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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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05 Jan 2013, 06:57
Shortcut solution: 3 is minimum and 18 is maximum as the sum of three die. Now, 18 will have same possibilities as 3 (1 possibility) 17 will have same possibilities as 4 (3 possibilites) 16 will have same possibilities as 5 (6 possibilites) . . . 10 will have same possibilities as 11 (27 possibilites) NOTE: 18+3 = 21, 17 + 4 = 21.... this helps to generate the series. Thus all possibilites are half  half divided between 310 and 1118. Thus if Joe has to get something from 11 to 18, his probability is 1/2 Kudos for this insight plz
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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04 Apr 2013, 13:21
What's about my approach ?
He needs a minimum score of 11:
The minimum throw must be: Dice 1: 4 or greater Dice 2: 4 or greater Dice 3: 3 or greater
3*( 3/6 * 3/6 * 4/6) = 1/2



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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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14 Apr 2013, 18:19
Quote: Can someone please explain what mistake i'm doing:
Total No. Of Possible Outcomes = 216
Outcomes where Joe scores 10 or less:
111 > 1 222 > 1 333 > 1 112 > 3, 113 > 3, 114 > 3, 115 > 3, 116 > 3, 221 > 3, 223 > 3, 224 > 3, 225 > 3, 226 > 3, 331 > 3, 332 > 3, 334 > 3, 441 > 3, 442 > 3,
Adding everything up = 48
Outcomes where Joe scores more than 10 = 216  48 = 168
Probability = 168/216 = 7/9 Quote: You are missing some cases: 123  6 ways; 124  6 ways; 125  6 ways; 126  6 ways; 134  6 ways; 135  6 ways; 136  6 ways; 145  6 ways; 234  6 ways; 235  6 ways.
So, total of 60 scenarios were missing. Together with the 48 cases you counted we would have 48+60=108 ways to get the sum of 10 or less, so the probability is 1108/216=1/2.
Hope it helps. My brain works in terms of combinatorics better than the "fast" approach. I feel it is much more intuitive. It is tough to come with the fast one immediately  it is good (My hats off to Bunuel and others who explain, get, and use this approach) but in my view it is more of a "you either come up with it or nothing" and therefore, completely unreliable (and just remembering solution to this problem will not get you the right answer once the similar problem is tweaked in a slightly different manner, as is the case with most GMAT problems). Combinations is a more conceptual and a step by step way that will and should eventually get you there. Now, I was wondering can a combinations way be summarized for this problem? Or what can be the most concise comb way to approach this? Ironically, to me this is one of the instances when shorter does not necessarily mean better, not to offend anyone here.
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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29 Mar 2014, 21:53
Bunuel wrote: noboru wrote: Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his? Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5. Expected value of three dice is 3*3.5=10.5. Mary scored 10 so the probability to get the sum more then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2. P=1/2. @Bunuel Plz help me with this. My method to solve was as below. Where did I go wrong? Joe should score more than 10 .Possible cases {4,3,4}* 3!/2!(ways of arrangement) , {3,4,5}*3! , {4,4,4}*1 ,{5,5,5}*1 ,{6,6,6}*1 Probability of getting any of the above case is (1/6)^3 .. Thus, [(1/6)^3 * 3!] + [(1/6)^3 * 3!/2!] + [(1/6)^3 * 1] + [(1/6)^3 * 1] +[(1/6)^3 * 1] = 1/18 .. Plz help !!!



