Daniel is playing a dice game in which a dice is rolled 5 times : Problem Solving (PS)

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

21 Sep 2015, 01:38

to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2

to have any number except 4 is 5/6*1/6*1/6=5/216

5/216+1/2=5/216+108/216=113/216

what is wrong?

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

21 Sep 2015, 03:01

Temurkhon wrote:

to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2

to have any number except 4 is 5/6*1/6*1/6=5/216

5/216+1/2=5/216+108/216=113/216

what is wrong?

You need 1 more 4 to make it exactly 3 4s. So probability of having only 1 '4' and 2 'not 4s' = 1/6 * 5/6*5/6 and not 1/6*1*1 as you have done.

Look at my solution above.

Hope this helps.

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

21 Sep 2015, 03:36

Hi Engr2012,

Please help me with this question. I did the following way..

4 4 X X X (X is the probability of obtaining another 4)

Attempt 1:

1st X = 4. the probability is 1/6 (Game Won- End of Game)

Attempt 2:

2nd X= 4: First X not 4, the probability 5/6 and second X = 4, the probability is 1/6 (Total probaility= 5/36) (End of Game, Game won)

Attempt 3

3rd X= 4 (First X not 4, Second X not 4, third X= 4) Probability = 25/216

Total probability for 4= 91/ 216.

The three XXX are a number (Say 5) probability = 1/216.

I am getting 92/216.

Where am I going wrong here.

Please explain kindly.

Engr2012 wrote:

Temurkhon wrote:

to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2

to have any number except 4 is 5/6*1/6*1/6=5/216

5/216+1/2=5/216+108/216=113/216

what is wrong?

You need 1 more 4 to make it exactly 3 4s. So probability of having only 1 '4' and 2 'not 4s' = 1/6 * 5/6*5/6 and not 1/6*1*1 as you have done.

Look at my solution above.

Hope this helps.

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

21 Sep 2015, 04:37

shriramvelamuri wrote:

Hi Engr2012,

Please help me with this question. I did the following way..

4 4 X X X (X is the probability of obtaining another 4)

Attempt 1:

1st X = 4. the probability is 1/6 (Game Won- End of Game)

Attempt 2:

2nd X= 4: First X not 4, the probability 5/6 and second X = 4, the probability is 1/6 (Total probaility= 5/36) (End of Game, Game won)

Attempt 3

3rd X= 4 (First X not 4, Second X not 4, third X= 4) Probability = 25/216

Total probability for 4= 91/ 216.

The three XXX are a number (Say 5) probability = 1/216.

I am getting 92/216.

Where am I going wrong here.

Please explain kindly.

Engr2012 wrote:

Temurkhon wrote:

to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2

to have any number except 4 is 5/6*1/6*1/6=5/216

5/216+1/2=5/216+108/216=113/216

what is wrong?

You need 1 more 4 to make it exactly 3 4s. So probability of having only 1 '4' and 2 'not 4s' = 1/6 * 5/6*5/6 and not 1/6*1*1 as you have done.

Look at my solution above.

Hope this helps.

Your assumption that the game "stops" once you get 3 of 4s is not correct. You still need to count for all 5 cases and then evaluate whether the person has won. Look at my solution above for scenario 1.

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

21 Sep 2015, 04:55

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BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?
Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216

Source: PerfectScores

Probability of getting 4 in first throw = 1/6
probability of getting 4 in second throw = 5/6*1/6 = 5/36
Probability of getting 4 in third throw = 5/6*5/6*1/6 = 25/216
Total probability of getting 1 more 4 will be
1/6+5/36+25/216 = 91/216

Probability of getting a numbers consecutively 3 times will be 1/6*1/6*1/6 = 1/216
there are 5 such numbers left.
so probability will be 5*1/216 = 5/216
Total probability will be
91/216+5/216 = 97/216

I dont understand why does he have to throw the dice in case of 4, after the game is already won.
The question does not specify whether he throws the dice after winning the game. Logically he shouldn't.
So, to consider or not to consider the events after winning the game in such questions ?

