The 23 people in a room, there will be

The Birthday ParadoxThe Birthday Paradox is a probability theory which concerns the probability of one pair in a set of randomly selected people sharing a birthday; the theory states that if there are 23 people in a room, there will be a 50% chance of one pair sharing a birthday. What first sparked my interest in this topic has been there my whole life, I share a birthday with my mother. When I tell people, they always think it’s really cool and rare and this has resulted in me developing an interest in probability and birthdays. I aim to investigate the birthday paradox through calculations, to investigate it in real life as well as apply it to an average Swedish family to see the likelihood of my own situation occurring. To investigate this, I will use the following softwares/devices:Texas Instruments 84 plus (Graphic calculator): I will use this for calculations.Desmos (Graphing software): I will use this to create visuals of functions which are relevant to this theory.LucidChart (Diagram software): To create probability tree. RealTimeBoard (Diagram software): To create the family treeGoogle Forms: To create a survey and gather dataWe have to keep in mind that the birthday paradox does not consider birthdays that occur on February 29th. Understanding the Birthday ParadoxIn order to find out how many people there has to be in a room for there to be a 50% chance of two people sharing a birthday, we must think of the amount of people in terms of possible pairs. The table below exemplifies this:Amount of peoplePossible pairsNumber of possible pairs2 people(1, 2)13 people(1, 2) (1, 3) (2, 3)34 people (1, 2) (1, 3) (1, 4) (2, 3) (2, 4) (3, 4)6To derive the number of pairs when N people are present, the following formula may be used: Possible pairs when N people are present= 0.5(N2-N)We then have to calculate the probability that given the number of pairs present (Represented by the letter P), what is the probability that at least one pair shares a birthday (Represented by the letter B)? This can be expressed as the following:P(B| P)Although this method is correct, it is easier to calculate the opposite which would be the probability of no people sharing a birthday, this can be done through the following formula:P(A)=Number of favourable outcomes (to A)Number of possible outcomesIf we apply the formula to the situation, the probability of no people sharing a birthday we apply the values below: P(No people share a birthday)=364365The number of favourable outcomes is 364 since it is the all the birthdays except the one (so there can’t be two birthdays on the same day) and the total amount of birthdays is 365. We then raise 365 to the power of P i.e. 365P. This mimics what we would see in a probability tree, however, creating a probability tree for this specific event would be very hard since there are so many outcomes. In the example below, I have used the colours blue, green and red as the possible outcomes. The events are equally probable i.e.:P(Blue)=13, P(Green)=13 and P(Red)=13The events are independent, this is to more accurately demonstrate the probabilities involved in the birthday paradox as they are independent as well. If you wanted to calculate P(Blue, Blue), you would calculate P(Blue) P(Blue)i.e. P(Blue)2 which is the same as: (13)2=19Returning to the birthday paradox, the formula would be: P(No people sharing a birthday in P pairs present)=(364365)PTo simplify this in terms of focusing on the amount of people present rather than pairs, we can insert the following formula: Possible pairs when N people are present= 0.5(N2-N)When applied, we can calculate the probability of no people sharing a birthday in a group of N people. P(No pair sharing a birthday when N people are present)=(364365)(0.5(N2-N))Convert the formula so it calculates the probability of one pair sharing a birthday in a group of N people. We do this through subtracting the current formula from 1, this is because the probability of these events combined is 1, there are only 2 possible outcomes which are: you share a birthday or you do not so if we subtract the probability of no pair sharing a birthday we will get the probability of one pair sharing a birthday. P(One pair sharing a birthday when N people are present)=1-(364365)(0.5(N2-N))The aim of the birthday paradox is to investigate how many people have to be present for there to be a 50% chance of two people sharing a birthday. We can apply this value to the formula to find the amount of people required. 