Slightly more than 1 in 2.
1:30
The probability of two people's birthday being the same is actually more likely than many would think. The key thing is to note that it doesn't matter what the first person's birthday is. All we need to work out is the probability that the second person has a birthday on any specific day. This probability is 1/365.25 The probability that they were born on June 10th is 1/365.25. The probability that they were born on February 2nd is 1/365.25 and the probability that they were born on the same day as you is 1/365.25
The probability, over presidents of all organisations, through all of time, is 1.
To select the first birthday, the probability is 1/30. Having gotten that, the conditional probability that the next birthday would be the same is (1/30)x(1/29) and that is 1/870----------------------------------------------------------------------------------------------I believe the question has to be rephrased to "What is the probability that two peoplein a group of 30 people share the same birthday?". Because in the way the questionis actually stated, "the probability that two persons selected randomly from a group of 30 have the same birthday", the event that "those two people would share theirbirthday" is independent of the size of the population they were selected from.In the case the actual question be "What is the probability that two people in a groupof 30 share the same birthday?, is given by the following expression that neglectsFebruary 29 (of the leap), but gives very good approximation to the expression thatconsiders February 29 and is a simpler one. [It has to be mentioned that the analysisleading to this expression considers birthdays a "random variable" where chances fora persons birthday are the same for any day of the year]:P(2 share bd out of n) = nC2 (1/365) Π1n-1[1-(i-1)/365]for n = 30, P(2 share bd out of 30) = 30C2 (1/365) Π1 29 [1-(i-1)/365] = 435∙(1/365)∙[1-1/365]∙[1-2/365]∙[1-3/365]∙ ∙∙∙ ∙[1-28/365] = 0.380215577... ≈ 0.380 ≈ 38.0%For the construction of the expression to calculate the probability of any two people sharing a birthday in a group of n people considering Feb 29 of the leap year see thequestion "What is the probability that in a room of 8 people 2 have the same birthday?"
If you assume that birth dates are uniformly distributed over the year (they are not), and you ignore leap years, then the probability of two people selected at random, share a birthday is 1/365.
Leaving aside leap years, the probability is 0.0137
1/365 = 0.00274
1:30
The probability of two people's birthday being the same is actually more likely than many would think. The key thing is to note that it doesn't matter what the first person's birthday is. All we need to work out is the probability that the second person has a birthday on any specific day. This probability is 1/365.25 The probability that they were born on June 10th is 1/365.25. The probability that they were born on February 2nd is 1/365.25 and the probability that they were born on the same day as you is 1/365.25
The probability, over presidents of all organisations, through all of time, is 1.
To select the first birthday, the probability is 1/30. Having gotten that, the conditional probability that the next birthday would be the same is (1/30)x(1/29) and that is 1/870----------------------------------------------------------------------------------------------I believe the question has to be rephrased to "What is the probability that two peoplein a group of 30 people share the same birthday?". Because in the way the questionis actually stated, "the probability that two persons selected randomly from a group of 30 have the same birthday", the event that "those two people would share theirbirthday" is independent of the size of the population they were selected from.In the case the actual question be "What is the probability that two people in a groupof 30 share the same birthday?, is given by the following expression that neglectsFebruary 29 (of the leap), but gives very good approximation to the expression thatconsiders February 29 and is a simpler one. [It has to be mentioned that the analysisleading to this expression considers birthdays a "random variable" where chances fora persons birthday are the same for any day of the year]:P(2 share bd out of n) = nC2 (1/365) Π1n-1[1-(i-1)/365]for n = 30, P(2 share bd out of 30) = 30C2 (1/365) Π1 29 [1-(i-1)/365] = 435∙(1/365)∙[1-1/365]∙[1-2/365]∙[1-3/365]∙ ∙∙∙ ∙[1-28/365] = 0.380215577... ≈ 0.380 ≈ 38.0%For the construction of the expression to calculate the probability of any two people sharing a birthday in a group of n people considering Feb 29 of the leap year see thequestion "What is the probability that in a room of 8 people 2 have the same birthday?"
The probability that 25 random people don't ALL share the same birthday is: 1 - (1/365)**24, or about 0.999999999999999999999999999999999999999999999999999999999999968 However, I suspect you meant to ask "What is the probability that 25 random people all have different birthdays?" That is: 1 * (364/365) * (363/365) * (362/365) * ... * (342/365) * (341/365) = 0.4313 So about 43% of the time nobody will share a birthday, and 57% of the time, two or more people will share a birthday.
The probability of at least 2 people in a group of npeople sharing a common birthday can be expressed more easily (mathematically) as 1 minus the probability that nobody in the group shares a birthday. Consider two people. The probability that they don't have a common birthday is 365/365 x 364/365. So the probability that they do share a birthday is 1-(365/365 x 364/365) = 1-365x364/3652 Now consider 3 people. The probability that at least 2 share a common birthday is 1-365x364x363/3653 And so on so that the probability that at least 2 people in a group of n people having the same birthday = 1-(365x363x363x...x365-n+1)/365n = 1-365!/[ (365-n)! x 365n ]In the case of 12 people this equates to 0.16702 (or 16.7%).
23. The probability that at least two people in a room share a birthday can be expressed more simply, mathematically, as 1 minus the probability that nobody in the room shares a birthday.Imagine a fairly simple example of a room with only three people. The probability that any two share a birthday is :1 - [ 365/365 x 364/365 x 363/365]i.e. 1-P(none of them share a birthday)=1 - [ (365x364x363) / 3653 ]=0.8%Similarly,P(any two share a birthday in a room of 4 people)= 1 - [ 365x364x363x362 / 3654 ] = 1.6%If you keep following that logic eventually you getP(any two share a birthday in a room of 23 people)=1 - [(365x364x...x344x343) / 36523 ] = 51%
It depends on how big the class is.
If you assume that birth dates are uniformly distributed over the year (they are not), and you ignore leap years, then the probability of two people selected at random, share a birthday is 1/365.
A birthday attack is a method of code decryption which exploits the birthday paradox - that which explains that within a class of 30 students, there is an assumed probability of two sharing the same birthday of 70 percent.