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As part of a game, four people each must secretly choose an
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06 Aug 2010, 22:40
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As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive.
What is the approximate likelihood that 2 people will choose same number? What is the approximate likelihood that 3 people will choose same number?




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Re: Probability  simple question.
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07 Aug 2010, 01:15
Financier wrote: As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that 2 people will choose same number? What is the approximate likelihood that 3 people will choose same number? Yes, this is a veeery simple question, but I want to understand in how many ways this question can be cracked. The more ways we know  the greater our confidence is When four people choose an integer between 1 and 4, inclusive 5 cases are possible: A. All choose different numbers  {a,b,c,d}; B. Exactly 2 people choose same number and other 2 choose different numbers  {a,a,b,c}; C. 2 people choose same number and other 2 also choose same number  {a,a,b,b}; D. 3 people choose same number  {a,a,a,b}; E. All choose same number  {a,a,a,a}. Some notes before solving: As only these 5 cases are possible then the sum of their individual probabilities must be 1: \(P(A)+P(B)+P(C)+P(D)+P(E)=1\) \(Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}\) As each person has 4 options, integers from 1 to 4, inclusive, thus denominator, total # of outcomes would be 4^4 for all cases. A. All choose different numbers  {a,b,c,d}:\(P(A)=\frac{4!}{4^4}=\frac{24}{256}\). # of ways to "assign" four different objects (numbers 1, 2, 3, and 4) to 4 persons is 4!. B. Exactly 2 people choose same number and other 2 choose different numbers  {a,a,b,c}:\(P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}\). \(C^2_4\)  # of ways to choose which 2 persons will have the same number; \(4\)  # of ways to choose which number it will be; \(P^2_3\)  # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters; C. 2 people choose same number and other 2 also choose same number  {a,a,b,b}:\(P(C)=\frac{{C^2_4*\frac{4!}{2!2!}}}{4^4}=\frac{36}{256}\). \(C^2_4\)  # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b}; \(\frac{4!}{2!2!}\)  # of ways to "assign" 4 objects out of which 2 a's and 2 b's are identical to 4 persons; D. 3 people choose same number  {a,a,a,b}:\(P(D)=\frac{C^3_4*4*3}{4^4}=\frac{48}{256}\). \(C^3_4\)  # of ways to choose which 3 persons out of 4 will have same number; \(4\)  # of ways to choose which number it will be; \(3\)  options for 4th person. E. All choose same number  {a,a,a,a}:\(P(E)=\frac{4}{4^4}=\frac{4}{256}\). \(4\)  options for the number which will be the same. Checking: \(P(A)+P(B)+P(C)+P(D)+P(E)=\frac{24}{256}+\frac{144}{256}+\frac{36}{256}+\frac{48}{256}+\frac{4}{256}=1\). Hope it's clear.
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Re: Probability  simple question.
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03 Sep 2010, 15:04
utkarshlavania wrote: In case 2 why does the order matter ...could you plz give a brief on that one B. Exactly 2 people choose same number and other 2 choose different numbers  {a,a,b,c}:\(P(B)=\frac{C^2_4*4*P^2_3}{4^4}=\frac{144}{256}\). \(C^2_4\)  # of ways to choose which 2 persons will have the same number; \(4\)  # of ways to choose which number it will be; \(P^2_3\)  # of ways to choose 2 different numbers out of 3 left for 2 other persons when order matters; Suppose the number which is repeated and chosen by 2 people is 4. Now, 2 other people must choose 2 different numbers out of 1, 2, and 3. XY 12 21 13 31 23 32 You can see that X choose 1 and Y choose 2 is different from Y choose 1 and X choose 2 (different scenario). That's why the order of the chosen numbers matters. Hope it's clear.
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03 Sep 2010, 19:27
Bunuel for the 2 case i.e., 2 people selecting the same number: I wanted to work the probability through the "selection" way  meaning taking the prob of of each and multiplying through, can you point doublecheck that this is correct too (the logic) 
So I can select 2 of the 4 in 4C2 ways. Now one of these 2 can select any of the 4 numbers and the second one has to select the same number, the probability of that happening is 4/4 * 1/4 The remaining 2 people can select any of the 3 numbers 3/4 x 3/4, so the total prob is 4C2 3/(4^3)  this is what you get too, I see the numbers match, but I wanted to make sure of the logic..
But for the case when each selects a different number, if I apply the same kind of logic Combinations of 4 people 4x3x2x1, for each combination the probability of selecting 4 different numbers = first person can choose 4/4 second can choose 1/3 third can choose 1/2 Multiplying them all 4!/(2 3 4) =1 , that does not make sense... I understand your answer, can you please tell me where I am wrong?



