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Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
24 May 2012, 00:52

5

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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.

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
24 May 2012, 01:26

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.

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
24 May 2012, 01:30

Expert's post

cyberjadugar wrote:

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.

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
06 Feb 2013, 03:14

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

LOOKING FOR TINA RINK BULMER, THE GIRL LIVING IN BRADFORD ENGLAND, VISITING HALONG BAY, VIETNAM ON 27 JAN 2014. ANYONE KNOW HER, PLS EMAIL TO: thanghnvn@gmail.com.

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
27 Feb 2013, 04:06

1

This post received KUDOS

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. _________________

LOOKING FOR TINA RINK BULMER, THE GIRL LIVING IN BRADFORD ENGLAND, VISITING HALONG BAY, VIETNAM ON 27 JAN 2014. ANYONE KNOW HER, PLS EMAIL TO: thanghnvn@gmail.com.

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
01 May 2015, 02:02

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).

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. _________________

Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
01 May 2015, 02:02

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.

I can not say any word for this excellency _________________

LOOKING FOR TINA RINK BULMER, THE GIRL LIVING IN BRADFORD ENGLAND, VISITING HALONG BAY, VIETNAM ON 27 JAN 2014. ANYONE KNOW HER, PLS EMAIL TO: thanghnvn@gmail.com.

gmatclubot

Re: Gambling with 4 dice, what is the probability of getting an
[#permalink]
01 May 2015, 02:02

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