Probabilities when Rolling 10 DiceDate: 07/29/99 at 19:25:41 From: Jeffrey Subject: A question of probability and odds. The house rolls 10 dice, one die at a time. What are the odds of the following: 1. A player rolling the same numbers in the same order as the house. 2. A player rolling the same numbers as the house without regard to the order in which they were rolled. 3. A player rolling 9 of 10 numbers the same as the house without regard to the order in which they were rolled. 4. A player rolling 8 of 10 numbers the same as the house without regard to the order in which they were rolled. Assumptions: 10 standard six-sided dice numbered 1 to 6. Thank you in advance. Date: 08/01/99 at 06:27:12 From: Doctor Anthony Subject: Re: A question of probability and odds. 1. A player rolling the same numbers in the same order as the house. Each of the 10 dice could produce 6 different numbers, so there are 6^10 different arrangements. The probability that a player will get the same numbers in the same order is: 1/6^10 2. A player rolling the same numbers as the house without regard to the order in which they were rolled. This is a more difficult calculation. We shall approach it by considering that we have 6 cells, representing the possible number on the face of any die, and that 10 balls are to be distributed into the 6 cells. The number of balls in any cell will represent the number of times that the cell number occurred when the 10 dice are rolled. It is possible of course that some cells (from 0 to 5) could be empty. We can represent the situation graphically as follows: |***| |**|****| |*| representing occupancy 3, 0, 2, 4, 0, 1 in the 6 cells. That is 3 1's, 0 2's, 2 3's, 4 4's, 0 5's and 1 6. With 6 cells we must start and end with a |, but there are five |'s and 10 *'s to permute in every possible way. That is 15 objects, 5 being alike of one kind and 10 being alike of a second kind. The number of arrangements is: 15! ------ = 3003 5! 10! So now the probability that the player gets the same distribution of numbers as the house is: 1/3003 3. A player rolling 9 of 10 numbers the same as the house without regard to the order in which they were rolled. There are 3003 ways in which the 10 balls could be placed in the cells. Imagine that all 10 are correct, that is, we have the same distribution of balls in the cells for the house and the player. We must now think of the number of ways that we could end up with just 9 correct. This can be achieved by taking one ball from one cell and transferring it to another cell. There will then be a shortage of balls in one cell but a surplus in another cell. So we must count the number of ways that we could make the transfer. Consider first the case with no empty cells. With no cells empty we can represent the situation as in the diagram below: |**|*|***|**|*|*| with occupancy 2, 1, 3, 2, 1, 1 Now we start and end with a |, but we cannot have any of the other 5 |'s from being together. We must choose 5 of the 9 gaps between the 10 *'s. This can be done in C(9,5) = 126 ways. If there are no empty cells we could take the ball in 6 different ways and place it in another cell in 5 ways - giving 30 different arrangements with one ball out of place. The probability of having one ball out of place in the case of no empty cells in the original house arrangement is: 30/126 = 5/21 However this must be multiplied by the probability that there are no empty cells. This is given by 126/3003. So total probability of having no empty cell and having 9 out of 10 correct is: 126/3003 x 30/126 = 30/3003 We can see that the number 126 has cancelled out, so we do not need to calculate the number of ways that we could have exactly, 0,1,2,...,5 empty cells. All we require is the number of ways that the transfers between cells can occur for the various configurations and put this total over 3003. If one cell is empty then the number of ways of making the transfer is calculated as follows. We can choose the cell from which to take a ball in 5 ways and place it in another cell also in 5 ways. (We can now use the empty cell into which to move the ball as one option). This gives 5 x 5 = 25 ways of having 9 out of 10 correct with one cell empty. The required probability is: 25/3003 With two cells empty the cell to take the ball from can be chosen in 4 ways, and the cell into which to place the ball in 5 ways, giving 4 x 5 = 20 ways of moving one ball. The probability that 9 out of 10 are correctly placed when two cells are originally empty is then: 20/3003 With three cells empty we can take one ball in 3 ways and place it in 5 ways = 15 ways. Required probability with 9 out of 10 correct and 3 cells empty is: 15/3003 With 4 cells empty we can take one ball in 2 ways and place it in 5 ways = 10 ways. Required probability that 9 out of 10 correct and 4 cells empty is 10/3003 With 5 cells empty we can take one ball in 1 way and place it in 5 ways = 5 ways. Required probability that 9 out of 10 correct and 5 cells empty is: 5/3003 So the total probability that 9 out of 10 are correct is: 30 + 25 + 20 + 15 + 10 + 5 105 5 = --------------------------- = ---- = --- 3003 3003 143 4. A player rolling 8 of 10 numbers the same as the house without regard to the order in which they were rolled. I will leave this for you to puzzle over. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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