Last visit was: 23 May 2024, 08:00 It is currently 23 May 2024, 08:00
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Senior Manager
Joined: 29 Mar 2012
Posts: 267
Own Kudos [?]: 1509 [115]
Given Kudos: 23
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Math Expert
Joined: 02 Sep 2009
Posts: 93418
Own Kudos [?]: 626049 [31]
Given Kudos: 81940
General Discussion
Senior Manager
Joined: 29 Mar 2012
Posts: 267
Own Kudos [?]: 1509 [0]
Given Kudos: 23
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Math Expert
Joined: 02 Sep 2009
Posts: 93418
Own Kudos [?]: 626049 [1]
Given Kudos: 81940
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
1
Bookmarks
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.
Director
Joined: 09 Jun 2010
Posts: 529
Own Kudos [?]: 524 [1]
Given Kudos: 916
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
very hard, I want to follow this post. this is 51/51 level , I think
Director
Joined: 09 Jun 2010
Posts: 529
Own Kudos [?]: 524 [2]
Given Kudos: 916
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
2
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.
Director
Joined: 09 Jun 2010
Posts: 529
Own Kudos [?]: 524 [1]
Given Kudos: 916
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
1
Bookmarks
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
Director
Joined: 17 Dec 2012
Posts: 588
Own Kudos [?]: 1531 [1]
Given Kudos: 20
Location: India
Gambling with 4 dice, what is the probability of getting an [#permalink]
1
Kudos
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.
Intern
Joined: 08 Mar 2016
Posts: 23
Own Kudos [?]: 11 [0]
Given Kudos: 13
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
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: https://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?
Math Expert
Joined: 02 Sep 2009
Posts: 93418
Own Kudos [?]: 626049 [1]
Given Kudos: 81940
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
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: https://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.
Current Student
Joined: 17 Jun 2016
Posts: 473
Own Kudos [?]: 948 [0]
Given Kudos: 206
Location: India
GMAT 1: 720 Q49 V39
GMAT 2: 710 Q50 V37
GPA: 3.65
WE:Engineering (Energy and Utilities)
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
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
Manager
Joined: 03 Apr 2013
Posts: 222
Own Kudos [?]: 243 [0]
Given Kudos: 872
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
Gambling with 4 dice, what is the probability of getting an [#permalink]
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: https://gmatclub.com/forum/a-box-contain ... 09279.html

Hope it helps.

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
Manager
Joined: 03 Sep 2018
Posts: 178
Own Kudos [?]: 90 [1]
Given Kudos: 924
Location: Netherlands
GPA: 4
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
1
Kudos
$$OOOO \implies$$ 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

$$\frac{(1+6+1)}{(1+6+1+4+4)}$$
Intern
Joined: 18 Oct 2018
Posts: 9
Own Kudos [?]: 6 [1]
Given Kudos: 39
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
1
Bookmarks
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
VP
Joined: 11 Aug 2020
Posts: 1258
Own Kudos [?]: 203 [1]
Given Kudos: 332
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
1
Kudos
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.
GMAT Club Legend
Joined: 03 Jun 2019
Posts: 5350
Own Kudos [?]: 4016 [1]
Given Kudos: 160
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
1
Kudos
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
Tutor
Joined: 05 Apr 2011
Status:Tutor - BrushMyQuant
Posts: 1779
Own Kudos [?]: 2106 [0]
Given Kudos: 100
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE:Information Technology (Computer Software)
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
Top Contributor
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 = $$\frac{1}{2}$$

Hope it helps!

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

Intern
Joined: 12 Jan 2022
Posts: 4
Own Kudos [?]: 0 [0]
Given Kudos: 431
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
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
Intern
Joined: 04 Sep 2020
Posts: 14
Own Kudos [?]: 4 [0]
Given Kudos: 95
Location: Uzbekistan
GMAT 1: 770 Q50 V44
GPA: 3.79
WE:Law (Consulting)
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
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.

Tutor
Joined: 11 May 2022
Posts: 1091
Own Kudos [?]: 712 [0]
Given Kudos: 81
Re: Gambling with 4 dice, what is the probability of getting an [#permalink]
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

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.