5,1,4,1),(
5,1,4,2),(5,1,4,3),(5,1,4,4),(5,1,4,5),(5,1,4,6),(5,1,5,1),(5,1,5,2),(5,1,5,3),(5,1,5,4),(5,1,5,5),(5,1,5,6),(5,1,6,1),(5,1,6,2)
,(5,1,6,3),(5,1,6,4),(5,1,6,5),(5,1,6,6),(5,2,1,1),(5,2,1,2),(5,2,1,3),(5,2,1,4),(5,2,1,5),(5,2,1,6),(5,2,2,1),(5,2,2,2),(5,2,2,
3),(5,2,2,4),(5,2,2,5),(5,2,2,6),(5,2,3,1),(5,2,3,2),(5,2,3,3),(5,2,3,4),(5,2,3,5),(5,2,3,6),(5,2,4,1),(5,2,4,2),(5,2,4,3),(5,2,
4,4),(5,2,4,5),(5,2,4,6),(5,2,5,1),(5,2,5,2),(5,2,5,3),(5,2,5,4),(5,2,5,5),(5,2,5,6),(5,2,6,1),(5,2,6,2),(5,2,6,3),(5,2,6,4),(5,
2,6,5),(5,2,6,6),(5,3,1,1),(5,3,1,2),(5,3,1,3),(5,3,1,4),(5,3,1,5),(5,3,1,6),(5,3,2,1),(5,3,2,2),(5,3,2,3),(5,3,2,4),(5,3,2,5),(
5,3,2,6),(5,3,3,1),(5,3,3,2),(5,3,3,3),(5,3,3,4),(5,3,3,5),(5,3,3,6),(5,3,4,1),(5,3,4,2),(5,3,4,3),(5,3,4,4),(5,3,4,5),(5,3,4,6)
,(5,3,5,1),(5,3,5,2),(5,3,5,3),(5,3,5,4),(5,3,5,5),(5,3,5,6),(5,3,6,1),(5,3,6,2),(5,3,6,3),(5,3,6,4),(5,3,6,5),(5,3,6,6),(5,4,1,
1),(5,4,1,2),(5,4,1,3),(5,4,1,4),(5,4,1,5),(5,4,1,6),(5,4,2,1),(5,4,2,2),(5,4,2,3),(5,4,2,4),(5,4,2,5),(5,4,2,6),(5,4,3,1),(5,4,
3,2),(5,4,3,3),(5,4,3,4),(5,4,3,5),(5,4,3,6),(5,4,4,1),(5,4,4,2),(5,4,4,3),(5,4,4,4),(5,4,4,5),(5,4,4,6),(5,4,5,1),(5,4,5,2),(5,
4,5,3),(5,4,5,4),(5,4,5,5),(5,4,5,6),(5,4,6,1),(5,4,6,2),(5,4,6,3),(5,4,6,4),(5,4,6,5),(5,4,6,6),(5,5,1,1),(5,5,1,2),(5,5,1,3),(
5,5,1,4),(5,5,1,5),(5,5,1,6),(5,5,2,1),(5,5,2,2),(5,5,2,3),(5,5,2,4),(5,5,2,5),(5,5,2,6),(5,5,3,1),(5,5,3,2),(5,5,3,3),(5,5,3,4)
,(5,5,3,5),(5,5,3,6),(5,5,4,1),(5,5,4,2),(5,5,4,3),(5,5,4,4),(5,5,4,5),(5,5,4,6),(5,5,5,1),(5,5,5,2),(5,5,5,3),(5,5,5,4),(5,5,5,
5),(5,5,5,6),(5,5,6,1),(5,5,6,2),(5,5,6,3),(5,5,6,4),(5,5,6,5),(5,5,6,6),(5,6,1,1),(5,6,1,2),(5,6,1,3),(5,6,1,4),(5,6,1,5),(5,6,
1,6),(5,6,2,1),(5,6,2,2),(5,6,2,3),(5,6,2,4),(5,6,2,5),(5,6,2,6),(5,6,3,1),(5,6,3,2),(5,6,3,3),(5,6,3,4),(5,6,3,5),(5,6,3,6),(5,
6,4,1),(5,6,4,2),(5,6,4,3),(5,6,4,4),(5,6,4,5),(5,6,4,6),(5,6,5,1),(5,6,5,2),(5,6,5,3),(5,6,5,4),(5,6,5,5),(5,6,5,6),(5,6,6,1),(
5,6,6,2),(5,6,6,3),(5,6,6,4),(5,6,6,5),(5,6,6,6),(6,1,1,1),(6,1,1,2),(6,1,1,3),(6,1,1,4),(6,1,1,5),(6,1,1,6),(6,1,2,1),(6,1,2,2)
,(6,1,2,3),(6,1,2,4),(6,1,2,5),(6,1,2,6),(6,1,3,1),(6,1,3,2),(6,1,3,3),(6,1,3,4),(6,1,3,5),(6,1,3,6),(6,1,4,1),(6,1,4,2),(6,1,4,
3),(6,1,4,4),(6,1,4,5),(6,1,4,6),(6,1,5,1),(6,1,5,2),(6,1,5,3),(6,1,5,4),(6,1,5,5),(6,1,5,6),(6,1,6,1),(6,1,6,2),(6,1,6,3),(6,1,
6,4),(6,1,6,5),(6,1,6,6),(6,2,1,1),(6,2,1,2),(6,2,1,3),(6,2,1,4),(6,2,1,5),(6,2,1,6),(6,2,2,1),(6,2,2,2),(6,2,2,3),(6,2,2,4),(6,
2,2,5),(6,2,2,6),(6,2,3,1),(6,2,3,2),(6,2,3,3),(6,2,3,4),(6,2,3,5),(6,2,3,6),(6,2,4,1),(6,2,4,2),(6,2,4,3),(6,2,4,4),(6,2,4,5),(
6,2,4,6),(6,2,5,1),(6,2,5,2),(6,2,5,3),(6,2,5,4),(6,2,5,5),(6,2,5,6),(6,2,6,1),(6,2,6,2),(6,2,6,3),(6,2,6,4),(6,2,6,5),(6,2,6,6)

