Gambling with 4 dice, what is the probability of getting an

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very hard, I want to follow this post. this is 51/51 level , I think

probability for EEEE to happen is

1/2*1/2*1/2*1/2=A

similarly probability for OOOO and OOEE, OEOE,...(there are 6 cases)

total 8 case

add 8 cases we have 8/A=1/2

is the result.

I can not say any word for this excellency

1.We can throw either an even or an odd with a die and so there are 16 possibilities .
2. No even and 4 odd, the sum of 4 odds being even and occurring once
3. 1 even and 3 odd, the sum being odd and occurring 4 times
4. 2 even and 2 odd the sum being even and occurring 6 times
5. 3 even and 1 odd the sum being odd and occurring 4 times
6. 4 even and no odd the sum being even and occurring once
7. 8 times , the sum is even and 8 times the sum is odd for a probability of 1/2.

Brunnel, can you help me understand how you got 4!/2!2! for EEOO cases? I understand to arrange we need to do 4!, but can you break the 2!2! part here? I intutively understand its because of the pair of O and E, but can you explain it further?

THEORY:

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).



So, the number of arrangements of four letters EEOO is 4!/(2!2!).

Hope it's clear.

What can a sum of two positive integers (or for that matter any integers) be ?
It can only be EVEN or ODD..
there is no other possibility ...(note that I said INTEGERS - not just any numbers)

So, a dice when rolled, can only produce integers ..
So summing the results of any number of dice can only produce an ODD or an EVEN output.


Hence, there are 50% chance that our output will be EVEN (and 50% chances that output will be ODD)
So, Option B 1/2

Hi Bunuel, Please help me with the concept that you have used in the second solution. You say that the probability will be 1/2 because the sum will either be even or be odd. 1/2 means that each of these events are equally likely. While solving Combinations problems, I have seen a similar approach from you in some questions. My question is, how can one infer whether the cases possible are equally likely? This will help save valuable time in exam once understood. Thank you for your help

\(OOOO \implies\) 1 way (and even)
\(EOOO \implies\) 4 ways
\(EEOO \implies\) 6 ways (and even)
\(EEEO \implies\) 4 ways
\(EEEE \implies\) 1 way (and even)

Therefore

\(\frac{(1+6+1)}{(1+6+1+4+4)}\)

This one we can solve easily by bernoulli combinations.
3 ways for even sum: (1) EEEE + (2) OOOO + (3) 2E2O
P(1)=P(2)=0.5^4
P(3)=2C4x(0.5^2)x(0.5^2)
Sum = 1/2

Gambling with 4 dice, what is the probability of getting an even sum?

A. 3/4
B. 1/2
C. 2/3
D. 1/4
E. 1/3

1. even + even + even + even: (1/2)^4
2. odd + odd + even + even: (1/2)^2 x (1/2)^2 x 4! / 2! x 2!
3. odd + odd + odd + odd : (1/2)^4

(1/2)^4 + (1/2)^4 + (1/2)^2 x (1/2)^2 x 4! / 2! x 2! = 1/2

B.

Asked: Gambling with 4 dice, what is the probability of getting an even sum?

Since there is equal probability of getting odd and even numbers in a dice, there is equal probability of getting even or odd sum in throw of 4 dice

IMO B

Given that we are Gambling with 4 dice and We need to find what is the probability of getting an even sum?

At each stage there is only two possibility of getting a sum. Either the sum can be even or odd.
Out of the siz numbers we have exactly 3 odd and exactly 3 even
=> Probability of getting an even or getting an odd sum at any stage will be same and will be equal to \(\frac{1}{2}\)

=> Probability of getting an even sum = \(\frac{1}{2}\)

So, Answer will be B
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

1. all e: (1/2)^4 = 1/16
2. all o: (1/2)^4 = 1/16
3. 2e + 2o = (1/2)^4*6 = 6/16 (6 ways to arrange)
P = 1/16 + 1/16 + 6/16 = 8/16 = 1/2

Finding 1 - prob (odd) may be easier and quicker. The sum will be odd if you get either 3 odd numbers and 1 even number or 3 even numbers and 1 odd.

Prob. of getting 3 odds and 1 even: (1/2)^4 * 4C3 (i.e. number of arrangements with 3 odd numbers and 1 even) = 4/16 = 1/4.

The same for 3 even numbers and 1 odd number, prob. = 1/4.

1 - Prob (3 odd and 1 even) - Prob (3 even and 1 odd) = 1/2.

Answer - b

Imagine we throw the four dice in order. Each time you throw, add the number to your running total.

The probability of getting an even on the first throw is 1/2, so there is a 1/2 probability that the running total after one throw is even.

To that running total, you are going to add either an even or an odd. And each time you add a new number, there's a 1/2 chance that you land on an even...if your running total was even before, it's 50/50 that the new running total will stay even...and it your running total was odd, it's 50/50 that the new running total will convert to even.

This continues forever regardless of the number of dice.

Answer choice B.