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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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30 Mar 2014, 10:39
sachinwar wrote: Bunuel wrote: noboru wrote: Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his? Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5. Expected value of three dice is 3*3.5=10.5. Mary scored 10 so the probability to get the sum more then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2. P=1/2. @Bunuel Plz help me with this. My method to solve was as below. Where did I go wrong? Joe should score more than 10 .Possible cases {4,3,4}* 3!/2!(ways of arrangement) , {3,4,5}*3! , {4,4,4}*1 ,{5,5,5}*1 ,{6,6,6}*1 Probability of getting any of the above case is (1/6)^3 .. Thus, [(1/6)^3 * 3!] + [(1/6)^3 * 3!/2!] + [(1/6)^3 * 1] + [(1/6)^3 * 1] +[(1/6)^3 * 1] = 1/18 .. Plz help !!! You are missing several other cases: {6, 6, 5} {6, 6, 4} {6, 6, 3} {6, 6, 2} ...
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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27 Apr 2014, 10:46
Bunuel wrote: mariyea wrote: How were you able to come up with (1+2+3+4+5+6)? I understand that one outcome out of six occurs when Joe rolls the dice but the other part... a bit puzzling??? Expected value of a roll of one die is 1/6*1+1/6*2+1/6*3+1/6*4+1/6*5+1/6*6=1/6*(1+2+3+4+5+6)=3.5. Sorry about this ..I m not too good in Probability. How did you come up with multiplied by 1 , 2 and so on .. To Roll a die .. possible outcome for each will be 1/6 .. but you multiplied this by 1 then 2 then 3 and so on .. means 1/6*1 .what is multiplied by 1 denotes here ? Thanks
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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28 Apr 2014, 02:50
himanshujovi wrote: MacFauz wrote: Bunuel wrote: Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?
Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5. Expected value of three dice is 3*3.5=10.5.
Mary scored 10 so the probability to get the sum more then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2.
P=1/2. Can someone please explain what mistake i'm doing: Total No. Of Possible Outcomes = 216 Outcomes where Joe scores 10 or less: 111 > 1 222 > 1 333 > 1 112 > 3, 113 > 3, 114 > 3, 115 > 3, 116 > 3, 221 > 3, 223 > 3, 224 > 3, 225 > 3, 226 > 3, 331 > 3, 332 > 3, 334 > 3, 441 > 3, 442 > 3, Adding everything up = 48 Outcomes where Joe scores more than 10 = 216  48 = 168 Probability = 168/216 = 7/9 I was also thinking on similar lines!! Flaw in this solution is explained here: maryandjoearetothrowthreediceeachthescoreisthe86407.html#p1101519Hope it helps.
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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29 May 2014, 22:10
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Hi Bunuel, Can you please explain your answer of (1/6)^3 = 1/216 in case mary scores 17.



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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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08 Jun 2014, 21:19
What about this approach?: For greater than 10 U need: 3 or more on 1 dice,4 or more on second,4 or more on third Thus total no of possibilities are 4*3*3=36 Now this case is three times as likely because u can get a 3 or more on the other 2 dies as well.. 36+36+36= 108 Total  216 P 1/2
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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08 Jun 2014, 22:38
If the problem were to be changed slightly and said that none of the first three digits repeat themselves, then the answer would become 3x2x2 = 12 going the the original method posted by Bunuel. Is that correct?
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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22 Jul 2014, 14:37
Bunuel wrote: noboru wrote: Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his? Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5. Expected value of three dice is 3*3.5=10.5. Mary scored 10 so the probability to get the sum more then 10 (11, 12, 13, ..., 18), or more then the average, is the same as to get the sum less than average (10, 9, 8, ..., 3) = 1/2. P=1/2. I think it is the only way to solve the problem in less than 2 minutes. When I tried to calculate the total probability of 11, 12, ...18, it took me more than 10 minutes.
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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20 Aug 2014, 02:01
Bunuel wrote: imania wrote: would you love to see how attacked it? if Joe is expected to outscore his friend, he should get these sums, 11,12,13...18 all possibilities are from 3 to 18 so : prob =8/16 equal to 1/2 PS. If you are wondering how I came to 3 as min because 1+1+1 and likewise 18 is max (6+6+6) Unfortunately this approach is not right though for this particular case it gave a correct answer. Consider this: if it were that Mary scored not 10 but 17 then Joe to outscore Mary should get only 18 and according to your approach as there are total of 16 scores possible then the probability of Joe getting 18 would be 1/16. But this is not correct, probability of 18 is (1/6)^3=1/216 not 1/16. This is because not all scores from 3 to 18 have equal # of ways to occur: you can get 10 in many ways but 3 or 18 only in one way (3=1+1+1 and 18=6+6+6). Hope it's clear. ======== Thanks Bunuel for the explanation however my 2 pence are as follows , The original explanation is perfectly sound, kindly correct me if I am wrong. P of Joe getting more than 10 is 8 ways / 16 ways =1/2 (yay!) 8= 11, 12,....18; 16= 3,4....18; Range of Summation of scores is 16 ; Max =18 and Min = 3 [ 3=1+1+1; 18=6+6+6 ] Prob of each =1/16. lets consider an Outcome of 15 [ 3+6+6, 4+5+6............] , Question 1, why should be the permutations of numbers within the group be considered as the outcome is always 15 , per my example and as far as the verbiage of the question goes.... so aint these are identical groups of three since the outcome is 15 always. So each of the Outcomes i.e 16 of them [per question] shud have an equal probability i.e 1 /16. Question 2, why are we considering the expected value when the question asks of Probability of Joe getting certain digits as outcomes. Regards..