[kedaman]

,(6,3,1,1),(6,3,1,2),(6,3,1,3),(6,3,1,4),(6,3,1,5),(6,3,1,6),(6,3,2,1),(6,3,2,2),(6,3,2,3),(6,3,2,4),(6,3,2,5),(6,3,2,6),(6,3,3,
1),(6,3,3,2),(6,3,3,3),(6,3,3,4),(6,3,3,5),(6,3,3,6),(6,3,4,1),(6,3,4,2),(6,3,4,3),(6,3,4,4),(6,3,4,5),(6,3,4,6),(6,3,5,1),(6,3,
5,2),(6,3,5,3),(6,3,5,4),(6,3,5,5),(6,3,5,6),(6,3,6,1),(6,3,6,2),(6,3,6,3),(6,3,6,4),(6,3,6,5),(6,3,6,6),(6,4,1,1),(6,4,1,2),(6,
4,1,3),(6,4,1,4),(6,4,1,5),(6,4,1,6),(6,4,2,1),(6,4,2,2),(6,4,2,3),(6,4,2,4),(6,4,2,5),(6,4,2,6),(6,4,3,1),(6,4,3,2),(6,4,3,3),(
6,4,3,4),(6,4,3,5),(6,4,3,6),(6,4,4,1),(6,4,4,2),(6,4,4,3),(6,4,4,4),(6,4,4,5),(6,4,4,6),(6,4,5,1),(6,4,5,2),(6,4,5,3),(6,4,5,4)
,(6,4,5,5),(6,4,5,6),(6,4,6,1),(6,4,6,2),(6,4,6,3),(6,4,6,4),(6,4,6,5),(6,4,6,6),(6,5,1,1),(6,5,1,2),(6,5,1,3),(6,5,1,4),(6,5,1,
5),(6,5,1,6),(6,5,2,1),(6,5,2,2),(6,5,2,3),(6,5,2,4),(6,5,2,5),(6,5,2,6),(6,5,3,1),(6,5,3,2),(6,5,3,3),(6,5,3,4),(6,5,3,5),(6,5,
3,6),(6,5,4,1),(6,5,4,2),(6,5,4,3),(6,5,4,4),(6,5,4,5),(6,5,4,6),(6,5,5,1),(6,5,5,2),(6,5,5,3),(6,5,5,4),(6,5,5,5),(6,5,5,6),(6,
5,6,1),(6,5,6,2),(6,5,6,3),(6,5,6,4),(6,5,6,5),(6,5,6,6),(6,6,1,1),(6,6,1,2),(6,6,1,3),(6,6,1,4),(6,6,1,5),(6,6,1,6),(6,6,2,1),(
6,6,2,2),(6,6,2,3),(6,6,2,4),(6,6,2,5),(6,6,2,6),(6,6,3,1),(6,6,3,2),(6,6,3,3),(6,6,3,4),(6,6,3,5),(6,6,3,6),(6,6,4,1),(6,6,4,2)
,(6,6,4,3),(6,6,4,4),(6,6,4,5),(6,6,4,6),(6,6,5,1),(6,6,5,2),(6,6,5,3),(6,6,5,4),(6,6,5,5),(6,6,5,6),(6,6,6,1),(6,6,6,2),(6,6,6,
3),(6,6,6,4),(6,6,6,5),(6,6,6,6)]
(90889 reductions, 162987 cells, 1 garbage collection)
Main>