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

21 Sep 2015, 04:58

kunal555 wrote:

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?
Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216

Source: PerfectScores

Probability of getting 4 in first throw = 1/6
probability of getting 4 in second throw = 5/6*1/6 = 5/36
Probability of getting 4 in third throw = 5/6*5/6*1/6 = 25/216
Total probability of getting 1 more 4 will be
1/6+5/36+25/216 = 91/216

Probability of getting a numbers consecutively 3 times will be 1/6*1/6*1/6 = 1/216
there are 5 such numbers left.
so probability will be 5*1/216 = 5/216
Total probability will be
91/216+5/216 = 97/216

I dont understand why does he have to throw the dice in case of 4, after the game is already won.
The question does not specify whether he throws the dice after winning the game. Logically he shouldn't.
So, to consider or not to consider the events after winning the game in such questions ?

Nowhere is it mentioned that the game is stopped as soon he has won the game. Thus you still need to consider the other cases and not assume things that are not given.

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

21 Sep 2015, 07:51

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Engr2012 wrote:

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216

Source: PerfectScores

Good question. +1

For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things.

Probability of any number from 1 to 6 = 1/6.

Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not.

2 scenarios for a win:

Scenario 1: 4 4 T N N , with T= 4 , N= not 4

Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots)

Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.

Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible)

Thus the final probability = 75/216 + 5/216 = 80/216

C is the correct answer.

Hi Engr2012,

I've also set up 2 correct scenarios, but had problenbs calculating them - particularly the second scenario:
1] (1/6)*(5/6)*(5*6)*3!/2!=75/216 --> Ok, here no questions Not 4*Not 4*4 (1/6)

BUT

2] 4 4 N N N --> 5/6*5/6*5/6 Not4*Not4*Not4, why do we have to choose 1/6 twice after 5/6 ?

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

21 Sep 2015, 08:02

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BrainLab wrote:

Engr2012 wrote:

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216

Source: PerfectScores

Good question. +1

For the sake of completeness, you need to mention that the dice is an unbiased die with numbers 1 to 6. You can not assume these things.

Probability of any number from 1 to 6 = 1/6.

Given that you already have 2 4s , leaving you with 3 more slots to determine if you will win or not.

2 scenarios for a win:

Scenario 1: 4 4 T N N , with T= 4 , N= not 4

Probability of this scenario = 1*1*(1/6)*(5/6)*(5*6)*3!/2!=75/216 .... (3!/2! to show the arrangement of TNN in 3 slots)

Scenario 2: 4 4 N N N , with N = any particular number but 4 and in this case for a 'win' all 'N's are the same number.

Probability of this scenario = 1*1*(5/6)*(1/6)*(1/6) = 5/216 ...(with 1/6 shows the probability of choosing that same number out of 6 possible)

Thus the final probability = 75/216 + 5/216 = 80/216

C is the correct answer.

Hi Engr2012,

I've also set up 2 correct scenarios, but had problenbs calculating them - particularly the second scenario:
1] (1/6)*(5/6)*(5*6)*3!/2!=75/216 --> Ok, here no questions Not 4*Not 4*4 (1/6)

BUT

2] 4 4 N N N --> 5/6*5/6*5/6 Not4*Not4*Not4, why do we have to choose 1/6 twice after 5/6 ?

That is because once you have 1 particular number out of (1,2,3,5,6), in order to "win" you MUST get that 1 particular number twice more in order to "win". Selecting 1 number out of 6 possible = 1/6. You can select 1 particular number of out of the remaining 5 numbers iwth probability of 5/6.

The winning combinations can be 44555 or 44111 or 44222 or 44333 or 44666

Thus for 4 4 N N N , in which N $\neq$4 = (5/6)*(1/6)*(1/6)

Hope this helps.