0.5=1-(364365)0.5(N2-N)We have to remember that this is used to calculate the probability of no people sharing a birthday. Our goal while calculating this will be to isolate N.We know that: 0.5=12 and, using simplification, we know that:1-(364365)0.5(N2-N)=1-2N2-N910.5(N2-N)3650.5(N2-N).We can equate them to each other so we get: 0.5=1-2(N2-N)910.5(N2-N)3650.5(N2-N)To solve for N, we start by simplifying the following section of the calculation: 2(N2-N)910.5(N2-N)3650.5(N2-N)=eln(2N2-N)eln(910.5(N2-N))eln(3650.5(N2-N))=e(N2-N)ln(2)e0.5(N2-N)ln(91)e0.5(N2-N)ln(365)=e(N2-N)ln(2)+0.5(N2-N)ln(91)+0.5(N2-N)ln(365)When we apply the simplification to the calculation we get the following:0.5=1-eln(2)(N2-N)+0.5ln(91)(N2-N)+0.5ln(365)(N2-N)Subtract 1 from both sides:-eln(2)(N2-N)+0.5ln(91)(N2-N)+0.5ln(365)(N2-N)=-0.5Take the natural logarithm of both sides:ln(2)(N2-N)+0.5ln(91)(N2-N)+0.5ln(365)(N2-N)=-ln(2)Expand and collect in terms of N:(-ln(2)-ln(91)2+ln(365)2)N+(ln(2)+ln(91)2-ln(365)2)N2=-ln(2)(-ln(2)-(ln(91)/2)-(ln(365)/2))N+(ln(2)+(ln(91)/2)+(ln(365)/2)N^2=-ln(2)Divide both sides by: ln(2)+ln(91)2-ln(365)2:(-ln(2)-ln(91)2+ln(365)2)Nln(2)+ln(91)2-ln(365)2+N2=-ln(2)ln(2)+ln(91)2-ln(365)2Add (-ln(2)-ln(91)2+ln(365)2)24(ln(2)+ln(91)2-ln(365)2)2 to both sides:(-ln(2)-ln(91)2+ln(365)2)Nln(2)+ln(91)2-ln(365)2+N2+(-ln(2)-ln(91)2+ln(365)2)24(ln(2)+ln(91)2-ln(365)2)2=(-ln(2)-ln(91)2+ln(365)2)24(ln(2)+ln(91)2-ln(365)2)2-ln(2)ln(2)+ln(91)2-ln(365)2Square the left side of the equation:(N+-ln(2)-ln(91)2+ln(365)22(ln(2)+ln(91)2-ln(365)2))2=(-ln(2)-ln(91)2+ln(365)2)24(ln(2)+ln(91)2-ln(365)2)2-ln(2)ln(2)+ln(91)2-ln(365)2Take the square root of both sides: N+-ln(2)-ln(91)2+ln(365)22(ln(2)+ln(91)2-ln(365)2)=((-ln(2)-ln(91)2+ln(365)2)24(ln(2)+ln(91)2-ln(365)2)2-ln(2)ln(2)+ln(91)2-ln(365)2)Add -ln(2)-ln(91)2+ln(365)2)22(ln(2)+ln(91)2-ln(365)2)to both sides:N=((-ln(2)-ln(91)2+ln(365)2)24(ln(2)+ln(91)2-ln(365)2)2-ln(2)ln(2)+ln(91)2-ln(365)2)–ln(2)-ln(91)2+ln(365)2)2(ln(2)+ln(91)2-ln(365)2)With decimal values applied we get the following: N=(3.54906628710-71.41962651510-6-0.3010299957-5.9574040410-4)-5.95740403810-4-0.0011914808N=(0.25–505.3039775)–0.5000000032N=505.5539775+0.5000000032N=±22.4845275168+0.5000000032N cannot be a negative value, hence:N=22.9845275223When there are 23 people present, there is a 50% chance of one pair sharing a birthday which is what the paradox states. In the future it would be interesting to investigate different ways of calculating this to develop my mathematical skills. One method I considered using to calculate this was quadratics, it began similarly to the calculation used but later changed and you had to find the “a”, “b” and “c” values. However, I found that this became very difficult due to the values it came down to containing several decimals.  A Visual Representation of the Birthday ParadoxThe graph above (figure 1) demonstrates the birthday paradox, it shows that regardless of whether you calculate the probability of no pair sharing a birthday or the probability of one pair sharing a birthday, you will get the same results as the answer is at the intersection of the two curves (23, 0.5).The two curves are:The probability of no pair sharing a birthday in the format:Y=(364365)0.5(x2-x)AndThe probability of one pair sharing a birthday in the format:Y=1-(364365)0.5(x2-x)The Birthday Paradox in real lifeI issued a survey on a group of IB students where I asked people when their birthdays were. I have made a summary of the answers which is figure 2. I received 27 responses. If we apply this value to the formula we can calculate the probability of two people sharing a birthday in a group where 27 people are present.P(One pair sharing a birthday when 27 people are present)=1-(364365)(0.5(272-27)The probability of one pair sharing a birthday is:1-(364365)(0.5(272-27))=0.61824016362%We can see that 2 people share a birthday on December 7th, the birthday paradox is supported by the data from my survey. In the future, it would be interesting to investigate this on an even larger scale, perhaps with people born at different times to compare and see how the birthday paradox could be more or less common. It would also be good to use a larger group of people as this would create a more representative sample where the results’ applicability would increase. The Familial Applications of the Birthday ParadoxI am going to be investigating the familial applications of the birthday paradox in