Before you are three urns. The first two each contain 4 white and 6 black balls. The third has 3 white and 6 black.
Take one ball from the first urn and add it to the second with out looking at it. Stir it in, then take one ball from the second and add it to the third without looking at it.
If you pick a ball from the third urn, what is the probability it will be white?
What is the least number of balls you can put in the urns (at least one black and one white in each) to make the probability at the end equal to exactly 1/3?
The following program computes the probability for part 1, 17/50 = 34%, and goes through all ways of distributing up to 10 extra balls beyond the 6 required (1 of each color in each of 3 urns). It does so by setting up 5 partitions between the 6 color/urn categories with digit 1 representing each extra ball before/between/after the respective partitions.
2000 W1=4:T1=10:W2=4:T2=10:W3=3:T3=9
2010 P1=W1//T1
2020 P2=(P1*(1+W2)+(1-P1)*W2)//(1+T2)
2030 P3=(P2*(1+W3)+(1-P2)*W3)//(1+T3)
2040 print P3,P3/1
2050 for Extra=1 to 10
2060 Lst="|||||"
2070 for I=1 to Extra:Lst=Lst+"1":next
2080 LstH=Lst
2085 repeat
2090 Luse=Lst
2092 dim Numb(6)
2093 Upto=1
2100 repeat
2110 C=left(Luse,1):Luse=mid(Luse,2,*)
2120 if C="1" then Numb(Upto)+=1
2130 if C="|" then Upto+=1
2140 until Luse=""
2150 W1=Numb(1)+1:T1=2+Numb(1)+Numb(2)
2160 W2=Numb(3)+1:T2=2+Numb(3)+Numb(4)
2170 W3=Numb(5)+1:T3=2+Numb(5)+Numb(6)
2180 erase Numb()
2190 P1=W1//T1
2200 P2=(P1*(1+W2)+(1-P1)*W2)//(1+T2)
2210 P3=(P2*(1+W3)+(1-P2)*W3)//(1+T3)
2220 ' if P3=1//3 then print Lst,P3;tab(50);P3/1
2225 if P3=1//3 then print Lst;tab(20);W1;T1-W1;W2;T2-W2;W3;T3-W3,Extra+6;tab(40);
2226 if P3=1//3 then print P1;P2;P3
2230 gosub *Permute(&Lst)
2240 until Lst=LstH
2250 next
9999 end
10010
10020 *Permute(&A$)
10025 local X$,I,L$,J
10030 X$=""
10040 for I=len(A$) to 1 step -1
10050 L$=X$
10060 X$=mid(A$,I,1)
10070 if X$<L$ then cancel for:goto 10100
10080 next
10090
10100 if I=0 then
10110 :for J=1 to len(A$)\2
10120 :X$=mid(A$,J,1)
10130 :mid(A$,J,1)=mid(A$,len(A$)-J+1,1)
10140 :mid(A$,len(A$)-J+1,1)=X$
10150 :next
10160 :else
10170 :for J=len(A$) to I+1 step -1
10180 :if mid(A$,J,1)>X$ then cancel for:goto 10200:endif
10190 :next
10200 :mid(A$,I,1)=mid(A$,J,1)
10210 :mid(A$,J,1)=X$
10220 :for J=1 to (len(A$)-I)\2
10230 :X$=mid(A$,I+J,1)
10240 :mid(A$,I+J,1)=mid(A$,len(A$)-J+1,1)
10250 :mid(A$,len(A$)-J+1,1)=X$
10260 :next
10270 :endif
10280 return
The answer for the second part is 9 balls: 1 white and 2 black in each urn, as shown in the first line of this table:
