/sci/ - Science & Math

>>5036948
math

>>5036967
yep.

I'm in a probability class right now and barely got my textbook an hour or so ago from someone here on sci.


https://rapidshare.com/files/1103488644/Probability and Statistics for Engineers - Anthony J. Hayter.pdf

Download. Read the first chapter.
Your question is addressed there.

>>5036974
thanks!

>>5036948
The reason that the probability is still 1/6 on each try is because all the previous trials don't influence the next trial. Each trial is an independent event. Also, it's not a paradox, it's a probabilistic rarity.

Your chance of never getting a 4 is very unlikely.
If you roll your dice a large amount of time (if you know calc, think about limits going to infinity)
you'll eventually get an equal amount of rolls for each number.

You can do a set proof for this, but I just suggest reading the book if you really want to get into it with proof >>5036974
Unless someone wants to type it out for you.

>>5036980
This.

If you were using a bingo machine that spits numered balls from 1 to 6, the odds of getting 4 would be higher.

>>5036986
If the numbers are removed from the ball spitter machine after they are rolled.

>>5036980
>previous trial doesn't influence the next
>never head of P(A|B)
Not directly, but it does change the probability.
It changes in a very similar way probability is changed when you're trying to roll the same number 3 times
(P(3 of the same in a row) = 1/6*1/6*1/6)

>>5036978
No prob.

>>5036978
Oh yeah, check out Chapter 3, too -- "Discrete Probability Distributions" -- if you don't get it after chapter 1.
I haven't read chapter 3, but it seems like it goes more into the math side of it.

>>5036948
The thing is, stop before rerolling and ask yourself how unlucky you were to never get a 4, but remember to reroll with the confidence that you have a static 1 chance on so many.

6 side die - odds of never getting a 4
(5 chance on 6)
1st turn odds: 5/6
2nd turn odds: 25/36 ~ 4.2/6
3rd turn odds: 125/216 ~ 3.5/6
4th turn odds: 625/1296 ~ 2.9/6
The 4th turn is less than 1 on 2 so the probability that you saw 4 by now is greater then not seeing it. The probability of you missing it or hitting it has nothing to do with the probability of hitting it or missing it in the next throw. There goes the paradox. The probability of hitting it next turn remains 1/6.

>>5037031
I must be thick or something... the odds of missing four decrease with each roll, yet they don't?

>>5037045
The more times you roll a die, the more likely it is that you will see every number.

>>5037045
it's more like of all the ways you could have rolled a die a bunch of times, the ways you can roll without getting a 4 take up a smaller percentage of the total ways you can roll

>>5037045
>the odds of missing four decrease with each roll
No, the odds of missing four never decrease.
They odds of rerolling up to x number of times decreases with times.
Or should I say the odds of rolling up to 3 times without a four are 3.5/6, the odds of rolling up to 4 times without a four are 2.9/6.

You have 5/6 chance to miss the 4 after 1 try
You have 4.2/6 chance to miss the 4 after 2 try
You have 3.5/6 chance to miss the 4 after 3 try
You have 2.9/6 chance to miss the 4 after 4 try

The fact you rolled 2 times and didnt get a four, thats 4.2 chance on 6, does not affect the chance of you not getting a four after 3 rolls. Everytime the probability of you continuing to roll has been modified by the probability of you rolling not four (5 / 6)

(5 / 6) = 5 / 6
(5 / 6)(5 / 6) = 25/36 ~= 4.2/6
(5 / 6)(5 / 6)(5 / 6) = 125/216 ~= 3.5/6
(5 / 6)(5 / 6)(5 / 6)(5 / 6) = 625/1296 ~ 2.9/6

The fact that the probability of your miss decrease with times you reroll only assures you that you will finally get a four.

>>5037096
Erreta

You have 5/6 chance to miss the 4 "using" 1 try
You have 4.2/6 chance to miss the 4 "using" 2 try
You have 3.5/6 chance to miss the 4 "using" 3 try
You have 2.9/6 chance to miss the 4 "using" 4 try

Hope this helps you master your will.
Lemme guess, newfag in Uni probability class?

Let's look at a coin toss instead, since it's easier to map out.
Take this same idea and extend it to dice.

P(N) is the probably of getting a result that has no heads.
It is found by K/T, where K = the number of outcomes with no heads,
and T = the number of total possible outcomes.
P(N) = K/T

(x y z) denotes results of a specific toss of 3 coins
{ (x y z), (x y z), (x y z), ... } denotes a set of all possible results.

Flip coin once:
{ (H), (T) } P(N) = 1 in 2, or P(N) = 0.5

Flip coin twice:
{ (H T), (T H), (H H), (T T) } 1 in 4, or P(N) = 0.25

Flip coin three times:
{ (H H H), (H H T),
. (H T H), (H T T),
. (T H H), (T H T),
(T T H), (T T T) } P(N) = 1 in 8, or P(N) = 0.125

In dice the numbers are difference, but it's the exact same idea.

>>5037110
H = Coin result is Heads
T = Coin result is Tails
I forgot to specify that.