a biological extended family. I will investigate the probability of sharing a birthday with an individual’s mum given that the individual shares a birthday with someone in their extended family. It is important to remember that this only concerns biological families as those are the statistics I will use to calculate it, this is due to a lack of resources since I could not find a lot of information concerning adopted children, step-parents, etc. in Sweden.First, I have to define what an average Swedish extended family is. I could not find specific data concerning this so I had to conclude one from other information I gathered. In Sweden, the average amount of children born per woman is 1.94. Since a man is required to conceive a child, there must be a biological family, hence, we can conclude that the average immediate Swedish family consists of 2 biological parents and 2 biological children. If we apply this to an extended family where the furthest back we go are grandparents (As those are usually the oldest relatives you have) we are able to conclude a family tree which looks like the one below2:For clarification, G1, G2 or G3 represents the generation the individual is in and the numbers are there to distinguish between the individuals. Finding the probability P(L | B) L represents the probability of someone being one of your parents.B represents the probability of a specific individual sharing a birthday with someone in their extended family. We will calculate P(L | B) from the perspective of a G3 individual, this will be G3#4. If we were to consider this in terms of my life, this would be the equivalent of me calculating the probability of me sharing a birthday with one of my parents. P(L)and P(B)are independent events which means that we use the following formula to calculate P(L | B):P(L | B)=P(L)P(B)This means that P(L | B)can be expressed as P(B | L)since the events are independent. Before we calculate this, we have to find the values of P(B)andP(L).Finding P(B)We will calculate P(B)through using the formula we developed earlier:P(One pair sharing a birthday when N people are present)=1-(364365)(0.5(N2-N))In this formula, N will be 15 as that is the amount of people in the extended family excluding the individual of interest i.e. G3#4. 1-(364365)(0.5(152-15))=0.2502879086Finding P(L)We will use the following formula to calculate this: P(A)=Number of favourable outcomes (to A)Number of possible outcomesSince all children have 2 biological parents the number of outcomes favourable to L will be 2, the total number of possible outcomes will be 15 since that is the amount of people in this extended family which the parents will obviously be part of. Hence:P(L)=215So, if we apply these values, we will get the following:0.2502879086215=0.03337172113%Thus, if I was to share a birthday with anyone in my extended family, there is a slightly higher than 3% chance that it would be one of my parents. Since I only have 2 parents, we can multiply this by 0.5 in order to see the probability of me sharing a birthday with specifically my father or mother:0.03337172110.5=0.01668586062%This means that given that I share a birthday with someone in my extended family, the probability of that being my mother is slightly less than 2%. The probability tree, which we investigated earlier, below essentially demonstrates what i just calculated: If we follow the multiplication process suggested by the probability tree we should get approximately 2%.0.250287908621512=0.01668586062%Hence, the probability of me sharing a birthday with my mum (given that I share a birthday with someone in my extended family) is 2%. In the future it would be interesting to investigate this in larger terms or investigating several different possible shared birthdays in an extended family, it would also be interesting to apply this to different countries in order to see how the probability is affected by different statistics. Conclusively, I demonstrated the birthday paradox and showed how it worked (mathematically) which I found very interesting, I also found a real life example of the birthday paradox occurring among IB students. From this information,  I can conclude that the birthday paradox is more prominent in real life than I and others expected. Additionally, I calculated the probability of someone sharing a birthday with their mum given that they share a birthday with someone in their extended family, I found this probability to be very low which was exciting.