Randomly Setting the TableDate: 1/25/99 at 15:11:44 From: Whitney Rooney Subject: Probability John has 4 place settings. Each consists of a cup, a saucer, and a plate; and each setting is a different color (red, green, blue, and yellow). If he places the settings randomly around a table, what is the probability that at least one place setting -- cup, saucer, and plate -- is of the same color? Most of us got 1/16 for an answer. We made a tree diagram and counted the possibilities that the colors would match. We assumed that the first layer is plates and it doesn't matter how you lay them out. Then there are 16 ways to lay out the saucers, (4 ways at each four plates), then there are 4 ways to lay out the cups at each of the 16 plate and saucer combinations. So there are 64 combinations, and 4 have the same color, so 4/64 = 1/16. I took squares of paper and colored the sides. For the first square (plates), the order of the colors doesn't matter. The next square is bigger and is the saucers. The colors can be in 6 different orders, and placed around the plates in 24 ways, since 6x4 = 24. The possibility that any match is 15/24. The next step was to realize that there is a different probability for third match when all 4 colors of the plates and saucers match, 2 of the colors match or just 1 color matches. It got real complicated, but out of 576 combinations, 123 should match at least one color. My friend and I spent hours Saturday trying to figure this out with our parents. I think the problem is like the marbles in a jar that you answered in the archives, because what you draw changes the probability of the next draw. Thanks for your help! Whitney Date: 01/25/99 at 16:27:23 From: Steve Rooney Subject: Re: probability I would like to expand a bit on our thinking above. I would appreciate any comments you may have time to give. I divided the problem into levels 1, 2, and 3. In the first level, plates, the order of placement doesn't affect the probability. Level 2 is the placement of the saucers, and level 3 is the placement of the cups. I divided level 2 into the probability that one and only one color match, that 2 colors match, and that all 4 colors match. The total probability that there is a match at level 2 is 15/24. This consists of the probability of one and only one match (8/24), of two matches only (6/24), and of all four matches (1/24). If one and only one color matches of the plates and saucers, the probability that the cup will match any one of these is 6/24. So the chance of all three matching is 8/24 * 6/24 = 48/576. If at level 2, there are 2 matches, (e.g. both the red and green plates and saucers match), we have 6 out of 24 ways to match two items, and 10 unique ways for each of those to match with a third item, so the probability of a complete match is 60/576. The possibility of three matches at level 2 is 0, for if 3 plates matched the saucers, then the fourth will also match. The probability that all four match at level 2 is 1/24. If all four match, then the probability that any one of the remaining cups completes a match is 15/24. So the probability of a complete match is 15/24 * 1/24 = 15/576. Thus the total probability is 48/576 + 60/576 + 15/576 = 123/576, which is approximately .21. Date: 01/26/99 at 20:19:05 From: Doctor Anthony Subject: Re: probability I will go through the calculation to find the probability that all three items match in at least one position. (This would also include 2, 3, or 4 positions being matched.) We can approach this in a different way, by considering what happens as we increase the number of positions at the table from 1 to 2 to 3 to 4. I will show the table of the calculations first and then explain how the entries are made. I am assuming that the colour of the cup is fixed and we then allocate a saucer and a plate at random. The number of colours is taken to equal the number of places. I will describe the process in detail below. Although the explanation is necessarily long- winded, making up the table was done very quickly. No. of Total 0 Correct 1 Correct 2 Correct 3 Correct 4 Correct Places Ways --------- --------- --------- --------- -------- ------- ---- 1 1 0 1 2 4 3 0 1 3 36 26 9 0 1 4 576 453 104 18 0 1 ---------------------------------------------------------------------- With 1 place, we initially have 1 colour and so the settings MUST be correct. With 2 places, we start with 2 colours. The