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Birthday Problem 10/6

I did the experiment of asking 31 people their birthdays, and I bet that would be a match, was I taking a big risk?
Try out the experiment yourselves with this applet: Birthday applet


Supposing that the probability of each day is $\frac{1}{365}$ (which is not quite true), we will use the complement trick, is is easier to compute the probability of no match rather than the probability of at least two matches, in which all kinds of possibilities have to be enumerated.

\begin{displaymath}P(at\; least \;\; one\;\; match)=
1-P(no\; match)=1\times\fra...
...mes \ldots \times \frac{(365-30)}{365} \;
=\;1- 0.2695=\;0.7305\end{displaymath}

I had a good chance of winning!

The general formula for k people in the room:

\begin{displaymath}log(P(no\; match))=\sum_{i=1}^{k-1} log(1-\frac{i}{365})
\end{displaymath}

We use the approximation for x small that $log(1-x) \approx -x$, so that this log is approximately

\begin{displaymath}log(P(no\; match))\approx \sum_{i=1}^{k-1} -\frac{i}{365}=
-\frac{1}{365}\sum_{i=1}^{k-1} i=
-\frac{1}{365}\frac{k(k-1)}{2}\end{displaymath}

Giving

\begin{displaymath}P(no\; match)\approx exp(-\frac{1}{365}\frac{k(k-1)}{2})\end{displaymath}

This approximation would have given




Susan Holmes
1998-12-07