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

21 Sep 2015, 13:37

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abhigoyal1989 wrote:

kunal555 wrote:

BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?
Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216

Source: PerfectScores

Probability of getting 4 in first throw = 1/6
probability of getting 4 in second throw = 5/6*1/6 = 5/36
Probability of getting 4 in third throw = 5/6*5/6*1/6 = 25/216
Total probability of getting 1 more 4 will be
1/6+5/36+25/216 = 91/216

Probability of getting a numbers consecutively 3 times will be 1/6*1/6*1/6 = 1/216
there are 5 such numbers left.
so probability will be 5*1/216 = 5/216
Total probability will be
91/216+5/216 = 97/216

I dont understand why does he have to throw the dice in case of 4, after the game is already won.
The question does not specify whether he throws the dice after winning the game. Logically he shouldn't.
So, to consider or not to consider the events after winning the game in such questions ?

Hi Kunal555,

I was also having the same question as you have. But look at the details of question. It says exactly 3 times. If it is not exactly 3 times, he will lose the game. Hence, the game should go on for total 5 throws, then only we can decide whether he won the game or not. If we stop the game at the third throw when he gets 4, then we are eliminating his chance of losing the game because we don't know the result of 4th and 5th throw. If he gets 4 in the fourth throw also, then it means he will lose the game because now, the total number of 4s are not exactly 3.

Please award kudos if you agree with my explanation.

You are right.
Also, first statement says "Daniel is playing a dice game in which a dice is rolled 5 times."
IMO it is simple and clear that he plays a game in which a dice HAS TO BE ROLLED 5 TIMES.
And if in 5 trials, a number shows up more than 3 times, he will loose.

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

21 Sep 2015, 16:13

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shriramvelamuri wrote:

Hi Engr2012,

Please help me with this question. I did the following way..

4 4 X X X (X is the probability of obtaining another 4)

Attempt 1:

1st X = 4. the probability is 1/6 (Game Won- End of Game)

Attempt 2:

2nd X= 4: First X not 4, the probability 5/6 and second X = 4, the probability is 1/6 (Total probaility= 5/36) (End of Game, Game won)

Attempt 3

3rd X= 4 (First X not 4, Second X not 4, third X= 4) Probability = 25/216

Total probability for 4= 91/ 216.

The three XXX are a number (Say 5) probability = 1/216.

I am getting 92/216.

Where am I going wrong here.

Please explain kindly.

Engr2012 wrote:

Temurkhon wrote:

to have 4 in the next 3 rolls is 1/6*1*1=1/6, but we have 3!/2!=3 such possibilities, (1/6)*3=3/6=1/2

to have any number except 4 is 5/6*1/6*1/6=5/216

5/216+1/2=5/216+108/216=113/216

what is wrong?

You need 1 more 4 to make it exactly 3 4s. So probability of having only 1 '4' and 2 'not 4s' = 1/6 * 5/6*5/6 and not 1/6*1*1 as you have done.

Look at my solution above.

Hope this helps.

Hi Shriram,

You are missing one crucial information that he has to throw dice 5 times. and he has to get same number exactly 3 times. If you are stopping at "attempt 1" where you are saying game will end and he will won because he has thrown same number 3 times, you are missing "exactly" term in the question. Thus to ensure he won the game, he has to throw a number other than 4. That is why we multiply 5/6 two times (4th throw and 5th throw). Your logic would have been fine in case the question says "he will win as soon as he gets three times a same number"

Please award kudos if you like my explanation. Thanks

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

27 Jun 2017, 21:34

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Very good and tricky question
The game is won when 1 number show up Exactly 3 times
We are given first two throw result in 4 so we need one more 4 to win , this can be done in 1*1*5/6*5/6*1/6*3
where 5/6 is probability of not getting 4 and they can be arranged in 3 ways.
Now we can also win the game if we get rest of numbers 3 times i.e 1,2,3,5,6
so probability will be =5*1/6*1/6*1/6
The probability of winning =5/216+75/216=80/216

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

27 Jun 2017, 22:02

Expert Reply

1. Daniel will win if he throws 4 once or any of the other numbers thrice in the next three throws.
2. Number of possibilities 4 can occur in the third throw out of 5 throws is 5^2=25. Similarly for the fourth and fifth throws for a total of 75
3. Number of possibilities that the other numbers are thrown thrice is 1 + 1+1+1+1=5
4. Probability is , (75+5)/216=80/216

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

17 May 2018, 05:54

Why is it assumed that after getting a 4 the other 2 numbers would be the same and this leads to divide by 2!. The 2 numbers can be 2 different numbers other than 4.
Pls help.

