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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

Looking for a one-liner answer...

Even sum can be obtained in following cases: EEEE - one case. Each E can take 3 values (2, 4, or 6), so total for this case is 3^4; OOOO - one case. Each O can take 3 values (1, 3, or 5), so total for this case is 3^4; EEOO - \(\frac{4!}{2!2!}=6\) cases (EOEO, OOEE, ...). Each E can take 3 values (2, 4, or 6) and each O can also take 3 values (1, 3, or 5), so total for this case is 6*3^2*3^2=6*3^4;

Total # of outcomes when throwing 4 dice is 6^4.

\(P=\frac{3^4+3^4+6*3^4}{6^4}=\frac{1}{2}\).

Answer: B.

Even without any math: the probability of getting an even sum when throwing 4 dice is the same as getting an even number on one die, so it must be 1/2.

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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24 May 2012, 02:26

Hi Bunuel,

The shortcut is perfect! Are you aware of any thread where we such approaches are discussed?

Similar solution can be applied to problem like no. of cases A occurs before B in a sequence, no. of cases where A & B are together In both cases half of the total no. of cases would be the answer. Since, either A comes after B or before B, either A & B are together or separated.

The shortcut is perfect! Are you aware of any thread where we such approaches are discussed?

Similar solution can be applied to problem like no. of cases A occurs before B in a sequence, no. of cases where A & B are together In both cases half of the total no. of cases would be the answer. Since, either A comes after B or before B, either A & B are together or separated.

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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01 May 2015, 03:02

1

This post was BOOKMARKED

Bunuel wrote:

cyberjadugar wrote:

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

Looking for a one-liner answer...

Even sum can be obtained in following cases: EEEE - one case. Each E can take 3 values (2, 4, or 6), so total for this case is 3^4; OOOO - one case. Each O can take 3 values (1, 3, or 5), so total for this case is 3^4; EEOO - \(\frac{4!}{2!2!}=6\) cases (EOEO, OOEE, ...). Each E can take 3 values (2, 4, or 6) and each O can also take 3 values (1, 3, or 5), so total for this case is 6*3^2*3^2=6*3^4;

Total # of outcomes when throwing 4 dice is 6^4.

\(P=\frac{3^4+3^4+6*3^4}{6^4}=\frac{1}{2}\).

Answer: B.

Even without any math: the probability of getting an even sum when throwing 4 dice is the same as getting an even number on one die, so it must be 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.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.
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Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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09 Jun 2017, 08:13

Bunuel wrote:

cyberjadugar wrote:

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

Looking for a one-liner answer...

Even sum can be obtained in following cases: EEEE - one case. Each E can take 3 values (2, 4, or 6), so total for this case is 3^4; OOOO - one case. Each O can take 3 values (1, 3, or 5), so total for this case is 3^4; EEOO - \(\frac{4!}{2!2!}=6\) cases (EOEO, OOEE, ...). Each E can take 3 values (2, 4, or 6) and each O can also take 3 values (1, 3, or 5), so total for this case is 6*3^2*3^2=6*3^4;

Total # of outcomes when throwing 4 dice is 6^4.

\(P=\frac{3^4+3^4+6*3^4}{6^4}=\frac{1}{2}\).

Answer: B.

Even without any math: the probability of getting an even sum when throwing 4 dice is the same as getting an even number on one die, so it must be 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?

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

Looking for a one-liner answer...

Even sum can be obtained in following cases: EEEE - one case. Each E can take 3 values (2, 4, or 6), so total for this case is 3^4; OOOO - one case. Each O can take 3 values (1, 3, or 5), so total for this case is 3^4; EEOO - \(\frac{4!}{2!2!}=6\) cases (EOEO, OOEE, ...). Each E can take 3 values (2, 4, or 6) and each O can also take 3 values (1, 3, or 5), so total for this case is 6*3^2*3^2=6*3^4;

Total # of outcomes when throwing 4 dice is 6^4.

\(P=\frac{3^4+3^4+6*3^4}{6^4}=\frac{1}{2}\).

Answer: B.

Even without any math: the probability of getting an even sum when throwing 4 dice is the same as getting an even number on one die, so it must be 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!).

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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25 Jun 2017, 11:22

cyberjadugar wrote:

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

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
_________________

Gambling with 4 dice, what is the probability of getting an [#permalink]

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03 Jul 2017, 01:51

Bunuel wrote:

cyberjadugar wrote:

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

Looking for a one-liner answer...

Even sum can be obtained in following cases: EEEE - one case. Each E can take 3 values (2, 4, or 6), so total for this case is 3^4; OOOO - one case. Each O can take 3 values (1, 3, or 5), so total for this case is 3^4; EEOO - \(\frac{4!}{2!2!}=6\) cases (EOEO, OOEE, ...). Each E can take 3 values (2, 4, or 6) and each O can also take 3 values (1, 3, or 5), so total for this case is 6*3^2*3^2=6*3^4;

Total # of outcomes when throwing 4 dice is 6^4.

\(P=\frac{3^4+3^4+6*3^4}{6^4}=\frac{1}{2}\).

Answer: B.

Even without any math: the probability of getting an even sum when throwing 4 dice is the same as getting an even number on one die, so it must be 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
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