I have 6 pieces of candy: two each of 3 different colors. They taste especially good if eaten two at a time, provided the colors are different.
These candies are in an opaque bag from which I pick two at a time. If they are different colors I eat them together (yum) but if they are the same I put them back in and draw again. I will repeat this process to eat two more.
What is the probability the last two candies will be of differing colors?
Repeat with two each of 4 colors.
Repeat with two each of 5 colors.
At any given point there will be an even number of singlets and a number of doublets in the bag. Let s be the number of singlets and d be the number of doublets.
Every time you choose two candies of differing color, you can either reduce the number of singlets by 2, reduce the number of doublets by 1 (by choosing a singlet and one member of a doublet) or increase the number of singlets by 2 while decreasing the number of doublets by 2 (by choosing 1 member of each of 2 doublets).
Let's try to build a table of probabilities, p, of success (ending up with 2 singlets), starting with the fact that no matter how many singlets there are (always an even number), if there are no doublets, p=1; and if there is nothing left but one doublet, p=0:
doublets:
singlets: 0 1 2 3 4 5 6
0 0
2 1
4 1
6 1
8 1
10 1
12 1
Let's expand the table.
If there are s singlets and d doublets, the probabilities of going to:
s-2 singlets and d doublets: (s/(s+2d) * (s-1)/(s+2d-1)) / (1 - (2d/(s+2d))*1/(s+2d-1))
s singlets and d-1 doublets: 2*(s/(s+2d) * 2d/(s+2d-1)) / (1 - (2d/(s+2d))*1/(s+2d-1))
s+2 singlets and d-2 doublets: ((2d/(s+2d)) * (2(d-1)/(s+2d-1)) / (1 - (2d/(s+2d))*1/(s+2d-1))
where the division by (1 - (2d/(s+2d))*1/(s+2d-1)) converts the overall probability into the conditional probability given that two different colors are definitely chosen.
These conditional probabilities are then used to fill in the table spot for s singlets and d doublets, by multiplying these conditional probabilities by the respective new configuration's probabilities, and adding the three products together. This should be done columnwise, so that the entry for s+2 singlets with d-2 doublets is available. With the 12 singlets, 0 doublets we started with, the successive columns will have one fewer entries capable of being calculated.
This looks like it calls for a computer program rather than going through the tedious and error-prone calculations by hand. BTW, this is known as dynamic program, where the table is built from the bottom up so that each entry is calculated but once. The opposite would be recursive programming, where the values of later-on (usually higher) entries are found by calling on the function for other (usually lower, but in this case involving higer s values) values, and them in turn calling on other values, until a stop is reached such as zero doubles or 0 singlets.
10 dim prob(12,6)
11 kill "candych.txt":open "candych.txt" for output as #2
20
30 for s=0 to 12 step 2
40 prob(s,0)=1
50 next
60 prob(0,1)=0
70
80 for d = 1 to 6
90 for s= 0 to 12-d step 2
91 p1=0
100 if s>0 then
110 :prob(s,d)=prob(s-2,d)*(s/(s+2*d) * (s-1)/(s+2*d-1)) / (1 - (2*d/(s+2*d))*1/(s+2*d-1))
