4

Assuming 30-day months, given 10 people in a room. What are the chances that 5 or more people are all born on the same day#? (i.e., 5 born on the 28th, or 5 born on the 6th, etc)

(EDIT: changed from chances of 5 to chances of 5 or more)

I have tried two answers so far.

In the first, you pick any person, and see what the chances are of the other 9, then 8, etc to match the first. This seems to be 10 * 9/30 * 8/30 * 7/30 * 6/30.

In the second, I suppose you could calculate the chances of 5 of the 10 having a birthday on day 1 + the chances of 5 having a birthday on day 2, etc.

These answers seem quite different. What do you all think?

joeslice
  • 143

5 Answers5

1

You use the Binomial distribution formula here

p = 1/30, q = 1-p = 29/30

Probability 5 born on the same day

$$P(X=5) = {10 \choose 5}p^5q^5=8.75*10^{-6}$$

Yau Kin Hoe
  • 299
  • That only gives the probability that exactly five are born on (say) the first. The OP wants the probability that five are born on any one of $30$ days. – Brian Tung May 12 '15 at 23:16
  • didn't he say "What are the chances that 5 people are all born on the same day#?" meaning not 6 not 7 but just 5? – Yau Kin Hoe May 12 '15 at 23:17
  • Fair enough—but one still has to account for all multiple days, and also for the possibility that five are born on one day, and five are born on another day. I think it's unclear whether OP meant exactly five or at least five, so I'll add to my answer accounting for both possibilities. – Brian Tung May 12 '15 at 23:19
  • I think when you have 5 born in 1 day, it would be considered 1 event. It does not matter whether the other 5 are born on a separate same day. Let me get my numerical result and compare it with you. (edit: yes we have the same answer) – Yau Kin Hoe May 12 '15 at 23:27
  • Your expression and mine are calculating two different things. You don't get into that issue unless you try to expand it to cover all the days of the month. (The last expression in my answer is, however, equivalent to yours.) – Brian Tung May 12 '15 at 23:29
  • You don't have to cover all the days of the month. I can explain it to you via stackexchange chat if you are still not convinced. – Yau Kin Hoe May 12 '15 at 23:33
  • Let us continue this discussion in chat. – Yau Kin Hoe May 12 '15 at 23:34
  • Probability 5 born on the same day is just (1/30)^5 – Hatim Nov 28 '17 at 04:17
1

The probability that at least five people were born on the first of the month (assuming uniform distribution) is

$$ p_1 = \sum_{k=5}^{10} \binom{10}{k} \left(\frac{1}{30}\right)^k \left(\frac{29}{30}\right)^{10-k} = \frac{886717229}{98415000000000} \doteq 0.0000090100 $$

The probability that at least five people were born on any of the thirty days of the month is almost

$$ 30p_1 = \frac{886717229}{3280500000000} \doteq 0.00027030 $$

However, this double-counts those cases where two sets of five people are born on two different days of the month. That happens with probability

$$ p_d = \frac{\binom{30}{2}\binom{10}{5}}{30^{10}} = \frac{203}{1093500000000} \doteq 0.00000000018564 $$

So the final probability is

$$ 30p_1-p_d = \frac{44335831}{164025000000} \doteq 0.00027030 $$

There is a difference made by $p_d$, but it's too small to show up in five significant digits.

ETA: If you want the probability that exactly five are born on the same day, then we'd set

$$ p_1 = \binom{10}{5} \frac{29^5}{30^{10}} = \frac{143578043}{16402500000000} \doteq 0.0000087534 $$

and then proceed as before.

Brian Tung
  • 34,160
  • Interesting. That's very close to what I have, but not exactly, and it suggests that we're subtracting different values from the basic value of $30p_1$. I wonder what the discrepancy is. – Brian Tung May 12 '15 at 23:58
  • Loss of precision isn't the issue; I evaluated your expression to the same value you did. If I have time I'll look deeper into it. – Brian Tung May 13 '15 at 00:07
  • It sure doesn't (format well, that is). But I figured it out OK. – Brian Tung May 13 '15 at 00:07
  • I found my mistake, I was slightly undercounting. When I added the missing term our answers became identical to every place. $0.00027029941441853377$ – Gregory Grant May 13 '15 at 00:33
  • Hunh! What's with the downvoting, Unknown Reader? – Brian Tung May 13 '15 at 02:07
  • I don't know who downvoted your answer, but that was really rude. I upvoted it myself. There must be some neg-happy degenerates around here. Maybe voting records should be public. – Gregory Grant May 14 '15 at 18:20
  • Sorry, I meant to say both answers are "0.0002702992287761012", I would edit my comment but the edit period has expired. – Gregory Grant May 14 '15 at 18:29
  • Re downvoting: No, I would not personally advocate for public voting records. I'm grousing about being downvoted without really wanting to out someone. :-) – Brian Tung May 14 '15 at 18:31
  • Well it is frustrating and happens to me too. Maybe they can develop an algorithm that analyzes the voting record and over time automatically detects such vandals and punishes them accordingly. – Gregory Grant May 14 '15 at 18:33
1

Suppose the max number of the ten who share one birthday is $n$. Then $n=5,6,7,8,9,10$ constitute a partition of the event of interest. To count the number of ways for fixed $n$, we first choose $n$ of the ten people to have one birthday, you can do that in ${10}\choose{n}$ ways. Those $n$ can be assigned one birthday in $30$ ways. The remaining $10-n$ can be assigned to other birthdays in $(29)^{10-n}$ possible ways (remember we're counting the number of ways to have exactly $n$ different days among the $10$ people). But we have to avoid, when $n=5$, the case that we assign the other $10-n=5$ people to one birthday (otherwise we will double count it). So for $n=5$ we have to correct by subtracting $29$ possibilities and then adding back all ways to have five for one day and the other five for another day, which happens in ${10\choose5}{30\choose2}$ ways.

Thus there are $${10\choose5}(30)((29)^{5}-29)+{10\choose5}{30\choose2}+{10\choose6}(30)(29)^{4}+{10\choose7}(30)(29)^{3}+{10\choose8}(30)(29)^{2}+{10\choose9}(30)(29)+{10\choose10}(30)$$ ways to assign ten people to birthdays so that five share one day. So just divide this by $30^{10}$ to get the probability.

This gives $0.0002702992287761012$ which agrees perfectly with Brian Tung's answer.

Gregory Grant
  • 14,874
1

Let us first calculate the chance that exactly five people share a birthday. There are ${10 \choose 5}=252$ ways to choose the people who will share a birthday, $30$ ways to choose the birthday, and $29^5$ ways to choose the birthdays of the other people, giving $\frac{252\cdot 30 \cdot 29^5}{30^{10}}\approx 0.000262$. This is not quite right because we have double counted the cases where there are two groups of five in the room, but that correction will be very small. Since the chance of five is small the chance of six is smaller yet, so I would just ignore it.

Ross Millikan
  • 374,822
0

it's simpler than you can imagine it's the probability that 5 were born on the same day (1/30 ^ 5) plus the probability that 6 were born on the same day (1/30 ^ 6) ... until you reach 10

p(5 <= x <= 10) = ((1/30)^5) + ((1/30)^6) + ((1/30)^7) + ((1/30)^8) + ((1/30)^9) + ((1/30)^10)
= 4.25713069e-8
Hatim
  • 103