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

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Joined: 29 Mar 2012
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Gambling with 4 dice, what is the probability of getting an [#permalink]

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24 May 2012, 01:31
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Question Stats:

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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
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 39673
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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24 May 2012, 01:52
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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

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.

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

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.

Similar problem to practice: a-box-contains-100-balls-numbered-from-1-to-100-if-three-b-109279.html

Hope it helps.
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Joined: 29 Mar 2012
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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.

Regards,
Math Expert
Joined: 02 Sep 2009
Posts: 39673
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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24 May 2012, 02:30
Expert's post
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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.

Regards,

DS questions on probability: search.php?search_id=tag&tag_id=33
PS questions on probability: search.php?search_id=tag&tag_id=54

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

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06 Feb 2013, 04:14
very hard, I want to follow this post. this is 51/51 level , I think
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Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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27 Feb 2013, 05:06
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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.
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Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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01 May 2015, 03:02
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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01 May 2015, 03:02
Bunuel 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

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.

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

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.

Similar problem to practice: a-box-contains-100-balls-numbered-from-1-to-100-if-three-b-109279.html

Hope it helps.

I can not say any word for this excellency
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Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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24 Jun 2016, 15:30
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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06 Jul 2016, 11:48
I got it wrong. Why am I wrong?
e+e=e
o+o=e
e+o=o

So it's even in 2/3 of the times.
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Posts: 39673
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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07 Jul 2016, 04:38
Mariwa wrote:
I got it wrong. Why am I wrong?
e+e=e
o+o=e
e+o=o

So it's even in 2/3 of the times.

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

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01 Jun 2017, 00:11
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:
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

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.

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

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.

Similar problem to practice: http://gmatclub.com/forum/a-box-contain ... 09279.html

Hope it helps.

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?
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Posts: 39673
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]

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10 Jun 2017, 09:33
SOUMYAJIT_ wrote:
Bunuel 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

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.

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

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.

Similar problem to practice: http://gmatclub.com/forum/a-box-contain ... 09279.html

Hope it helps.

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

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25 Jun 2017, 11:22
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
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Re: Gambling with 4 dice, what is the probability of getting an   [#permalink] 25 Jun 2017, 11:22
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