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Starting a new thread to avoid hi-jacking my previous thread:

https://www.physicsforums.com/forumdisplay.php?f=73

There are 7 items, and 4 containers. A 4 sided die (tetrahedron) is used as a random number generator and is thrown 7 times. After each toss, an item is dropped into a container corrersponding to the bottom side of the die.

There are (n+r-1)c(r) = 10c7 = 120 possible outcomes. Here is a table showing the possible outcomes. Each digit of each 4 digit number represents the number of items in a container:

0007 0016 0025 0034 0043 0052 0061 0070 0106 0115

0124 0133 0142 0151 0160 0205 0214 0223 0232 0241

0250 0304 0313 0322 0331 0340 0403 0412 0421 0430

0502 0511 0520 0601 0610 0700 1006 1015 1024 1033

1042 1051 1060 1105 1114 1123 1132 1141 1150 1204

1213 1222 1231 1240 1303 1312 1321 1330 1402 1411

1420 1501 1510 1600 2005 2014 2023 2032 2041 2050

2104 2113 2122 2131 2140 2203 2212 2221 2230 2302

2311 2320 2401 2410 2500 3004 3013 3022 3031 3040

3103 3112 3121 3130 3202 3211 3220 3301 3310 3400

4003 4012 4021 4030 4102 4111 4120 4201 4210 4300

5002 5011 5020 5101 5110 5200 6001 6010 6100 7000

Out of these 120 outcomes, (7-1)c(4-1) = 6c3 = 20 of them have no empty containers.

1114 1123 1132 1141 1213 1222 1231 1312 1321 1411

2113 2122 2131 2212 2221 2311 3112 3121 3211 4111

The goal here is to determine the probability of occurrence for each of the 120 possible outcomes.

One method that could be used to calculate the probablities would be to consider the total number of possible outcomes of the 7 throws of the 4 sided die as a 7 digit number in base 4, going from {0,0,0,0,0,0,0} (all items in first container), to {3,3,3,3,3,3,3} (all items in last container), so total number of cases = 4

A program could cycle through all 16384 cases, creating a histogram for the 120 possible outcomes. For examples of the output of using such a program, each of the 4 permutations for {0,0,0,7} occurs in 1 of the 16384 cases, each of the 4 permutations of {1,1,1,4} occurs in 210 of the 16384 cases, each of the 12 permutations of {1,1,2,3} occurs in 420 of the 16384 cases, and each of the 4 permutations of {1,2,2,2} occurs in 630 of the 16384 cases.

I'm wondering if this is a valid approach for determining the probabilities of the outcomes.

https://www.physicsforums.com/forumdisplay.php?f=73

There are 7 items, and 4 containers. A 4 sided die (tetrahedron) is used as a random number generator and is thrown 7 times. After each toss, an item is dropped into a container corrersponding to the bottom side of the die.

There are (n+r-1)c(r) = 10c7 = 120 possible outcomes. Here is a table showing the possible outcomes. Each digit of each 4 digit number represents the number of items in a container:

0007 0016 0025 0034 0043 0052 0061 0070 0106 0115

0124 0133 0142 0151 0160 0205 0214 0223 0232 0241

0250 0304 0313 0322 0331 0340 0403 0412 0421 0430

0502 0511 0520 0601 0610 0700 1006 1015 1024 1033

1042 1051 1060 1105 1114 1123 1132 1141 1150 1204

1213 1222 1231 1240 1303 1312 1321 1330 1402 1411

1420 1501 1510 1600 2005 2014 2023 2032 2041 2050

2104 2113 2122 2131 2140 2203 2212 2221 2230 2302

2311 2320 2401 2410 2500 3004 3013 3022 3031 3040

3103 3112 3121 3130 3202 3211 3220 3301 3310 3400

4003 4012 4021 4030 4102 4111 4120 4201 4210 4300

5002 5011 5020 5101 5110 5200 6001 6010 6100 7000

Out of these 120 outcomes, (7-1)c(4-1) = 6c3 = 20 of them have no empty containers.

1114 1123 1132 1141 1213 1222 1231 1312 1321 1411

2113 2122 2131 2212 2221 2311 3112 3121 3211 4111

The goal here is to determine the probability of occurrence for each of the 120 possible outcomes.

One method that could be used to calculate the probablities would be to consider the total number of possible outcomes of the 7 throws of the 4 sided die as a 7 digit number in base 4, going from {0,0,0,0,0,0,0} (all items in first container), to {3,3,3,3,3,3,3} (all items in last container), so total number of cases = 4

^{7}= 16384.A program could cycle through all 16384 cases, creating a histogram for the 120 possible outcomes. For examples of the output of using such a program, each of the 4 permutations for {0,0,0,7} occurs in 1 of the 16384 cases, each of the 4 permutations of {1,1,1,4} occurs in 210 of the 16384 cases, each of the 12 permutations of {1,1,2,3} occurs in 420 of the 16384 cases, and each of the 4 permutations of {1,2,2,2} occurs in 630 of the 16384 cases.

I'm wondering if this is a valid approach for determining the probabilities of the outcomes.

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