Wednesday

January 18, 2017
Posted by **Michael** on Saturday, March 16, 2013 at 8:56pm.

- MATH -
**Reiny**, Saturday, March 16, 2013 at 9:23pmproceed with the following argument

Whatever number you pick, the prob that the

second person picks a different number is 89/90

the prob that the 2nd and 3rd person will pick a different number = (89/90)(88/90)

the prob that the 2nd, the 3rd and the 4th pick different numbers = (89/90)(88/90)(87/90)

etc

for 29 calculations

= (89x88x87x86x...x61)/90^29 = 1 - P(89,28)/90^29

= .000069048

so the prob that somebody picks the same

= 1 - .000069048

= .99993095 - go with bob MATH -
**Reiny**, Saturday, March 16, 2013 at 9:28pmGo with bobpursley's solution

I found the probabilty that at least two people in the class will have picked the same number, not just the number that you picked. - MATH -
**bobpursley**, Saturday, March 16, 2013 at 9:30pmOk, there are 29 other students.

the probability that someone will pick your number is 1-noOnewillpick

Pr(NoOnepicks)=(98/99)^29

Put this in your google search engine

(98/99)^29=

then subtract it from one.

I get about .25, or a quarter of the time, someone will pick you number.

There is an old question, in a class of 25 kids, what is the probability that two kids in the class celebrates their birthday on the same day?

Pr:1-pr(nokidshavesame birthday)

pr(no kids)=360/360*359/360*....

= 360!/360-25)! * 1/360^25

put this in your wolfram calculator:

(360!/(360-25)!) * 1/360^25

http://www.wolframalpha.com/input/?i=%28360!%2F%28360-25%29!%29+*+1%2F360^25&dataset=

I get about .43

which means the probility of two kids celebrating their birthdays on the same day in a clss of 25 is 1-.43=.57, better than even. - MATH -
**bobpursley**, Saturday, March 16, 2013 at 9:31pmReiny: I read your post, then glanced at mine, said oh my, erased my post, then said oh my again, and reposted. Such a life when one is tired.

- Michael - MATH -
**Reiny**, Saturday, March 16, 2013 at 9:38pmDon't even look at my solution, I really messed up

Somehow I read it as 90 not 99

so all the calculations are bogus. - at bob -MATH -
**Reiny**, Saturday, March 16, 2013 at 9:42pmAnd I did the same thing, glad you were able to get yours back.

I also don't even know why I started my series as

89x88 ...

when it should have been 98x97x...

looks like I read the 99 as 90

I usually make the font bigger, my reading glasses are a bit too weak.

BTW, aren't there 365 days in a year ??

look at your birthday problem solution, lol - MATH -
**bobpursley**, Saturday, March 16, 2013 at 9:59pmI think I actually computed with 365 first, the calculator wouldn't accept it, then redid it, not thinking.

- MATH -
**bobpursley**, Saturday, March 16, 2013 at 10:09pmHere it is with 365 days...

pr(no one)=.43

which leads to the same result.