Probability - FAQ

The chance that all 20 individuals have unique birthdays (not counting leap years) is computed as (364/365) multiplied by (363/365) and so on, leading to a total of 58.8562%. Thus, the chance of at least one pair sharing a birthday stands at 41.1438%. Interestingly, it takes just 23 people for the chances of a match to exceed 50%.

Envision those 30 individuals arranged in a line. The probability that the second person doesn’t share a birthday with the first is 364/365. If that’s the case, the chance that the next person also doesn't share a birthday with either of the first two drops to 363/365, and so on. The total probability that no one shares a birthday is (364/365) multiplied by (363/365) continuing down to (346/365), which results in 29.3684%. It’s often a point of inquiry about the minimum number of people needed for the likelihood of a birthday match to surpass 50%, and the answer is 23 individuals, where the probability of at least one match increases to 50.7297%.

1 in 635,241.

In addressing your inquiry, I've constructed a detailed matrix representing the probabilities of all possible combinations regarding the years of death for both you and the 28-year-old female. I won't delve into the specifics, but the key takeaway is that, according to calculations, one of you will pass away in approximately 41.8 years, and the other will pass around 57.3 years. These figures are rounded down—not accounting for partial years.

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I recently learned about a woman in the UK who had her first and second children born on the exact same dates as Prince George and Princess Charlotte. What do you think are the chances of that occurring? Wizard of Vegas .

To recap, Prince George was born on July 22, 2013, and Princess Charlotte arrived on May 2, 2015—this means there's a gap of 649 days. If we account for a typical nine-month gestation period, that leaves us with 379 days from George's birth to Charlotte's conception.

Based on personal observation, let us assume that the average interval between siblings is about three years. This would imply there are 825 days from the birth of the first child to the conception of the second. Applying the exponential distribution to this, I calculate that the likelihood of a gap of precisely 379 days is approximately 0.0442%.

Next, we'll assume that any woman aged between 20 and 39 is a possible candidate. According to the statistics, there are about 16,924,000 women in the United Kingdom who fall into this age bracket. If we divide this number by two to focus only on women, we arrive at 8,462,000 potential mothers in the UK.

The fertility rate in the UK, which signifies the average number of offspring born to women of childbearing age, is roughly 1.92. Using the Poisson distribution, I find that the probability of having two or more children is about 69.83%. This means the estimated number of women in the UK of childbearing age who will have two or more children is 8,462,000 multiplied by 69.83%, which equals approximately 5,909,015. Wikipedia Considering that women typically give birth at younger ages, let’s average the age of first-time mothers across the range from 20 to 37. Therefore, the estimated count of women in the UK who give birth on exactly Prince George's birthday is calculated to be 5,909,015 divided by (17 times 365), yielding about 952.32.

We have previously established that the probability of a precise age difference of 379 days between the first and second child stands at 0.0442%. Consequently, the expected number of women whose second child was born on the same day as Princess Charlotte, given that their first child shared the same birthday as Prince George, is approximately 952.32 multiplied by 0.000442, which equals roughly 0.421.

Utilizing the exponential distribution, with a mean of 0.421, the chance that at least one woman gives birth to children on the exact same days as Prince George and Princess Charlotte is 34.36%.

Additionally, I discovered that the equivalent probability for the same set of circumstances occurring in the United States is estimated to be 86.32%.

On average, how many random draws would you need from a uniform distribution ranging from 0 to 1 for their sum to equal 1?

What is the expected number of draws needed from a uniform distribution between 0 and 1 before the sum exceeds 1?

Answer:

Answer:

A solution can be found in my Math Problems site, problem 230.

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