#21
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Re: Monty at it again
why does this graph of possible results produce different results?
1 2 3 4 5 6 7 8 9 10 1 . . AB AB AC AC AD AD AC AD 2 . . AB AB AC AC AD AD AC AD 3 AB AB . . BC BC BD BD BC BD 4 AB AB . . BC BC BD BD BC BD 5 AC AC BC BC . . CD CD . CD 6 AC AC BC BC . . CD CD . CD 7 AD AD BD BD CD CD . . CD . 8 AD AD BD BD CD CD . . CD . 9 AC AC BC BC . . CD CD . CD 10 AD AD BD BD CD CD . . CD . AB = 8/74 AC = 12/74 AD = 12/74 BC = 12/74 BD = 12/74 CD = 18/74 <paste into notepad (with courier font) for the columns to line up> |
#22
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Re: Monty at it again
I don't think I can explain this any clearer .
The probability 1 or 2 gets selected on the first draw is 2/10 , agree ? Given that 1 or 2 has been selected , the other number gets removed from the vault so that there are 8 numbers left . The probability it's B should be 2/8 since he has two good choices out of 8 . Agree ? However , B may hit 3 or 4 on the first draw rather than the second with probability 2/10 . Agree ? Given that B hits a 3 or a 4 , we would remove the other number so that there are 8 numbers left . The probability A hits a 1 or a 2 from 8 possible choices is 2/8 . Agree ? Now we add 2/10*2/8 + 2/10*2/8 = 0.1 Do the other examples the same way . |
#23
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Re: Monty at it again
[ QUOTE ]
I don't think I can explain this any clearer . The probability 1 or 2 gets selected on the first draw is 2/10 , agree ? Given that 1 or 2 has been selected , the other number gets removed from the vault so that there are 8 numbers left . The probability it's B should be 2/8 since he has two good choices out of 8 . Agree ? However , B may hit 3 or 4 on the first draw rather than the second with probability 2/10 . Agree ? Given that B hits a 3 or a 4 , we would remove the other number so that there are 8 numbers left . The probability A hits a 1 or a 2 from 8 possible choices is 2/8 . Agree ? Now we add 2/10*2/8 + 2/10*2/8 = 0.1 Do the other examples the same way . [/ QUOTE ] I completely agree with your math there. That's the exact same results I posted earlier: Given that Albert has 2 tickets therefore P(A|) = 2/10 Bill has 2 tickets therefore P(B|) = 2/10 Charley has 3 tickets therefore P(C|) = 3/10 Dennis has 3 tickets therefore P(D|) = 3/10 P(B|A) = 2/8 P(C|A) = 3/8 P(D|A) = 3/8 P(A|B) = 2/8 P(C|B) = 3/8 P(D|B) = 3/8 P(A|C) = 2/7 P(B|C) = 2/7 P(D|C) = 3/7 P(A|D) = 2/7 P(B|D) = 2/7 P(C|D) = 3/7 P(AB) = P(A|)*P(B|A) + P(B|)*P(A|B) = 2/10*2/8 + 2/10*2/8 = 1/10 P(AC) = P(A|)*P(C|A) + P(C|)*P(A|C) = 2/10*3/8 + 3/10*2/7 = 9/56 P(AD) = P(A|)*P(D|A) + P(D|)*P(A|D) = 2/10*3/8 + 3/10*2/7 = 9/56 P(BC) = P(B|)*P(C|B) + P(C|)*P(B|C) = 2/10*3/8 + 3/10*2/7 = 9/56 P(BD) = P(B|)*P(D|B) + P(D|)*P(B|D) = 2/10*3/8 + 3/10*2/7 = 9/56 P(CD) = P(C|)*P(D|C) + P(D|)*P(C|D) = 3/10*3/7 + 3/10*3/7 = 9/35 But what I'm having trouble with is why my alternate method of counting rather than calculating is giving different results. |
#24
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Re: Monty at it again
Hey Mykey , you're right ...the total sample space should be 74 .
Your second solution is correct and the first solution is flawed . For P(AB) we said that the total sample space is 80 so that the probability is 8/80 but this is wrong . The 80 elements in our sample space includes multiplicities of equal prizes . There are clearly 10 choices for the first prize , but there is not always 8 choices for the second . If the first one belongs to c , then there are 7 choices for the second draw . There are 8 choices for the second prize precisely 40% of the time . There are 7 choices for the second prize 60% of the time . 10*8*0.4 + 10*7*0.6 = 74 . |
#25
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Re: Monty at it again
So, what method of using P(A|), P(B|A) etc. will give the proper answer to the problem?
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#26
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Re: Monty at it again
Let Z be the event that no player can win twice .
