Probability question... # of people needed to have your EXACT birthday?

Reply Mon 13 Nov, 2017 12:54 pm
I'm writing an IB math paper on the birthday paradox... At first, I wanted to answer the question,

How many friends would you have tomato to have a 90+% probability that for everyday of the year, you have at least one friend with a birthday ?
The answer is 2153, (saw off a math/science-y youtube channel) but I have no idea how they came to that, and I need to show the work.

discouraged, I decided to try a different question, "How many people would you need (#n) to reach a probability of 90+% that someone shares you EXACT birthday, including not just the month and day, but also the year and hour.

I know these are really complicated and I've likely bitten off more than I could chew, but is there anyone out there that can help me?? I would REALLY appreciate it!
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Reply Mon 13 Nov, 2017 02:40 pm
The simpler version of the question: "How many friends would you need to have a 90+% probability that for everyday of the year, you have at least one friend with a birthday?" If we assume (incorrectly btw*) that people are born with equal chances on any day of the year, then we can treat them as 1/365th chance of being born on any given day.

This is easier to conceptualize with the familiarity of dice models, so let's start by rolling successively more dice. If we roll two dice, then there are 6x6 = 36 possible outcomes. If we roll three dice there are 6x6x6 = 216 outcomes. If we roll four dice there are 6x6x6x6 = 1,296,000 and so on.

This pattern can be summarized as numOutcomes = numSides ^ numDice

The sum of the dice has different probabilities of occurrence; with two dice there is only one way to get an outcome of two, both dice have to be 1, so that's 1 of the 36 possible outcomes, 1/36 chance of becoming the reality we experience. But for example rolling a sum of four can happen three different ways ( {1,3}, {2,2}, {3,1} ) so 3 of the 36 outcomes total four, 3/36 chance of being four.

This is useful because we can look at the negative background and say if 3/36 chances are four then it is also the case that 33/36 chances are not four.

At the beginning we saw how the number of outcomes grew with the addition of dice, and so too how the probability of getting an outcome with the weight of only one chance shrunk. As we add more dice, the chances of everything but the example grows.

The approach to take is to work on setting the chances of not getting a particular outcome.

* It turns out that registered births are significantly less on the weekends, possibly due to doctor availability and the use of birth-delaying medications.
Reply Mon 13 Nov, 2017 02:50 pm
Thank you for your analogy, but I gotta say I still have no idea how that would pertain to my birthday problem. Specifically how would I set up the problem I previously described? I understand about the negative space subtraction...but its the setting up of the problem that is getting me. I need some more specific detailed assistance if you are able to, please!!
Reply Mon 13 Nov, 2017 09:07 pm
How many friends would you have tomato to have a 90+% probability

Red or green?


Change 365 (and adjust associated calculations accordingly) to 365 x 24 x 50 in the birthday equation and solve for n with a probability of 90%.

1/50 is an approximation to the percentage of all births that occur in each calendar year cf. 130 million current births p.a. in a total population of 7.4 billion.
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Reply Tue 14 Nov, 2017 06:29 am
1) Google "how many people are born each minute". That is a pretty good guess for how many people were born the same minute you were world wide.
2) You can assume they are all alive or use mortality tables to figure out how many have likely died.
3) Google the population of the world. The ratio of the number above divided by the population of the world shows the probability of meeting someone randomly born the same minute as you.
4) Do the calculation for 90% probability.
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Reply Fri 17 Nov, 2017 04:40 pm
Each person is a 365-sided dice that will result in one of the days.

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