I didn't understand Birthday Paradox completely >>>> is it that in every 23 person there is a 50% chance to find a two persons have the same birthday??(the day without the year)

Guest Sep 18, 2014

#1**+8 **

Let me see.

Okay lets choose the people one at a time SO order does count.

The number of ways that the 23 people can all have different birthdays is 365C23 I think

The total number of ways 23 people can have birthdays = 365^23

The chance of finding 2 or more with the same birthday would be 1-(365C23)/(365^23)

$${\mathtt{1}}{\mathtt{\,-\,}}{\frac{{\left({\frac{{\mathtt{365}}{!}}{({\mathtt{365}}{\mathtt{\,-\,}}{\mathtt{23}}){!}}}\right)}}{{{\mathtt{365}}}^{{\mathtt{23}}}}} = {\mathtt{0.507\: \!297\: \!234\: \!323\: \!985\: \!4}}$$

which is slightly more than a 50% chance.

Now to my mathematician colleagues. I think that this is correct but can this be done using combinations instead of Permutations?

I canot think of a way

Melody
Sep 19, 2014

#1**+8 **

Best Answer

Let me see.

Okay lets choose the people one at a time SO order does count.

The number of ways that the 23 people can all have different birthdays is 365C23 I think

The total number of ways 23 people can have birthdays = 365^23

The chance of finding 2 or more with the same birthday would be 1-(365C23)/(365^23)

$${\mathtt{1}}{\mathtt{\,-\,}}{\frac{{\left({\frac{{\mathtt{365}}{!}}{({\mathtt{365}}{\mathtt{\,-\,}}{\mathtt{23}}){!}}}\right)}}{{{\mathtt{365}}}^{{\mathtt{23}}}}} = {\mathtt{0.507\: \!297\: \!234\: \!323\: \!985\: \!4}}$$

which is slightly more than a 50% chance.

Now to my mathematician colleagues. I think that this is correct but can this be done using combinations instead of Permutations?

I canot think of a way

Melody
Sep 19, 2014

#2**+5 **

First ask what's the probability that *none* of them have a birthday in common. Put the individuals in order from first to 23rd.

The probability that the 2nd person doesn't have the same birthday as the first is 364/365.

The probability that the 3rd doesn't have the same birthday as either of the first two is 363/365.

The probability that the 4th doesn't have the same birthday as any of the first three is 362/365.

...

...

The probability that the 23rd doesn't have the same birthday as any of the others is 343/365.

Therefore the overall probability that *no one* has a birthday in common is obtained by multiplying all the above probabilities together.

This is 364*363*362*...*343/365^{22 }= 364!/((365-23)!*365^{22}) = 365!/((365-23)!*365^{23})

The probability that at least two people have the same birthday is therefore 1 - 365!/((365-23)!*365^{23}) as Melody has noted.

Alan
Sep 19, 2014