P(AB|Z ) =P(ABZ)/P(Z) P(Z)=1 since eventually we can continue this until no player can win twice . P(ABZ) = n(ABZ)/s(ABZ) = 8/(4/10*8/10 + 6/10*7/10) =8/74 There are 8 ways that the event ABZ occur together . Also 4 out of 10 times there will be no match 8 out of 10 times ; 6 out of 10 times , there will be no match 7 out of 10 times . This gives us 74 in the denominator . |
#27
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Re: Monty at it again
[ QUOTE ]
Let Z be the event that no player can win twice . P(AB|Z ) =P(ABZ)/P(Z) P(Z)=1 since eventually we can continue this until no player can win twice . P(ABZ) = n(ABZ)/s(ABZ) = 8/(4/10*8/10 + 6/10*7/10) =8/74 There are 8 ways that the event ABZ occur together . Also 4 out of 10 times there will be no match 8 out of 10 times ; 6 out of 10 times , there will be no match 7 out of 10 times . This gives us 74 in the denominator . [/ QUOTE ] Sorry but it looks to me like you've got an answer (8/74) that you are trying to wedge into a formula. 8/(4/10*8/10 + 6/10*7/10) = 400/37 |
#28
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Re: Monty at it again
It was a typo !!!
The numerator is 8 . The denominator is (4*8 + 6*7 )=74 The denominator should list the total number of elements . I'm sure you get the idea . 4 times the first number will belong to A or B . Given that the first card belongs to A or B , the second card will be a non-matching card 8 times . Similarly , 6 out of 10 times the first card will belong to C or D .Given that the first card belongs to C or D , the second card will be a non-matching card 7 times . |
#29
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Re: Monty at it again
I used this code to simulate the drawings.
<font class="small">Code:</font><hr /><pre> var AB, AC, AD, BC, BD, CD, n : integer; p1, p2, w1, w2 : integer; begin AB := 0; AC := 0; AD := 0; BC := 0; BD := 0; CD := 0; n := 0; randomize; while n < 1000000 do begin p1 := trunc(random*10); case p1 of 0, 1 : w1 := 1; 2, 3 : w1 := 2; 4..6 : w1 := 3; 7..9 : w1 := 4; end; repeat p2 := trunc(random*10); case p2 of 0, 1 : w2 := 1; 2, 3 : w2 := 2; 4..6 : w2 := 3; 7..9 : w2 := 4; end; until w1 <> w2; case w1 of 1 : case w2 of 2 : inc(AB); 3 : inc(AC); 4 : inc(AD); end; 2 : case w2 of 1 : inc(AB); 3 : inc(BC); 4 : inc(BD); end; 3 : case w2 of 1 : inc(AC); 2 : inc(BC); 4 : inc(CD); end; 4 : case w2 of 1 : inc(AD); 2 : inc(BD); 3 : inc(CD); end; end; inc(n); end; writeln('AB = ',AB/N:8:5); writeln('AC = ',AC/N:8:5); writeln('AD = ',AD/N:8:5); writeln('BC = ',BC/N:8:5); writeln('BD = ',BD/N:8:5); writeln('CD = ',CD/N:8:5); readln; end. </pre><hr /> Gave results of AB = 0.09988 AC = 0.16108 AD = 0.16107 BC = 0.16050 BD = 0.16093 CD = 0.25653 |
#30
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Re: Monty at it again
Finally this version assumes if the same person won twice, both drawings would be invalid and redrawn.
<font class="small">Code:</font><hr /><pre> var AB, AC, AD, BC, BD, CD, n : integer; p1, p2, w1, w2 : integer; begin AB := 0; AC := 0; AD := 0; BC := 0; BD := 0; CD := 0; n := 0; randomize; while n < 1000000 do begin repeat p1 := trunc(random*10); case p1 of 0, 1 : w1 := 1; 2, 3 : w1 := 2; 4..6 : w1 := 3; 7..9 : w1 := 4; end; p2 := trunc(random*10); case p2 of 0, 1 : w2 := 1; 2, 3 : w2 := 2; 4..6 : w2 := 3; 7..9 : w2 := 4; end; until w1 <> w2; case w1 of 1 : case w2 of 2 : inc(AB); 3 : inc(AC); 4 : inc(AD); end; 2 : case w2 of 1 : inc(AB); 3 : inc(BC); 4 : inc(BD); end; 3 : case w2 of 1 : inc(AC); 2 : inc(BC); 4 : inc(CD); end; 4 : case w2 of 1 : inc(AD); 2 : inc(BD); 3 : inc(CD); end; end; inc(n); end; writeln('AB = ',AB/N:8:5); writeln('AC = ',AC/N:8:5); writeln('AD = ',AD/N:8:5); writeln('BC = ',BC/N:8:5); writeln('BD = ',BD/N:8:5); writeln('CD = ',CD/N:8:5); readln; end. </pre><hr /> Giving results of AB = 0.10866 AC = 0.16282 AD = 0.16197 BC = 0.16147 BD = 0.16279 CD = 0.24230 So the flaw was in the counting method, not the probability method. |
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