plates could be chosen in 2 different orders and also the saucers, giving 4 possibilities with the settings. With just 2 colours, there is 1 way only for 2 to be correct. There cannot be just 1 correct with only 2 colours, so the column under 1 correct shows 0. By subtraction, the number of ways with 0 correct is 3. With 3 places, we start with 3 colours. The plates could be arranged in 3! = 6 ways and also the saucers in 6 ways, giving 36 ways with 3 colours. Using the fact that 3 arrangements existed, giving 0 correct places when only 2 colours were used, we can think of adding a complete correct place to these 3 to get 1 correct place. And the correct colour could be chosen in 3 ways giving 3 x 3 = 9 with 1 correct. We cannot have 2 correct, and then 3 correct can clearly be chosen in 1 way only. By subtraction, the number under 0 correct is 26. Finally, we start with 4 colours. Now the 26 under the 0 correct column for 3 places can be added to by a correct placing and then entered under the 1 correct column. But the correct place can be chosen in 4 ways, giving 4 * 26 = 104 under the 1 correct column. There is only 1 possibility under the 4 correct column and zero under the 3 correct column. For 2 correct, we can see that 18 = 3 * 6 is the number. The following should illustrate why. G and Y are to be the incorrect placings, R and B are correct. RB|GY RB|GY RB|GY RB|GY RB|YG RB|YG RB|YG RB|YG RB|GY This shows 3 ways that the incorrect placings can be arranged. The two correct colours, R and B in example above, can be chosen in C(4,2) = 6 ways. So the total for 2 correct colours is 3 * 6 = 18. Therefore, we have the total with 4 placings giving at least one correct setting: 104 + 18 + 1 = 123 And the required probability is 123/576 = 41/192 = 0.21354. This answer agrees with the calculation (though not using the same method) as that which you developed. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 01/30/99 at 10:08:16 From: Doctor Anthony Subject: Re: probability Returning to the problem of the cups, saucers and plates, I think that a general method can be developed such that the probabilities in the cases of 6, 8 or 12 place settings can easily be calculated, especially with the aid of a spreadsheet like Excel. I will go through a table for a total of six place settings, but clearly the method can be extended indefinitely without invoking advanced theorems of combinatorial theory. The basic idea is very simple. If you have the results for n place settings, then the results for n+1 settings is obtained from the previous line for, say, r correct settings by imagining that you add a correct setting to the existing arrangements (thus increasing the number of correct settings to r+1), but you multiply the previous answer by the ratio: number of ways of choosing r+1 from n+1 C(n+1,r+1) ---------------------------------------- = ---------- number of ways of choosing r from n C(n,r) I reproduce the table I gave you before: the numbers fit the calculation just described. The number for 0-correct is always found by subtraction from the total for that line after the other values have been filled in. The number for n correct is always 1 and the number for n-1 correct is always 0. No. of Total 0 Correct 1 Correct 2 Correct 3 Correct 4 Correct Places Ways --------- --------- --------- --------- -------- ------- ---- 1 1 0 1 2 4 3 0 1 3 36 26 9 0 1 4 576 453 104 18 0 1 ---------------------------------------------------------------------- The 9 in line for 3 places is obtained from 3 in line 2 by multiplying 3 by the factor C(3,1)/C(2,0) = 3. The 104 in line line 4 is obtained from 26 in previous line by multiplying 26 by the factor C(4,1)/C(3,0) = 4. The 18 in line 4 is obtained from 9 in line 3 by multiplying 9 by the factor C(4,2)/C(3,1) = 2. I shall now extend the table to six places: No. of Total 0-Corr. 1-Corr 2-Corr 3-Corr 4-Corr 5-Corr 6-Corr Places Ways ----- ------- ------ ------- ----- ------ ------ ------- ---- 1 1 0 1 2 4 3 0 1 3 36 26 9 0 1 4 576 453 104 18 0 1 5 14400 11844 2265 260 30 0 1 6 518400 439975 71064 6795 520 45 0 1 ----------------------------------------------------------------- For example, 520 = 260 x C(6,3)/C(5,2) 6795 = 2265 x C(6,2)/C(5,1) Now the probability of at least one correct setting is 78425/518400 = 0.15128 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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