Monty at it again

[mykey1961]

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>

[jay_shark]

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 .

[mykey1961]

[ 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.

[jay_shark]

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 .

[mykey1961]

So, what method of using P(A|), P(B|A) etc. will give the proper answer to the problem?

[jay_shark]

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 .

[mykey1961]

[ 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

[jay_shark]

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 .

[mykey1961]

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 &lt; 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 &lt;&gt; 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

[mykey1961]

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 &lt; 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 &lt;&gt; 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.