[Spajeoly]

Eh.
You nut.

[dolor]

Originally posted by kedaman
count

Main> [(a,b,c,d)|a<-[1..6],b<-[1..6],c<-[1..6],d<-[1..6],a>4||b>4||c>4||d>4]
[(1,1,1,5),(1,1,1,6),(1,1,2,5),(1,1,2,6),(1,1,3,5),(1,1,3,6),(1,1,4,5),(1,1,4,6),(1,1,5,1),(1,1,5,2),(1,1,5,3),(1,1,5,4),(1,1,5,
5),(1,1,5,6),(1,1,6,1),(1,1,6,2),(1,1,6,3),(1,1,6,4),(1,1,6,5),(1,1,6,6),(1,2,1,5),(1,2,1,6),(1,2,2,5),(1,2,2,6),(1,2,3,5),(1,2,
3,6),(1,2,4,5),(1,2,4,6),(1,2,5,1),(1,2,5,2),(1,2,5,3),(1,2,5,4),(1,2,5,5),(1,2,5,6),(1,2,6,1),(1,2,6,2),(1,2,6,3),(1,2,6,4),(1,

whatever happened to some of the ones w/o a 5 or 6 like (1,1,1,1) ?

[kedaman]

they were excluded from the list with the condition a>4||b>4||c>4||d>4

[DiGiTaIErRoR]

Originally posted by Spajeoly
He must make his own dice with all 5's & 6's, not to mention, one that will somehow display 2 sides up @ the same time.

1/3 * 4 = 4/3? Wouldn't it really equal 4/12?

Uhm. 4/12 = 1/3. Rofl.

1/3 * 4/1

1*4   4
- - = -
3*1   3

[DiGiTaIErRoR]

Originally posted by sw_is_great
probability is 4/3

how can it be more than 1 (100%) ?????

Consider my example, 100 coin flips, what's the probability that you'll get heads atleast once? Well, heads will occur 50 times(on average), so I suppose you can say 100%.

You people are using the wrong method to figure out this particular probility. Your setups are calculating a desired result for each roll. But the problem only needs it to happen on one roll.

[Spajeoly]

Originally posted by DiGiTaIErRoR
Uhm. 4/12 = 1/3. Rofl.

1/3 * 4/1

1*4   4
- - = -
3*1   3

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.