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Re: Probability  simple question.
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12 Oct 2010, 07:57
anilnandyala wrote: c) [a,a,b,b]
4C2 4 3C2 3/256
4C2 is no of ways of selecting 2 persons who have same number 4 is the no of ways in which no it will be 3C2 is the no of ways of selcting 2 persons who have same number 3 is the no of ways in which no it will be
can some one explain whats wrong with this? thanks in advance If you doing this way then after first selection of 2 persons there are only 2 more left not 3, so 2C2 not 3C2. Also you should divide this by 2! to get rid of duplications, so you would have 4C2*4*2C2*3/2=36.
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Re: As part of a game, four people each must secretly choose an
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31 Oct 2017, 13:13
kpali wrote: pinchharmonic wrote: Quote: Think of AABC has a word with repeating alphabets A. So AABC can be arranged in 4!/2! ways Now as ordering is important here we have 4!/2! * (4/4*1/4*3/4*2/4)*1/2 as final answer where we can say 4!/2! * 1/2 is nothing but 4C2 Hope this helps thanks, but i'm on point with everythign you said, but where is the 1/2 from? I also rearranged the 4 letters using (4!/2!), but I can't figure out why you have the 1/2 at the end. bumping for an explanation of one of these approaches, thanks Hi Bunuel, Regarding case b. (aabc) This seems to have never been answered. And i am very much under the same dilemma. even after doing a 4!/2! where is this additional factor of 1/2 coming from? you can rearrange everything since the order in this case matters, since the order represents different people [(4/4) * (1/4) * (3/4) * (2/4)] * (4! / 2!) where 4! allows me to rearrange everything and 2! removes the duplicates for person 1 and 2. Also if there is a factor then why 2? since we have aabc over here, which is three different no's out of four. Please please help me understand. Also, in case c. (aabb) I am calculating probability in the following fashion : 4/4 * 1/4 * 3/4 * 1/4 * 4!/(2!*2!) divided by 4^4 which is equal to 72/4^4 again i seem to be missing a factor of 2 and i cannot at all understand what i am doing wrong here as i am even removing duplicates introduced by duplicate a and b. i am able to get the correct answer for all the other scenarios using my approach (aaab, abcd,aaaa) except for the two posted above(aabc,aabb) which is baffling me even to a greater degree. it would be really great if you could help me out here. Cheers, Kriti Just my 2 cents regarding the aabc case: Basically there are 2 choices. 1) which 2 people are going to have the same number and 2) which number are these people going to take and which other 2 numbers the other 2 people are going to take. When you do: (4/4) * (1/4) * (3/4) * (2/4)] * (4! / 2!), (4/4) * (1/4) * (3/4) * (2/4) part takes care of the second part, the "which number are these people going to take and which other 2 numbers the other 2 people are going to take part" ie, if the first guy and second guy choose one number(the same number), the other 2 guys can choose the 2 different numbers in 3C2=6 ways. Since the first 2 guys have four options for choosing the "same number" , total ways =6*4=24. Now you need to take care of the "which 2 people are going to have the same number" part. So we have 2 people with same number and 2 people with different number. Sort of like 2 groups. The order in the 2 groups doesn't matter. when you are multiplying with (4! / 2!).. you are only taking care of the lack of order in one group. so you need to divide by an additional 2! to take care of lack of order in other group also. That is where the additional 2! in the denominator is coming from. If the question was how many ways can A &B have same number and C&D have different, then the answer would have been (4/4) * (1/4) * (3/4) * (2/4)]. This already takes care of the different number options. But here, you have to also choose which 2 among a,b,c,d get to have the same number. So to select which 2 people get the same number, (4! / 2!) is not enough coz then you are considering the order in one of the groups. That's why you get a larger answer. I dunno if that makes sense. But I hope it helps Hopefully some expert will explain more clearly.



