Answer
See the explanation
Work Step by Step
To find the probability of at least two of three arbitrary records hashing to the same bucket, we can use the concept of complementary probability, which is 1 minus the probability of all records hashing to different buckets.
Probability of no collision:
$(\text{ no collision})=\frac{10\cdot 9\cdot 8}{10^3}=0.72$
Probability of at least one collision:
$P(\text{collision})=1-0.72=0.28$
Given 10 buckets, the probability of a record not hashing to a particular bucket is 9/10. Therefore, the probability of all three records not hashing to the same bucket is (9/10)^2 = 81/100.
So, the probability of at least two of the three records hashing to the same bucket is 1 - (81/100) = 19/100.
To determine when it's more likely for collisions to occur than not, we need to find the point where the probability of collisions exceeds 50%.
Let's denote the number of records as N. The probability of no collisions happening with N records can be calculated using the formula:
(1 - 1/10) * (1 - 2/10) * ... * (1 - (N-1)/10)
We want this probability to be less than 50%, so:
(1 - 1/10) * (1 - 2/10) * ... * (1 - (N-1)/10) < 0.5
You would then solve this equation for N to find the minimum number of records needed. However, solving this equation can be complex, so typically, you would use approximations or simulations to find a practical answer.
For $N=2$ we have:
$\left(1-\frac{1}{10}\right)=0.9>0.5$
For $N=3$ we have:
$\left(1-\frac{1}{10}\right)\left(1-\frac{2}{10}\right)=0.72>0.5$
For $N=4$ we have:
$\left(1-\frac{1}{10}\right)\left(1-\frac{2}{10}\right))\left(1-\frac{3}{10}\right)=0.504>0.5$
For $N=5$ we have:
$\left(1-\frac{1}{10}\right)\left(1-\frac{2}{10}\right))\left(1-\frac{3}{10}\right)\left(1-\frac{4}{10}\right)=0.3024<0.5$
So the smallest number of records where collisions are more likely than not is 5.