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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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07 Mar 2015, 00:31
What would be the approach to be followed if the question asked is to find the probability of Joe hitting more than 15??



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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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19 Nov 2016, 05:46
I solved by listing down options and finding patterns in the listing. The Possible ways in which the total can be above 10 can be calculated as follows ( keeping Die1 = 1 and Die 3 reducing from 6 downwards get different options of D2)
D1 D2 D3 1 4 6 1 5 6 1 6 6 1 5 5 1 6 5 1 6 4
These combinations can be repeated for 6 values of D1 ( 1,2,3,4,5,6)
Thus we have 6*6 ways . Now the above method can be employed for D2 and D3 as well giving 6*6 ways each. So the number of possibilities where the throw of 3 dies will give us more than 10 = 6*6 + 6*6+6*6 Total Possibilities of 3 throws = 6*6*6
So the probability = ( 6*6 +6*6+6*6 ) / (6*6*6) = 3*6*6/6*6*6 = 1/2



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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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29 Mar 2017, 06:16
Bunuel wrote: imania wrote: would you love to see how attacked it? if Joe is expected to outscore his friend, he should get these sums, 11,12,13...18 all possibilities are from 3 to 18 so : prob =8/16 equal to 1/2 PS. If you are wondering how I came to 3 as min because 1+1+1 and likewise 18 is max (6+6+6) Unfortunately this approach is not right though for this particular case it gave a correct answer. Consider this: if it were that Mary scored not 10 but 17 then Joe to outscore Mary should get only 18 and according to your approach as there are total of 16 scores possible then the probability of Joe getting 18 would be 1/16. But this is not correct, probability of 18 is (1/6)^3=1/216 not 1/16. This is because not all scores from 3 to 18 have equal # of ways to occur: you can get 10 in many ways but 3 or 18 only in one way (3=1+1+1 and 18=6+6+6). Hope it's clear. Hi Bunuel, I have a doubt if that's the question which you created (i.e Joe to outscore marry on 17) then how would you solve and what would be the answer?



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Mary and Joe are to throw three dice each. The score is the [#permalink]
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29 Mar 2017, 06:32
jamescath wrote: Bunuel wrote: imania wrote: would you love to see how attacked it? if Joe is expected to outscore his friend, he should get these sums, 11,12,13...18 all possibilities are from 3 to 18 so : prob =8/16 equal to 1/2 PS. If you are wondering how I came to 3 as min because 1+1+1 and likewise 18 is max (6+6+6) Unfortunately this approach is not right though for this particular case it gave a correct answer. Consider this: if it were that Mary scored not 10 but 17 then Joe to outscore Mary should get only 18 and according to your approach as there are total of 16 scores possible then the probability of Joe getting 18 would be 1/16. But this is not correct, probability of 18 is (1/6)^3=1/216 not 1/16.This is because not all scores from 3 to 18 have equal # of ways to occur: you can get 10 in many ways but 3 or 18 only in one way (3=1+1+1 and 18=6+6+6). Hope it's clear. Hi Bunuel, I have a doubt if that's the question which you created (i.e Joe to outscore marry on 17) then how would you solve and what would be the answer? The answer would be 1/216.
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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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29 Mar 2017, 06:40
Hi Bunuel, I have a doubt if that's the question which you created (i.e Joe to outscore marry on 17) then how would you solve and what would be the answer?[/quote]
The answer would be 1/18.[/quote]
How you approached it?
Like as explained: Expected value of one die is 1/6*(1+2+3+4+5+6)=3.5. Expected value of three dice is 3*3.5=10.5.
Mary scored 17 so the probability to get the sum more then 17 (18) is 1/18 (how ? I got stuck in last step though I understood the 1/2 part in original question)



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Re: Mary and Joe are to throw three dice each. The score is the [#permalink]
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