Probability - FAQ

I believe I remember reading that if there is a group of twenty people in a room the odds of two of them sharing the same birthday is less than 50/50. Is this true?

Ginny from Seattle, Washington

The probability of 20 different people all having different birthdays (ignoring leap day) is (364/365)*(363/365)*(362/365)*...*(346/365) = 58.8562%. So the probability of at least one birthday match is 41.1438%. Also, 23 is the fewest number of people needed for the probability of a match to be greater than 50%.

If you have 30 people, all born in the same 365-day calendar year, what is the probability that any two of them will have the same birthday? Please explain the formula in your response.

Scott from Madison, Indiana

Think of the 30 people as lined up. The probability the second person doesn’t match the first person is 364/365. Then, assuming they didn’t match, the probability the next person does not match either of the first two is 363/365. Then keep going one person at a time. The overall probability no two people match is (364/365)*(363/365)*...*(346/365) = 29.3684%. It is often asked what is the fewest people you need for the probability of a match to be at least 50%. The answer is that with 23 people the probability of at least one match is 50.7297%.

There are 75 multiple choice questions in an exam. Each question contains 4 possible answers only 1 is correct. The exam pass mark is 50%. What are the chances of passing the exam by guessing each answer?

Wendy from London

1 in 635,241.

Life expectancy for people of various ages has been calculated and summarized with data at the Social Security web site. However, I want to know the life expectancy of two people. Say I have two people: a thirty-year-old male (me) and a twenty-eight-year-old female (my gf). According to the chart, I will live another 46.89 years and she will live another 53.22 years. But, how long is it expected until we both are dead? How do I calculate this?


First, it would be appropriate to use cohort life tables, as opposed to the period life table you linked to. I tried to find cohort life tables online but was unsuccessful. However, we can still use the table provided. It may underestimate how long you will live slightly, because it won’t take into account future increases in life expectancy.

Answering your question involved creating a large matrix of the probability of each combination of year of death for you and the 28-year-old female. Forgive me if I don’t get into the details. The bottom line is that I show that first one of you will die in 41.8 years, and the latter death will be in 57.3 years. Both figures round down; in other words, you don’t get credit for partial years.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

What is i^i?


I don't want to just tell you the answer without giving you the opportunity to solve it yourself.

First, here is a hint to help. If you don't already know this equation, you're unlikely to solve it.

Otherwise, I admit I'm bad. Just show me the solution.

For discussion about the equation in the hint, please visit my forum at Wizard of Vegas.

I heard there was a woman in the United Kingdom who gave birth to her first and second children on exactly the same dates as Prince George and Princess Kate. What are the odds of that?


I'll have to make some rough assumptions to answer this one.

To review, Prince George was born on July 22, 2013 and Princess Charlotte on May 2, 2015. That is a difference of 649 days. If we allow for a nine-month gestation, that is 379 days between the birth of George and the conception of Charlotte.

Just going off of personal observation, let's assume the mean time between siblings is three years. That would mean 825 days between birth and conception of the next child. Using the exponential distribution, I find the probability of a difference of exactly 379 days is 0.0442%.

Next, let's assume that any female between the ages of 20 and 39 is a potential candidate. According to Wikipedia, the population of the United Kingdom in that age range is 16,924,000. Let's divide that by two to get rid of the men for 8,462,000 women of childbearing age in the UK.

The fertility rate in the United Kingdom, which is the average number of children born to each woman of childbearing age, is 1.92. Using the Poisson distribution, I find the probability of two or more children is 69.83%. So, the number of women in the UK of childbearing age who will have two or more children is 8,462,000 × 69.83% = 5,909,015.

Since women generally have children closer to age 20 than 40, let's roughly say that the mother's age at the time of the first born will be evenly distributed between the ages of 20 and 37. So, the number of women who will have their first child in the UK on exactly the birthday of Prince George is 5,909,015/(17×365) = 952.32.

We already established that the probability of an exact age difference between the first and second child of 379 days is 0.0442%. Thus, the expected number of women who had their second child on exactly the same day Princess Charlotte was born, who already had their first born on the exact day Prince George was born, is 952.32 × 0.000442 = 0.421.

Using the exponential distribution, given a mean of 0.421, the probability of at least one woman having her first and second children on the exact same days as Prince George and Princess Charlotte were born is 34.36%.

By the way, I find the probability of the same thing in the United States is 86.32%.

What is the average number of draws you would need from a uniform distribution from 0 to 1 for the sum to equal 1?



The answer is e = 2.7182818....


Here is the solution.

What is the expected number of random numbers drawn from the uniform distribution between 0 and 1 for the sum to exceed 1?



Here is solution.

Two players, Sam and Dan, each have five coins. Both must choose to place one to five coins in his hand. At the same time, each must reveal the number of coins played. If both choose the same number of coins, then Sam will win collect all coins played. If both choose different numbers of coins, then Dan will collect all coins played. Assuming both players are prefect logicians, what is the optimal strategy for Dan?


Dan should randomize his strategy as follows:

  • Probability of picking one coin = 77/548.
  • Probability of picking one coin = 107/548.
  • Probability of picking one coin = 117/548.
  • Probability of picking one coin = 122/548.
  • Probability of picking one coin = 125/548.

With this strategy, Dan can expect to win 3.640510949 coins every turn, regardless of how many coins Sam picks.

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

A related question, which led to this one, can be found in my forum at Wizard of Vegas.