Resolve this probability "paradox"

Factual Questions
1

I looked at what I thought was a trivial probability question, and got a quick answer. Then I did a more rigorous approach and got a different answer. I’m pretty sure my second answer is correct but can’t figure out what’s wrong with the reasoning of the first answer.

Given a day of the week, find the probability that at least one of two randomly chosen people was born on that day of the week (assuming that births are distributed equally across days of the week).

The chance of a single person being born on that day would be 1/7. If you have two people, I leaped to the conclusion that the chance of at least one of them being born on that day was 1/7 + 1/7 = 2/7. But when I extended this to seven people, I realized the chances of at least one person out of seven being born on that day would be 1, which isn’t right (it’s never exactly 1 no matter how many people you include).

For two people, there are 7 x 7 = 49 permutations for them to be born on particular days of the week. Given a particular day, 13 of those permutations will include that day (one permutation is when they are both born on that day, 6 where one of them is born on that day, and another 6 where the other one is born on that day). So the chance of at least one person of two being born on a given day of the week is 13/49.

A heuristic of probability is that the probability of at least one of two independent events happening is the sum of the probability of the two events, and the probability of both events happening is the product of the probability of two events. But that doesn’t work here.

I am certain that I am missing something that will turn out to be embarrassingly obvious, but what is it?

2

I think that’s where you go wrong, but I can’t remember my probability classes from school! But isn’t the pool of people being reduced with each subsequent person (since you’ve eliminated the first) or something like that? Sorry, I realy can’t remember; I’m sure someone will be alopng imminently who knows how it works.

3

This is not correct. The probably of at least one of two independent events happening is: 1 - (probability of neither event happening), i.e.

1- [(6/7)*(6/7)] = 13/49

Alternatively, given PersonA and PersonB,

P (PersonA was born on that weekday, PersonB wasn’t) = (1/7)(6/7) = 6/49
P (Person A wasn’t born on that weekday, PersonB was) = (6/7)(1/7) = 6/49
P (Both people were born on that weekday) = (1/7)*(1/7) = 1/49

P (at least one was born on that weekday) = 6/49 + 6/49 + 1/49 = 13/49

4

You are double-counting the event of both people being born on the same day.

The event “A is born on Monday and B is born on Monday” is the same as the event “B is born on Monday and A is born on Monday”.

If you’ve got two events that are statistically independent (chance of one happening doesn’t impact the chance of the other happening, which also means that the events are not mutually exclusive), then the probability of “A or B or both” = P(A)+P(B)-P(A and B).

In your example, that’s 1/7 + 1/7 - (1/7 * 1/7) = 13/49.

5

So, you’re looking for the probability that at least one of the two people were born on that day. We could do this two ways.

First, the hard way:
The probability that at least one of the two people were born on that day is the probability that just the first person was born on that day or just the second person was born on that day or both were born on that day.

Since these are disjoint, we can just add the three. Since the birthday of each person is independent, we can multiply them. So, just the first person = (1/7)(6/7). Just the second person = (6/7)(1/7) and both = (1/7)(1/7). Add them together to get 6/49+6/49+1/49 = 13/49.

The easy way:
The probability that at least one person was born on that day is 1 - the probability that neither was born on that day. Again, they’re independent, so the probability of neither being born on that day is (6/7)(6/7), so the probability of at least one being born is 1 - (36/49) = 13/49.

Where you went wrong in your first attempt is that the two events are not disjoint. That is, one person being born on the day doesn’t preclude the other person also from being born on that day.

6

You can’t add those types of probabilities like that because you encounter the result you found and it obviously isn’t right.

Here is the full answer if you really want to understand it:

7

Easiest way to do this: Either at least one of them was born on that day, or neither of them was. So, the probabilities of those two events are complementary (sum to 100%). The probability neither of them was born on the given day, assuming that their birth dates are known to be independent (that you know, for example, that they’re not each other’s twin siblings) is the product of the probabilities of each one not being born on that day. This is 6/7; so the probability that neither of them was born on the given day is 36/49, and the probability that at least one was is 13/49, a shade less than the 2/7 that you originally estimated.

8

That’s a reasonable approximation when the probabilities are low. For instance, if I buy two lottery tickets, both with randomly-determined numbers, it’s a good approximation to say that my chances of winning have doubled, since there’s such a low chance of them both being the same. Even for your days-of-the-week example, 1/7 is a reasonably small probability, so it still gets you into the right ballpark (14/49, instead of the correct 13/49). But when you go to the case of 7 people, the answer you get is 7*1/7, or 1, which is not a small probability, so the approximation fails horribly there.

9

Reminds me of the game I’ve seen at carnivals “Crown and Anchor” (I believe it is sometimes called Chuck-o-Luck too). In that game, you have three dice each with six sides. Sometimes numbers are used, sometimes symbols.

The three dice are thrown. Let’s say I bet on “6”. If one dice lands on a six, I get $1 (i.e., I become a dollar richer). Since there are three dice and each has a 1/6 chance, those should be fair odds (1/6 + 1/6 + 1/6 = 1/2). But, it’s even better than that! If two “6” show up, I get 2:1 payment, and if all dice show up as “6”, I get 3:1 payment! (Each payment assumes I keep my original $1 bet if I win).

Of course, by this thinking, if you have six dice, one should always land on a “6”, but we know that’s not true.

We need to first calculate out having one and only one of the three dice showing a six. If die #1 shows a six, the other two dice can only show 25 possible combinations (we are leaving out the possibility of them showing a six for now). That’s only 75:216 possible winning combinations which is pretty close to 1:3 while you’re being paid as if it is a 1:2 possibility.

Next, we can calculate out the possibilities of two dice landing on a “6”. There are three possibilities of which two dice can show a six (1st and 2nd, 2nd and 3rd, and 1st and 3rd). And, there are five possible combinations of each of those or 15:216 in total. Yet, you’re being paid as if this is a 3:1 possibility. And of course, there’s only one out of a possible 216 possibility that all three dice will land on a six. Yet, you’re being paid as if it was a 4:1 possibility.

Pretty much, there are only 91 possible winning combinations. If you played 216 times, and each possible roll came up, you’d have paid $216, yet only would win $198.

10

This is a case of what’s known as the inclusion-exclusion principle.

11

Or to put it another way, there is a 5/6 chance that a die will NOT roll a six, so the probability that none of three dice will roll six is (5/6)^3, or 125/216, slightly more than half the time. This is one of many types of bet that are based on seeming to have 50/50 odds while actually being just a bit in the house’s favor. Not a lot but enough that the more you play the more you lose.

12

Actually, compared to “honest” gambling like in a casino, those odds are a lot more than “just a bit” in the house’s favor. Most games, the house edge is only 2-3%, and some it’s less than 1%, but that’s over 8%.