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What is the chance that at least two people were born on the same day of the week if there are 3 people in the room?

I'm wondering if my solution is accurate, as my answer was different than the solution I found:

Probability that there are at least 2 people in the room born on the same day = 1 - (No one was born on the same day) - (Exactly one person was born on the same day)

There are (3 choose 2) different pairs of couples. Each couple has the same birthday as another couple with the chances of 1/7 and different with chances 6/7. Thus:

$$1 – (6/7)^3 – 3(1/7)(6/7)^2 = 0.0553$$

Thanks for any help!

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Intuitively, $$P=\frac1{7^3}\left[\binom71\binom32\binom61+\binom71\right]=0.388$$ –  wangdw May 6 '12 at 7:58
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2 Answers

up vote 3 down vote accepted

Label the three people $A,B$, and $C$. Suppose that no two were born on the same day of the week. $A$ can be born on any day of the week. The probability that $B$ was born on a different day is $\frac67$. (We are of course assuming that the seven days are equally likely, though I believe that in fact this isn't the case.) Given that $A$ and $B$ were born on different days of the week, the probability that $C$ was born on one of the remaining $5$ days of the week is $\frac57$. Thus, the probability that they were born on three different days of the week is $\frac67\cdot\frac57=\frac{30}{49}$, and the probability that at least two of them were born on the same day of the week is $1-\frac{30}{49}=\frac{19}{49}$.

You can check this as follows. There are $7^3=343$ possible ways of assigning days of the week to $A,B$, and $C$. $7\cdot6\cdot5=210$ of these result in $A,B$, and $C$ being assigned different days of the week, so the remaining $343-210=133$ assignments give at least two of the three people the same day of the week. Since all assignments are (assumed to be) equally likely, the probability of getting one of those is $\frac{133}{343}=\frac{19}{49}$.

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Another possible approach: name the guys $A,B,C$. Look at the ordered triple $d_1,d_2,d_3$ of days on which $A,B,C$ were born respectively. You have $7^3$ such triples. Now, count the "good" triples. i.e triples in which at least two guys have birthday on the same day:
1) exactly two people have birthday on the same day: choose two days of the week: $\frac{7!}{5!}=42$ (ordered) and choose the two guys that were born on the first day: $\binom{3}{2}=3$. Total of $3\cdot 42=126$ triples.
2) all the guys were born on the same day: choose the day: 7 options.
So you got $126+7=133$ good triples out of $7^3$. So the probability is $\frac{133}{7^3}\sim0.39$

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Edited. apparently $6\cdot7=42$ and not $21$ :-) –  Dennis Gulko May 6 '12 at 8:47
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