111 :p1=p1+(s/(s+2*d) * (s-1)/(s+2*d-1)) / (1 - (2*d/(s+2*d))*1/(s+2*d-1))
115 if (1 - (2*d/(s+2*d))*1/(s+2*d-1)) > 0 then
120 :prob(s,d)= prob(s,d)+prob(s,d-1)*2*(s/(s+2*d) * 2*d/(s+2*d-1)) / (1 - (2*d/(s+2*d))*1/(s+2*d-1))
121 :p1=p1+2*(s/(s+2*d) * 2*d/(s+2*d-1)) / (1 - (2*d/(s+2*d))*1/(s+2*d-1))
130 if d>=2 and (1 - (2*d/(s+2*d))*1/(s+2*d-1)) > 0 then
140 :prob(s,d)= prob(s,d)+prob(s+2,d-2)*(2*d/(s+2*d)) * (2*(d-1)/(s+2*d-1)) / (1 - (2*d/(s+2*d))*1/(s+2*d-1))
141 :p1=p1+(2*d/(s+2*d)) * (2*(d-1)/(s+2*d-1)) / (1 - (2*d/(s+2*d))*1/(s+2*d-1))
142 print p1
150 next
160 next
170
180 for s=0 to 12 step 2
190 for d=0 to 6
200 print using(3,7),prob(s,d);
205 print #2, using(3,7),prob(s,d);
210 next
220 print
225 print #2,
230 next
231
232 erase prob()
233 dim prob(12,6)
234 for s=0 to 12 step 2
240 prob(s,0)=1
250 next
260 prob(0,1)=0
270
280 for d = 1 to 6
290 for s= 0 to 12-d step 2
291 p1=0
300 if s>0 then
310 :prob(s,d)=prob(s-2,d)*(s//(s+2*d) * (s-1)//(s+2*d-1)) // (1 - (2*d//(s+2*d))*1//(s+2*d-1))
311 :p1=p1+(s//(s+2*d) * (s-1)//(s+2*d-1)) // (1 - (2*d//(s+2*d))*1//(s+2*d-1))
315 if (1 - (2*d//(s+2*d))*1//(s+2*d-1)) > 0 then
320 :prob(s,d)= prob(s,d)+prob(s,d-1)*2*(s//(s+2*d) * 2*d//(s+2*d-1)) // (1 - (2*d//(s+2*d))*1//(s+2*d-1))
321 :p1=p1+2*(s//(s+2*d) * 2*d//(s+2*d-1)) // (1 - (2*d//(s+2*d))*1//(s+2*d-1))
330 if d>=2 and (1 - (2*d//(s+2*d))*1//(s+2*d-1)) > 0 then
340 :prob(s,d)= prob(s,d)+prob(s+2,d-2)*(2*d//(s+2*d)) * (2*(d-1)//(s+2*d-1)) // (1 - (2*d//(s+2*d))*1//(s+2*d-1))
341 :p1=p1+(2*d//(s+2*d)) * (2*(d-1)//(s+2*d-1)) // (1 - (2*d//(s+2*d))*1//(s+2*d-1))
342 print p1
350 next
360 next
370
380 for s=0 to 12 step 2
390 for d=0 to 6
400 print prob(s,d);" ";
405 print #2, prob(s,d);" ";
410 next
420 print
425 print #2,
430 next
470
777 close #2
The p1 counts verify that the weights assigned for the three possible sources of each of the probabilities add up to 1. They do.
doublets
singlets 0 1 2 3 4 5 6
0 1.0000000 0.0000000 1.0000000 0.8000000 0.8769231 0.8917802 0.9072638
2 1.0000000 0.8000000 0.8769231 0.8917802 0.9072638 0.9183540 0.9270402
4 1.0000000 0.9142857 0.9188504 0.9245634 0.9308443 0.9365120 0.9414675
6 1.0000000 0.9523810 0.9451139 0.9449034 0.9468746 0.9494783 0.9521768
8 1.0000000 0.9696970 0.9608358 0.9582462 0.9580923
10 1.0000000 0.9790210 0.9707691
12 1.0000000
doublets
singlets 0 1 2 3 4 5 6
0 1 0 1 4/5 57/65 10144/11375 5500627/6062875
2 1 4/5 57/65 10144/11375 5500627/6062875 7132435372/7766542875 54385624175677/58665876030125
4 1 32/35 1087/1183 2871116/3105375 157981084171/169718059875 10700625458543232/11426042090337875 302289395380852218263/321083208780584625375
6 1 20/21 48077/50869 1943079856/2056379325 71769386294183/75796085540175 34113210274641100556/35928373207203852375 752482546890087526979034523/790276099985519578276720875
8 1 32/33 230419/239811 3096527828/3231453225 19635660282206457/20494539662581985
10 1 140/143 6906433/7114393
12 1
Answers:
2 each of 3 colors: p = 4/5 = 0.8
2 each of 4 colors: p = 57/65 ~= 0.8769231
2 each of 5 colors: p = 10144/11375 ~= 0.8917802
2 each of 6 colors: p = 5500627/6062875 ~= 0.9072638
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Posted by Charlie
on 2016-05-25 11:37:33 |
No Subject
