Question

Short probability question:

I may have gotten "lucky", but what are the chances that if I roll 2 (fair) 6-sided die, that I would correctly guess both numbers three out of four times?

Answer 1

same idea to @UnkleRhaukus Qn last night :)

Answer 2

binomial distribution wid p = 1/36, n = 4, k = 3 :

\[\large \binom{4}{3} \left(\frac{1}{36}\right )^3 \left(\frac{35}{36}\right)^1 \]

Answer 3

Had to post this question after watching Through the Wormhole, is luck real :)

Answer 4

that number looks very unlikely ^

Answer 5

luck is real ;)

Answer 6

0.0000277841284

Answer 7

very lucky to me lol

Answer 8

or 35/1259712


thank you

Answer 9

Another question, that is related

Answer 10

what a nice display pic !
i sow something like this in a song
but i forget what it is xD

Answer 11

wait, wolfram says :

http://www.wolframalpha.com/input/?i=%5Cbinom%7B4%7D%7B3%7D+%5Cleft%28%5Cfrac%7B1%7D%7B36%7D%5Cright+%29%5E3+%5Cleft%28%5Cfrac%7B35%7D%7B36%7D%5Cright%29%5E1

Answer 12

check ur calculation once

Answer 13

Oh I did (4/3)*(1/36)^3 * (35/6), not

( (4/3)*(1/36)^3 ) * (35/36)

Answer 14

\[\large \binom{4}{3} = ^4C_3 = 4\]

Answer 15

its a combination symbol :)

Answer 16

Oh ok lol

Answer 17

If you did it with C, I would of remembered

Answer 18

Would the prob. change if I guessed the two numbers correctly 3 out of the 4 times, but the numbers for each die was switched?

Like: I roll die 1 and die 2. I guess die 1 to be 1, and die 2 to be 2. The numbers are still 1 and 2, but die 2 is 1, and die 1 is 2

Answer 19

*It would change, so:

Answer 20

Would would be the prob. that I roll 2 fair 6-sided die, and guessed the numbers of each pair of die correctly 3 out of 4 times?

Answer 21

The first question, either die could be one of the two numbers. What would be the prob. that each die I guessed would be specifically one of the two numbers?

Answer 22

still here?

Answer 23

So now the order of prediction doesnt matter is it ?

Answer 24

if u predict (1, 2) then both below give u success :
(1, 2) and (2, 1)

Answer 25

Is that the right ?

Answer 26

Yes. So instead of saying what are the chances of (1,2) or (2,1); (3,4) or (4,3), and (5,6) or (6,5)... what are the chances of die (*1,2), (3,4), and then (5,6)

Answer 27

just find the probability for success (p) and plug that in the binomial formula ?

Answer 28

ok, but I haven't taken statistics in years. This is just a random thing I wanted to figure out :)

Answer 29

we can work it using combinatorics

Answer 30

total equivalent outcomes = 6(6+1)/2 = 21
so, p = 1/21

Answer 31

the probability for predicting 3 out of 4 times correctly, when order of dies doesnt matter is :

\[ \large \binom{4}{3} \left(\frac{1}{21}\right )^3 \left(\frac{20}{21}\right)^1 \]

Answer 32

let me knw if the number 21 is throwing u off :)

Answer 33

so 80/194481

Answer 34

Yep !

http://www.wolframalpha.com/input/?i=%5Cbinom%7B4%7D%7B3%7D+%5Cleft%28%5Cfrac%7B1%7D%7B21%7D%5Cright+%29%5E3+%5Cleft%28%5Cfrac%7B20%7D%7B21%7D%5Cright%29%5E1

Answer 35

I remember some things about stat, just not too much :)

Answer 36

So the first answer you gave was when order mattered?

Answer 37

yes !

Answer 38

for the first problem :

(1, 2) is different from (2, 1)

so u get 36 total outcomes

Answer 39

So the chances of predicting correctly one time is 1/36

Answer 40

for ur second problem,
the chance of predicting correctly one time is 1/21

Answer 41

I understand a little more now. thanks again. ...Since when I did do it 2 years ago, when the order was messed up...I wasn't all that lucky.

Answer 42

haha cool :)
btw that derivation of binomial formula is trivial - it directly follows from permutations

Answer 43

look it up in google when u r free...

Answer 44

ok, thanks!

Answer 45

Always good to get a refresher

Answer 46

this textbook is a real nice refresher on statistics :

http://www.openintro.org/stat/textbook.php

Answer 47

click on "Download the Second Edition PDF"

Enjoy...

Answer 48

`chapter 3.4` covers the binomial distribution