Probability debate / experiment

[mortgageman]

I think I have it….and used Excel to get there!!

I created a 25 x 25 grid and in each cell entered “=1/365” being the probability that each person will share a birthday with a specific other person. (obviously cells which refer to a person sharing with themselves is left blank).

Summing each row will give you 24/365 which is the probability of one person sharing a birthday with anyone else in the room.

Summing the whole 25 x 25 grid gives you 25(24/365) or 600/365 which needs to be divided by 2 to remove duplicates (i.e. person 2 sharing with person 3 and person 3 sharing with person 2).

So the final answer is:
In a room of 25 people, the probability that any 2 of them share the same birthday is 600/730 (or 60/73), which equates to approximately 0.82192

I didn't try to follow your post, but as I hope you have seen from the links provided in the past few days, your answer of 82% for n=25 is incorrect.

 

[erik.van.geit]

I think I have it….and used Excel to get there!!

Summing each row will give you 24/365 which is the probability of one person sharing a birthday with anyone else in the room.

Summing the whole 25 x 25 grid ...

... which equates to approximately 0.82192

I cannot agree with this explanation
there are at least two posts in this thread which have links to complete explanations
if you apply your trick to 50 persons, what would be the result ?

EDIT: for some reason missed Gens post: Hi, Gene

 

[Lewiy]

I see the error of my ways. I think what's happened is I have failed to take into account that each individual themselves has a 1/365 chance of having a birthday on any given day. Will have to re-think this.

 

[Lewiy]

Ok, taking a different angle now.

Probability equals the number of ways an event can occur divided by the number of possible outcomes.

The number of possible outcomes in this example is given by 365^25 as each person could have a birthday on one of 365 days (if we ignore leap years).

The number of ways that 365 people could all have different birthdays is given by 365*364*363*362……etc. Or 365! (factorial 365). Thus the number of ways 25 people could all have different birthdays is 365!-340!

Consequently the number of ways in which 25 people do not have the same birthday is (365^25)-(365!-340!) which is the number of possible outcomes less the combinations where all birthdays are different.

Hence, I put it to you that the probability of at least 2 people in a room of 25 have the same birthday is:
((365^25)-(365!-340!))/(365^25)

I have yet to evaluate this as the numbers are huge. However, you can test the maths with a simpler example:

You have 3 regular dice (with numbers 1 to 6 on each) the probability of any 2 dice having the same number when rolled (using the above logic) would be:
((6^3)-(6!-3!))/(6^3) which is 0.697478992.

 

[EE2006]

Ok, taking a different angle now.

Probability equals the number of ways an event can occur divided by the number of possible outcomes.

The number of possible outcomes in this example is given by 365^25 as each person could have a birthday on one of 365 days (if we ignore leap years).

The number of ways that 365 people could all have different birthdays is given by 365*364*363*362……etc. Or 365! (factorial 365). Thus the number of ways 25 people could all have different birthdays is 365!-340!

Consequently the number of ways in which 25 people do not have the same birthday is (365^25)-(365!-340!) which is the number of possible outcomes less the combinations where all birthdays are different.

Hence, I put it to you that the probability of at least 2 people in a room of 25 have the same birthday is:
((365^25)-(365!-340!))/(365^25)

I have yet to evaluate this as the numbers are huge. However, you can test the maths with a simpler example:

You have 3 regular dice (with numbers 1 to 6 on each) the probability of any 2 dice having the same number when rolled (using the above logic) would be:
((6^3)-(6!-3!))/(6^3) which is 0.697478992.

I don't really remember a whole lot from the statistics courses I've done in the past, but something about your method seems familiar. However, I'm pretty sure there is a problem with the factorial terms you are using. There are a couple of issues that I can see with your response right away.


(1) As Eric already detailed on page 1 of this thread, the probability of any 2 dice having the same number when 3 are rolled is 0.44444.

(2) The equation you have posted doesn't actually work out to 0.697.... It calculates to -2.30556.


If you do figure out the correct formula for your method, be sure to post it. I've been tryinig to figure it out but haven't been able to.

 

[Lewiy]

Yeah you're right, I should be dividing the factorials rather than taking them away:

Corrected formula:

((365^25)-(365!/340!))/(365^25)

And for the dice example:

((6^3)-(6!/3!))/(6^3) which is now 0.4444444

 

[Lewiy]

Additional:

Have now managed to evaluate using the following:

365! / (340!(365^25)) gives the probability that no one has the same birthday. This equals 0.4313

Therefore 1-0.4313 = 0.5687 so the answer is 56.87%

 

[mortgageman]

Big Snip of material addressed elsewhere anyway

EDIT: for some reason missed Gens post: Hi, Gene

Hi Eric - I see that you have turned you face to the right. Was that a political move?

 

[erik.van.geit]

Big Snip of material addressed elsewhere anyway

EDIT: for some reason missed Gens post: Hi, Gene

Hi Eric - I see that you have turned you face to the right. Was that a political move?

to stay on topic
the probability to turn it to the left is equal the right: no statistical conclusions can be retreived from this "choice"
also the entire picture explains why I'm looking to the LEFT (you missed the correct angle !! )

best regards,
Erik