Can you figure out what the odds of this happening are?

For the last 6 holidays, my family has drawn a piece of paper from a hat to see who is randomly chosen to say grace. Everyone draws a piece of paper, but only one piece says, ';You say grace'; on it. There were 9 people at the first holiday, 9 at the next, 9 at the third, 8 at the fourth, 9 at the fifth and 10 at the last. I have drawn this ';grace'; piece of paper 4 out of the 6 holidays! What are the mathematical chances or probability that I would have drawn this piece of paper 4 out of the 6 holidays with that many people involved? Thanks!Can you figure out what the odds of this happening are?
This is a bit of a long problem to solve since the probability for each holiday is not the same. If you got ';grace'; 4 out of 6 times, that means you didn't get it 2 out of the 6 times. Let's look at how that can be arranged:


You can not say grace on the 1st and 2nd, or 1st and 3rd, etc.


1,2 1,3 1,4 1,5 1,6 2,3 2,4 2,5 2,6 3,4 3,5 3,6 4,5 4,6 5,6





So, you've got these 15 arrangements of times for you not to say grace (and to say grace). Look at the first outcome: you did not say grace on the 1st and 2nd holiday. The probability of that is (8/9)*(8/9), but you did say grace on the other holidays, and that probability is (1/9)*(1/8)*(1/9)*(1/10). So, the probability of you saying grace on just the last 4 holidays is (8/9)*(8/9)*(1/9)*(1/8)*(1/9)*(1/10) = 64/524880 = 4/32805 = about .000122 and you've got 14 other possibilities:


1,3: (8/9)*(1/9)*(8/9)*(1/8)*(1/9)*(1/10) = 64/524880 = 4/32805


1,4: (8/9)*(1/9)*(1/9)*(7/8)*(1/9)*(1/10) = 56/524880 = 7/65610


1,5: (8/9)*(1/9)*(1/9)*(1/8)*(8/9)*(1/10) = 64/524880


1,6: (8/9)*(1/9)*(1/9)*(1/8)*(1/9)*(9/10) = 72/524880


2,3: (1/9)*(8/9)*(8/9)*(1/8)*(1/9)*(1/10) = 64/524880


2,4: (1/9)*(8/9)*(1/9)*(7/8)*(1/9)*(1/10) = 56/524880


2,5: (1/9)*(8/9)*(1/9)*(1/8)*(8/9)*(1/10) = 64/524880


2,6: (1/9)*(8/9)*(1/9)*(1/8)*(1/9)*(9/10) = 72/524880


3,4: (1/9)*(1/9)*(8/9)*(7/8)*(1/9)*(1/10) = 56/524880


3,5: (1/9)*(1/9)*(8/9)*(1/8)*(8/9)*(1/10) = 64/524880


3,6: (1/9)*(1/9)*(8/9)*(1/8)*(1/9)*(9/10) = 72/524880


4,5: (1/9)*(1/9)*(1/9)*(7/8)*(8/9)*(1/10) = 56/524880


4,6: (1/9)*(1/9)*(1/9)*(7/8)*(1/9)*(9/10) = 63/524880


5,6: (1/9)*(1/9)*(1/9)*(1/8)*(8/9)*(9/10) = 72/524880


Add all of this up and I think you get 959/524880 = about .001827 or roughly 1 in 500.


Can you figure out what the odds of this happening are?
1/9 is the average so 1/9^4=1 in 6561 for 4 64/81 for the other 2 not to be chosen. Somewhere between 1 in 5184 and 1 in 6561.