Also do we need to learn conditional probability and Bayes theorem etc for GMAT?

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

23 May 2018, 10:35

Probability of Daniel not winning the game = 5/6 x 5/6 x 5/6 = 125/216
Probability of win = 1- 125/216 = 91/216.

Do not understand why the solution has to be so complicated. Please explain if it's wrong.
Thanks in advance.

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

23 May 2018, 10:59

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Expert Reply

nidhiprasad wrote:

Why is it assumed that after getting a 4 the other 2 numbers would be the same and this leads to divide by 2!. The 2 numbers can be 2 different numbers other than 4.
Pls help.

Also do we need to learn conditional probability and Bayes theorem etc for GMAT?

Hey nidhiprasad ,

We are no where assuming that case. When I am saying 5 possibilities for each of 444XX, I can say it could be anything(same or different). Look at the solution posted by me here and let me know in case of further concerns.

Also, there is no need to learn probability and Bayes theorem etc for GMAT.

duttarupam2344 wrote:

Probability of Daniel not winning the game = 5/6 x 5/6 x 5/6 = 125/216
Probability of win = 1- 125/216 = 91/216.

Do not understand why the solution has to be so complicated. Please explain if it's wrong.
Thanks in advance.

Hey duttarupam2344 ,

You have missed a scenario when your last three numbers are infact the same. In that case also, the player will win. Hence, you cannot directly talk 5/6 for all the three scenarios.

Does that make sense?

L

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

07 Jun 2019, 10:45

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BrainLab wrote:

Daniel is playing a dice game in which a dice is rolled 5 times. If a number turns up exactly 3 times then the game is won. Daniel has thrown the dice two times and got number 4 both times. What is the probability that Daniel will win the game ?
Note: the dice is an unbiased die with numbers 1 to 6

a) 1/216
b) 75/216
c) 80/216
d) 90/216
e) 100/216

Source: PerfectScores

we basically need to find P of winning for last three throws
so
getting one time 4 in 3rd throw; 1/6 and not getting is 5/6
so it can be , 5/6*1/6*1/6 ; 5/216
or 1/6*5/6*5/6; 25/216
or 5/6*5/6*1/6 ; 25/216
or 5/6*1/6*2/6; 25/216
total ; 75/216+5/216; 80/216
IMO C

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

29 Nov 2019, 00:17

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Total no. of throws = 5

Game is won only if "a number turn up exactly 3 times"

First throw: Number 4 pops

Second throw: Number 4 pops again

Required: Probability that Daniel will the game in the remaining 3 throws

Daniel can win the game if and only if these two scenarios occur

Scenario I: Number 4 turn up only once in the reaming three throws

$\frac{1}{6} * \frac{5}{6} * \frac{5}{6} + \frac{5}{6} * \frac{1}{6} *\frac{ 5}{6} + \frac{ 5}{6} * \frac{5}{6} * \frac{1}{6} = \frac{75}{216}$

Scenario II: Any number other than 4 turns up thrice in the remaining three throws

$\frac{5}{6} * \frac{1}{6} * \frac{1}{6} = \frac{5}{216}$

Adding Scenario I and II equals $\frac{80}{216}$

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Re: Daniel is playing a dice game in which a dice is rolled 5 times [#permalink]

29 Nov 2019, 00:17

GMAT Problem Solving (PS) Questions

Daniel is playing a dice game in which a dice is rolled 5 times