What is the smallest value of n that assures at least a 50% chance that at least two students select the same integer?

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SUMMARY

The smallest value of n that ensures at least a 50% chance that at least two students select the same integer from a range of 1 to 10 is 5. This conclusion is derived from the principles of the Birthday problem, where the probability of unique selections decreases as more individuals are added. Specifically, with 10 possible integers, the calculations show that with 5 students, the probability of at least one duplicate selection exceeds 50%.

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Each individual in a group of n students is asked to pick an integer at random between 1 and 10 (inclusive). What is the smallest value of n that assures at least a 50% chance that at least two students select the same integer?

Show your work! Intuition is not enough for this problem and might be wrong. :)

Hint: [sp]This problem is based off of the Birthday problem. You can use the same method to solve it.[/sp]
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Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone

Solution (from anemone):
$\displaystyle \text{ P(at least two students select the same integer)} \ge 0.5$

$\displaystyle1-\text{P(all of the students select different integer)} \ge 0.5$

$\displaystyle 1-0.5 \ge \text{P(all of the students select different integer)}$

$\displaystyle 0.5\ge \text{P(all of the students select different integer)}$

$\displaystyle 0.5 \ge \frac{10!}{(10-n)!(10^n)} $

Now, by making a two-column table for both the values for $n$ and $\displaystyle \frac{10!}{(10-n)!(10^n)} $ and start calculating from $n=1$, we find that the smallest value of n that assures at least a 50% chance that at least two students select the same integer occurs when $n=5$.
 

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