Probability Help

Question

What is the probability that at least three of the nine justices of the U.S. Supreme Court have the same birthday? Round your answer to the nearest ten thousandth.

Answer 1

1. This problem is similar to the Von Mises Paradox Birthday problem, which asks how many people need to be in a room in order for at least two to have the same birthday.  (Answer: 23).   And there is a simple formula that can be derived, and the numbers plugged in.  However, the problem gets considerably more complicated (especially in terms of the number of calculations that must be performed) when the problem description changes from asking for at least two people to three people.
2. Assumption will be that a year is 365 days, i.e., ignore the impact of leap years.
3. P[at least 3 Justices having same Bday] =  1 -  P[none have the same Bday] - P[exactly one pair have the same Bday, and 3 do not] - P[exactly two pair have the same Bday, and 3 do not] - P[exactly three pair have the same Bday, and 3 do not] - P[exactly four pair have the same Bday and 3 do not]  {Eqn 1}
4. Notice in Eqn 1 that the approach is to find the probability that 3 Justices have the same birthday by subtracting the sum of all other possibilities from 1.  The reason we stop at four pair is because there are 9 Justices, so at most, only four pair can have the same BDay.
5. I will try (time permitting) to use additional posts to get a numerical answer to this problem by solving Eqn. 1.

------------------------------------------------------------------------- Part III (note Part II appears as a separate Comment section) [STEP 1]The next step is to find out the number of ways Bdays can be distributed so that k pairs have the same birthday, and everyone else in the group of N size has some different Bday. --------------------------------------------------------------------------------- [STEP 2] The equation for this is: 1. # of Bday possibilities for k pairs: (365 * ...(365-N+k+1))/ k!{Eqn 3} -------------------------------------------------------------------------------- [STEP 3] For example, when k = 2 and N = 9 1. # of Bday possibilities for 2 pairs: (365*... (365-9+2+1)) / 2! 2. # of Bday possibilities for 2 pairs: (365*...(359)) / 2 3. # of Bday possibilities for 2 pairs: (365*364*363*362*361*360*359) / 2 4. # of Bday possibilities for 2 pairs: 8.14542E+17/2 5. # of Bday possibilities for 2 pairs: 4.07271E+17 -------------------------------------------------------------------------------- [STEP 4] Calculating Eqn 3 for all values of k: 1. K = 0: # of Bday possibilities = 1.04103#+23 2. k = 1: # of Bday possibilities = 2.91606E+20 3. k = 2: # of Bday possibilities = 4.07271E+17 4. k = 3: # of Bday possibities = 3.78153E+14 5. k = 4: # of Bday possibilities = 2.62606E+11 -------------------------------------------------------------------------------- [STEP 5] So, what we have calculated in step 4 above is the number of ways that exactly k pairs of Justices can have the same birthday while everyone else does not. -------------------------------------------------------------------------------- [STEP 6] Now to figure out the Probability of this occurring, we need to divide the # of "favorable" Bday possibilities by the total number of possibilities. The total number is simply 365^N. In this problem, N = 9, so the total number of possibilities is 365^9. And all this means is that Justice 1 can have a Bday on any one of 365 days, and so can Justice 2, Justice 3, etc. --------------------------------------------------------------------------------- [STEP 7] Dividing each value from step 4 by 3659 yields the following: 1. k = 0: P[no Justices have the same Bday] = 0.90537617 2. k = 1: P[one pair of Justices have the same Bday] = .00253607 3. k = 2: P[two pairs of Justices have the same Bday] = 3.542E-06 4. k = 3: P[three pairs of Justices have the same Bday] = 3.2888E-09 5. k = 4: P[fours pairs of Justices have the same Bday] = 2.2839E-12 -------------------------------------------------------------------------------- [STEP 8] Now, if you recall from Part II of this answer, we calculated the number of different ways that k pairs can exist. So, now we must multiply the results from part 2 by the results obtained in Step 7 above for each value of k: 1. k = 0: 0.90537617* 1 = 0.90537617 2. k = 1: 0.00253607*36 = 0.09129844 3. k = 2: 3.542E-06 * 756 = 0.00267775 4. k = 3: 3.2888E-09 * 7560 = 2.4863E-05 5. k = 4: 2.2839E-12 * 22680 = 5.1798E-08 -------------------------------------------------------------------------------- [STEP9] What we now have in Step 8 is the probability of each term which appeared in Eqn 1. Now the last step is to sum up all the values from Step 8, and subtract this sum from 1.0: 1. The sum of all the values from Step 8 is: 0.90537616 + 0.09129844 + 0.00267775 + 2.4863E-05 + 5.1798E-08 = 0.99937727 2. P[at least 3 Justices having the same Bday] = 1- 0.99937727 = 0.000622735 3. Since the problem description asked for rounding to the nearest ten thousandth, our final answer is 0.0006. So, there is roughly a 6 in 10,000 chance that 3 of the 9 Justices will have the same Bday.