distr of urn1 urn2 urn3 tot
extra balls w b w b w b balls p1 p2 p3
|1||1||1 1 2 1 2 1 2 9 1//3 1//3 1//3
1||1|||11 2 1 2 1 1 3 10 2//3 2//3 1//3
1|||11||1 2 1 1 3 1 2 10 2//3 1//3 1//3
|1|11|||11 1 2 3 1 1 3 11 1//3 2//3 1//3
1|111||1||1 2 4 1 2 1 2 12 1//3 1//3 1//3
|1|1|111||1 1 2 2 4 1 2 12 1//3 1//3 1//3
|1||1|1|111 1 2 1 2 2 4 12 1//3 1//3 1//3
111|1|1|||11 4 2 2 1 1 3 13 2//3 2//3 1//3
111|1||11||1 4 2 1 3 1 2 13 2//3 1//3 1//3
1||111|1||11 2 1 4 2 1 3 13 2//3 2//3 1//3
1||1|1111||1 2 1 2 5 1 2 13 2//3 1//3 1//3
1||1||1|1111 2 1 2 1 2 5 13 2//3 2//3 1//3
1|||11|1|111 2 1 1 3 2 4 13 2//3 1//3 1//3
1|111|11|||11 2 4 3 1 1 3 14 1//3 2//3 1//3
|1|1111|1||11 1 2 5 2 1 3 14 1//3 2//3 1//3
|1|11||1|1111 1 2 3 1 2 5 14 1//3 2//3 1//3
11|11111||1||1 3 6 1 2 1 2 15 1//3 1//3 1//3
1|111|1|111||1 2 4 2 4 1 2 15 1//3 1//3 1//3
1|111||1|1|111 2 4 1 2 2 4 15 1//3 1//3 1//3
|1|11|11111||1 1 2 3 6 1 2 15 1//3 1//3 1//3
|1|1|111|1|111 1 2 2 4 2 4 15 1//3 1//3 1//3
|1||1|11|11111 1 2 1 2 3 6 15 1//3 1//3 1//3
11111|11|1|||11 6 3 2 1 1 3 16 2//3 2//3 1//3
11111|11||11||1 6 3 1 3 1 2 16 2//3 1//3 1//3
111|1|111|1||11 4 2 4 2 1 3 16 2//3 2//3 1//3
111|1|1|1111||1 4 2 2 5 1 2 16 2//3 1//3 1//3
111|1|1||1|1111 4 2 2 1 2 5 16 2//3 2//3 1//3
111|1||11|1|111 4 2 1 3 2 4 16 2//3 1//3 1//3
1||11111|11||11 2 1 6 3 1 3 16 2//3 2//3 1//3
1||111|1|1|1111 2 1 4 2 2 5 16 2//3 2//3 1//3
1||11|111111||1 2 1 3 7 1 2 16 2//3 1//3 1//3
1||1|1111|1|111 2 1 2 5 2 4 16 2//3 1//3 1//3
1||1||11|111111 2 1 2 1 3 7 16 2//3 2//3 1//3
1|||11|11|11111 2 1 1 3 3 6 16 2//3 1//3 1//3
Where p1 is the probability of choosing a white ball from urn 1 to place in urn 2; p2 is the probability of choosing a white ball from urn 2 to place in urn 3; p3 is the probability of choosing a white ball from urn 3. Note that UBASIC places two slashes in its representation of rational numbers.
For curiosity, if you want to use the same 11 white and 18 black balls you started out with, any of these will work to produce a final probablity of 1/3:
urn1 urn2 urn3
w b w b w b p1 p2 p3
111111|1111111111111|11|||11 7 14 3 1 1 3 29 1//3 2//3 1//3
11111|11111111111|11||1|1111 6 12 3 1 2 5 29 1//3 2//3 1//3
1111|111111111|11||11|111111 5 10 3 1 3 7 29 1//3 2//3 1//3
111|1111111|11||111|11111111 4 8 3 1 4 9 29 1//3 2//3 1//3
11|11111|11||1111|1111111111 3 6 3 1 5 11 29 1//3 2//3 1//3
1|111|11||11111|111111111111 2 4 3 1 6 13 29 1//3 2//3 1//3
|1|11||111111|11111111111111 1 2 3 1 7 15 29 1//3 2//3 1//3
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Posted by Charlie
on 2007-02-20 17:03:26 |
part 2: brute force