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Re: Probability  simple question.
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03 Sep 2010, 14:49
In case 2 why does the order matter ...could you plz give a brief on that one



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Re: Probability  simple question.
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04 Sep 2010, 10:29
thanks once again bunuel



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Re: Probability  simple question.
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04 Sep 2010, 22:27
Thanks for the question as well as for the answer



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12 Oct 2010, 06:08
c) [a,a,b,b]
4C2 4 3C2 3/256
4C2 is no of ways of selecting 2 persons who have same number 4 is the no of ways in which no it will be 3C2 is the no of ways of selcting 2 persons who have same number 3 is the no of ways in which no it will be
can some one explain whats wrong with this? thanks in advance



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Re: Probability  simple question.
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31 May 2011, 22:58
mainhoon wrote: Bunuel for the 2 case i.e., 2 people selecting the same number: I wanted to work the probability through the "selection" way  meaning taking the prob of of each and multiplying through, can you point doublecheck that this is correct too (the logic) 
But for the case when each selects a different number, if I apply the same kind of logic Combinations of 4 people 4x3x2x1, for each combination the probability of selecting 4 different numbers = first person can choose 4/4 second can choose 1/3 third can choose 1/2 Multiplying them all 4!/(2 3 4) =1 , that does not make sense... I understand your answer, can you please tell me where I am wrong? For the case(a,b,c,d)= 1st chooses 4/4 2nd chooses 3/4 3rd chooses 2/4 4th chooses 1/4 So we have 4/4*3/4*2/4*1/4=4!/4^4 Similarly for (a,a,b,c) = 4C2*4/4*1/4*3/4*2/4 For (a,a,b,b) = 4C2*4/4*1/4*3/4*1/4*1/2! For(a,a,a,b) = 4C3*4/4*1/4*1/4*3/4 For (a,a,a,a)= 4/4*1/4*1/4*1/4



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29 Aug 2011, 20:27
someonear wrote: mainhoon wrote: Bunuel for the 2 case i.e., 2 people selecting the same number: I wanted to work the probability through the "selection" way  meaning taking the prob of of each and multiplying through, can you point doublecheck that this is correct too (the logic) 
But for the case when each selects a different number, if I apply the same kind of logic Combinations of 4 people 4x3x2x1, for each combination the probability of selecting 4 different numbers = first person can choose 4/4 second can choose 1/3 third can choose 1/2 Multiplying them all 4!/(2 3 4) =1 , that does not make sense... I understand your answer, can you please tell me where I am wrong? For the case(a,b,c,d)= 1st chooses 4/4 2nd chooses 3/4 3rd chooses 2/4 4th chooses 1/4 So we have 4/4*3/4*2/4*1/4=4!/4^4 Similarly for (a,a,b,c) = 4C2*4/4*1/4*3/4*2/4 For (a,a,b,b) = 4C2*4/4*1/4*3/4*1/4*1/2! For(a,a,a,b) = 4C3*4/4*1/4*1/4*3/4 For (a,a,a,a)= 4/4*1/4*1/4*1/4 where did you get the 4c2 in the AABC case? i had the following in place of the 4c2 which gives the wrong answer, I can't for the life of me figure out why it's 4c2 then you can rearrange everything since the order in this case matters, since the order represents different people [(4/4) * (1/4) * (3/4) * (2/4)] * (4! / 2!) where 4! allows me to rearrange everything and 2! removes the duplicates for person 1 and 2. am i doing something wrong?



