There are four integer numbers chosen purely randomly and added up. You are told that the resulting number is even. What is the probability that all the original numbers were even?

Probability Puzzles: The Four Numbers Puzzle

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- Feb 6th 2013, 12:05 PMbroccoli7The four numbers puzzle
There are four integer numbers chosen purely randomly and added up. You are told that the resulting number is even. What is the probability that all the original numbers were even?

Probability Puzzles: The Four Numbers Puzzle - Dec 15th 2014, 05:56 AMmetalworkerRe: The four numbers puzzle
some may say the chances are 3 in 5 favouring the additions as being even, unfortunately, long term statistics don't control singular events, so maybe it's just a 1 in 2 chance, I have a gut feeling there may be a large difference in opinion as to the answer to this, that maybe why no one has had a go at it

- Dec 15th 2014, 06:11 AMmetalworkerRe: The four numbers puzzle
we do have long term statistics on coin tosses but they cannot control a single event, since this is a single event, surely the answer would be purely random as the original numbers were random

- Dec 15th 2014, 09:50 AMSorobanRe: The four numbers puzzle
Hello, broccoli7!

Quote:

There are four numbers chosen randomly and added.

You are told that the resulting number is even.

What is the probability that all the original numbers were even?

There areways to get an even sum.*eight*

$\quad\{E,E,E,E\} \quad \text{1 way}$

$\quad\{E,E,O,O\} \quad {4\choose2,2} = 6\text{ ways}$

$\quad\{O,O,O,O\} \quad \text{1 way}$

Therefore: $\:\text{Prob} \;=\;\dfrac{1}{8}$

- Dec 15th 2014, 10:12 AMebainesRe: The four numbers puzzle
Another approach is to use Bayes Theorem:

P(All even | Add to even) = P(All even & Add to even)/P(add to even).

The numerator is the same as P(all even), since if they are all even it's a given that hey add to an even number. The denominator is 1/2, so:

P(All even|Add to even) = (1/16)/(1/2) = 1/8. - Dec 15th 2014, 10:43 AMmetalworkerRe: The four numbers puzzle
I actually read that problem as, what are the chances of a randomly chosen 4 digit number having the sum of the digits being an even number, but it is very hard to see how singular random events are influenced by statistics gained from long-term observation, am I right?

- Dec 15th 2014, 11:04 AMebainesRe: The four numbers puzzle
metalworker - the problem stated that the sum is even, so that's a given. Now, given that the sum is even what is the probability that all four numvbers are even? That's the question that Soroban and I answered.

Probability has to do with the likelihood that something will occur, but we cannot say for sure whether that something does in fact occur for any individual trial. Regarding wherther the sum of 4 random digits is odd or even: we can't predict the results of any one trial. But we can do many trials of randomly-selecting 4 digits to see how often their sum adds to an even number, and we would expect to find that the percentage of trials that results in even numbers is typically close to 50% (but rarely exactly 50%). For example if you ran 100 trials you may find perhaps 48 times the nunbers add to an even result. Then do 100 more trials and you may find that this time 55 of the sums are even. We say the probability is 50%, meaning that as you do more and more trials the result should approach 50%. Stated another way - it you were to bet on the outcome a fair payoff for winning the bet would be 2:1. - Dec 15th 2014, 11:17 AMmetalworkerRe: The four numbers puzzle
thank you ebaines, of the five different ways of adding 4 digits, surely 3 of them favour an even outcome?

- Dec 15th 2014, 01:14 PMebainesRe: The four numbers puzzle
Yes, that's true, but here are multiple ways to have, say, 2 odd and 2 evens. There are actually 16 different ways to select 4 digits, and 8 of those 16 add to an even number, while 8 also add to odd:

All odd:

O+O+O+O = E

3 odd + 1 even:

O+O+O+E = O

O+O+E+O = O

O+E+O+O = O

E+O+O+O = O

2 odd + 2 even:

O+O+E+E = E

O+E+O+E = E

E+O+O+E = E

O+E+E+O = E

E+O+E+O = E

E+E+O+O = E

1 odd + 3 even:

O+E+E+E = O

E+O+E+E = O

E+E+O+E = O

E+E+E+O = O

All even:

E+E+E+E = E

Note that there are 6 ways to have 2 odds and 2 evens, versus only one way to have 4 evens. Hence the probability of drawing two odds plus two evens is 6 times greater than the probability of drawing four evens.