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The Odds Gods smile on birthday/card matches

The classic birthday problem asks how many people are required to ensure a greater than 50% chance of having at least one birthday match, meaning that two or more people share a birthday. The surprisingly small answer, assuming that all birthdays are equally likely and ignoring leap years like 2012, is 23 people.

The standard argument goes like this:

\[ \begin{align}
\operatorname{Prob}(\textrm{at least one match}) &= 1 – \operatorname{Prob}(\textrm{no match}) \\
&= 1 – \operatorname{Prob}(\textrm{all 23 birthdays are different}) \\
&= 1 – \frac{365}{365} \times \frac{364}{365} \times \frac{363}{365} \times \dots{} \times \frac{343}{365}
\end{align} \]

which comes out a tad over 0.5, or 50%.

The same assumptions and arguments show that with 60 people selected at random one is over 99% likely (“almost certain”) to get a match.

The MathWorld page on the birthday problem mentions that with only 14 people selected at random, one has a (slightly) greater than 50% chance of getting an “almost match” — namely birthdays within a day of each other.

About twenty years ago, Persi Diaconis & Susan Holmes showed that if one takes into account that in modern times a significant proportion of births are induced, and tend not to be scheduled at weekends, then the resulting Mon-Fri clustering throws off the uniform distribution. This “lumpy logic” leads to an even higher chance of birthday coincidences.

For instance, with merely 16 or 17 people selected at random, one has a (slightly) greater than 50% chance of getting a birthday match under this distribution. And the more the merrier: with 25 people the chance of getting a match is notably higher than it was under the unrealistic uniform distribution assumption.

Now let’s switch attention from birthdays to cards, namely to a standard deck of 52, which has been thoroughly shuffled.

How many cards need we pick at random so that there is a greater than 50% chance of getting at least one pair with the same value?

Once again, let’s turn things around, and focus on the chances of there being no match.

If $k$ cards are picked at random from a full deck, where $3 \leq k \leq 13$, then since there are four cards of each value, it follows that:

\[ \operatorname{Prob}(\textrm{at least one match}) = \large 1 – \frac{52}{52} \times \frac{48}{51} \times \frac{44}{50} \times \dots{} \times \frac{52-4(k+1)}{52-k+1}. \]

Plugging in various values for $k$, we find that we need to pick at least 6 cards to be at least 50% sure of a card value match. Given 8 or 9 cards, there is a high probability (89% or 95%) of a match, and with 10 cards, it’s very likely indeed (a 98% chance) to occur.

Try it out with ten randomly chosen cards; it’s amazing how often one gets a lot more than a matching pair. Pairs of pairs, or even triple pairs, or three of a kind, are far from unusual.

We can take advantage of this card match principle in two ways, using different procedures, when dividing up 10 random cards into two poker hands in seemingly fair ways.

That’s exactly what we did in the June 2006 Card Colm at MAA Online: Better Poker Hands Guaranteed


3 Responses to “The Odds Gods smile on birthday/card matches”

  1. Mike Williams

    During one of my undergraduate mathematics lectures, our professor decided to teach us about this. Despite there being ~300 people in the lecture theatre, there was not a single pair with the same birthday. The mounting disbelief from the poor professor as he went through the days of the year was hilarious.

  2. Kevoboyle

    I like the induction variation. There’s a problem, however. This would only hold true if all subjects were born in the same year, as in a class in school for example. Weekend dates would be different every year and have no bearing if subjects were different ages.

  3. Kevoboyle

    Just scribbled a few calculations. Even if we assume that every one of the 17 people born in a year were born Mon – Fri, it’s still only 41.3% chance of a shared birthday. In other words, I’ll have a fiver on no matches for that one if anyone’s offering.


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