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Re: Probability  simple question.
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30 Aug 2011, 07:20
pinchharmonic wrote: someonear wrote: mainhoon wrote: Bunuel for the 2 case i.e., 2 people selecting the same number: I wanted to work the probability through the "selection" way  meaning taking the prob of of each and multiplying through, can you point doublecheck that this is correct too (the logic) 
But for the case when each selects a different number, if I apply the same kind of logic Combinations of 4 people 4x3x2x1, for each combination the probability of selecting 4 different numbers = first person can choose 4/4 second can choose 1/3 third can choose 1/2 Multiplying them all 4!/(2 3 4) =1 , that does not make sense... I understand your answer, can you please tell me where I am wrong? For the case(a,b,c,d)= 1st chooses 4/4 2nd chooses 3/4 3rd chooses 2/4 4th chooses 1/4 So we have 4/4*3/4*2/4*1/4=4!/4^4 Similarly for (a,a,b,c) = 4C2*4/4*1/4*3/4*2/4 For (a,a,b,b) = 4C2*4/4*1/4*3/4*1/4*1/2! For(a,a,a,b) = 4C3*4/4*1/4*1/4*3/4 For (a,a,a,a)= 4/4*1/4*1/4*1/4 where did you get the 4c2 in the AABC case? i had the following in place of the 4c2 which gives the wrong answer, I can't for the life of me figure out why it's 4c2 then you can rearrange everything since the order in this case matters, since the order represents different people [(4/4) * (1/4) * (3/4) * (2/4)] * (4! / 2!) where 4! allows me to rearrange everything and 2! removes the duplicates for person 1 and 2. am i doing something wrong? Think of AABC has a word with repeating alphabets A. So AABC can be arranged in 4!/2! ways Now as ordering is important here we have 4!/2! * (4/4*1/4*3/4*2/4)*1/2 as final answer where we can say 4!/2! * 1/2 is nothing but 4C2 Hope this helps



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30 Aug 2011, 12:21
Quote: Think of AABC has a word with repeating alphabets A. So AABC can be arranged in 4!/2! ways Now as ordering is important here we have 4!/2! * (4/4*1/4*3/4*2/4)*1/2 as final answer where we can say 4!/2! * 1/2 is nothing but 4C2 Hope this helps thanks, but i'm on point with everythign you said, but where is the 1/2 from? I also rearranged the 4 letters using (4!/2!), but I can't figure out why you have the 1/2 at the end.



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31 Aug 2011, 14:00
pinchharmonic wrote: Quote: Think of AABC has a word with repeating alphabets A. So AABC can be arranged in 4!/2! ways Now as ordering is important here we have 4!/2! * (4/4*1/4*3/4*2/4)*1/2 as final answer where we can say 4!/2! * 1/2 is nothing but 4C2 Hope this helps thanks, but i'm on point with everythign you said, but where is the 1/2 from? I also rearranged the 4 letters using (4!/2!), but I can't figure out why you have the 1/2 at the end. bumping for an explanation of one of these approaches, thanks!



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26 May 2013, 22:37
Bunuel wrote: anilnandyala wrote: c) [a,a,b,b]
4C2 4 3C2 3/256
4C2 is no of ways of selecting 2 persons who have same number 4 is the no of ways in which no it will be 3C2 is the no of ways of selcting 2 persons who have same number 3 is the no of ways in which no it will be
can some one explain whats wrong with this? thanks in advance If you doing this way then after first selection of 2 persons there are only 2 more left not 3, so 2C2 not 3C2. Also you should divide this by 2! to get rid of duplications, so you would have 4C2*4*2C2*3/2=36. Hi Bunnel, Here why do we have to divide by 2? I didn't get this part. No. of cases for 2 pairs choosing two different nos. = 4c2(no. of 2's pairs)*4c1(no. which the first pair can choose)*3c1(no which second pair can choose) = 72 I am not making arrangements here these are only combinations, why is there repetition here?



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Re: Probability  simple question.
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26 May 2013, 23:23
cumulonimbus wrote: Bunuel wrote: anilnandyala wrote: c) [a,a,b,b]
4C2 4 3C2 3/256
4C2 is no of ways of selecting 2 persons who have same number 4 is the no of ways in which no it will be 3C2 is the no of ways of selcting 2 persons who have same number 3 is the no of ways in which no it will be
can some one explain whats wrong with this? thanks in advance If you doing this way then after first selection of 2 persons there are only 2 more left not 3, so 2C2 not 3C2. Also you should divide this by 2! to get rid of duplications, so you would have 4C2*4*2C2*3/2=36. Hi Bunnel, Here why do we have to divide by 2? I didn't get this part. No. of cases for 2 pairs choosing two different nos. = 4c2(no. of 2's pairs)*4c1(no. which the first pair can choose)*3c1(no which second pair can choose) = 72 I am not making arrangements here these are only combinations, why is there repetition here? Suppose you choose numbers 1 and 2. Now, in how many ways can you assign 1, 1, 2, 2, to W, X, Y, Z? In 12 ways or in 6? The correct answer is 6: 4!/(2!2!)=6. Listing the cases might help: W  X  Y  Z1  1  2  2 2  2  1  1 1  2  1  2 1  2  2  1 2  1  1  2 2  1  2  1
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Re: Probability  simple question.
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27 May 2013, 19:38
Hi Bunnel,
Here why do we have to divide by 2? I didn't get this part.
No. of cases for 2 pairs choosing two different nos. = 4c2(no. of 2's pairs)*4c1(no. which the first pair can choose)*3c1(no which second pair can choose) = 72
I am not making arrangements here these are only combinations, why is there repetition here?[/quote]
Suppose you choose numbers 1 and 2. Now, in how many ways can you assign 1, 1, 2, 2, to W, X, Y, Z? In 12 ways or in 6?
The correct answer is 6: 4!/(2!2!)=6. Listing the cases might help: W  X  Y  Z 1  1  2  2 2  2  1  1 1  2  1  2 1  2  2  1 2  1  1  2 2  1  2  1[/quote]
Hi Bunnel,
I listed out the possible teams, as you said their are only 3 ways to make 6 pairs. wx yz wy xz wz xy
That is all, only 3 ways to make 6 distinct pairs. Looks like 4c2 giving me total no of pairs and we are picking 2 pairs at a time to assign 2 number out of 4 to give 12 ways to pick numbers.
But this is tricky, unless I list the cases this might not be very apparent.
Thanks for the extra questions.



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Re: Probability  simple question.
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28 May 2013, 00:00
cumulonimbus wrote: Hi Bunnel,
I listed out the possible teams, as you said their are only 3 ways to make 6 pairs. wx yz wy xz wz xy
That is all, only 3 ways to make 6 distinct pairs. Looks like 4c2 giving me total no of pairs and we are picking 2 pairs at a time to assign 2 number out of 4 to give 12 ways to pick numbers.
But this is tricky, unless I list the cases this might not be very apparent.
Thanks for the extra questions. Not sure I understand correctly what you mean there, anyway: \(C^2_4=6\) in my solution for C here: aspartofagamefourpeopleeachmustsecretlychoosean98684.html#p760687 refers to the number # of ways to choose which 2 numbers out of 4 will be used in {a,a,b,b}. NOT to the number of ways we can choose 2 people out of 4. Does this make sense?
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17 May 2014, 06:55
pinchharmonic wrote: pinchharmonic wrote: Quote: Think of AABC has a word with repeating alphabets A. So AABC can be arranged in 4!/2! ways Now as ordering is important here we have 4!/2! * (4/4*1/4*3/4*2/4)*1/2 as final answer where we can say 4!/2! * 1/2 is nothing but 4C2 Hope this helps thanks, but i'm on point with everythign you said, but where is the 1/2 from? I also rearranged the 4 letters using (4!/2!), but I can't figure out why you have the 1/2 at the end. bumping for an explanation of one of these approaches, thanks! Hi Bunuel, Regarding case b. (aabc) This seems to have never been answered. And i am very much under the same dilemma. even after doing a 4!/2! where is this additional factor of 1/2 coming from? you can rearrange everything since the order in this case matters, since the order represents different people [(4/4) * (1/4) * (3/4) * (2/4)] * (4! / 2!) where 4! allows me to rearrange everything and 2! removes the duplicates for person 1 and 2. Also if there is a factor then why 2? since we have aabc over here, which is three different no's out of four. Please please help me understand. Also, in case c. (aabb) I am calculating probability in the following fashion : 4/4 * 1/4 * 3/4 * 1/4 * 4!/(2!*2!) divided by 4^4 which is equal to 72/4^4 again i seem to be missing a factor of 2 and i cannot at all understand what i am doing wrong here as i am even removing duplicates introduced by duplicate a and b. i am able to get the correct answer for all the other scenarios using my approach (aaab, abcd,aaaa) except for the two posted above(aabc,aabb) which is baffling me even to a greater degree. it would be really great if you could help me out here. Cheers, Kriti




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