maths my current amusement :)

[raaaid]

how many times does the number 11 appear in a two times lotery with ten digits

one when the first lottery gets one and when the second gets one

but in how many oportunities does the number eleven appear in a 3 times lottery? and here the caught:

19

and how many times does appear the number 21?

20 :)

the caught is when the lottery turns out 111

you win 10 times on *11 and ten times on 11* but 111 its shared so there are 19 numbers in the lottery with the number eleven while 20 with the number 21

this is similar thats easier to get 2-1 in a two dice throw than 1-1

likes its easier to get in a two same time coin toss to get head- tails than heads-heads

well now im making a program to study this farther i hope you find this amusing as i do :)

edit:

does anyone know if theres any game in which you bet about the chances of a certain number appearing in an undefined succesion of numbers?

bet for the number 9876 would be better than betting for 1111 or 2222 check it here search in your browser simetrical numbers vs no simetrical ones:

http://www.eveandersson.com/pi/digits/1000000

edit:

times the number appears:

11=8964
22=8994
33=8987
44=8906
55=9132
66=8807
77=8743
88=9037
99=9035
00=8889


12=9541
23=9721
34=9735
45=9811
56=9809
67=9664
78=9782
89=9844
90=9697
01=9699

edit:

you may wonder why it doesnt come out a closer to a round number like 10000

its because every 50 cahrachters the line cuts so the search doesnt search on that one

if its one char every 50 so of one million youre playing 980k so the average sholud be getting 9800

edit:

just imagine a casino turns out that pi succesion:

a person betting on 11 would lose badly while the person betting at 21 would keep

the caught is that for chance rules then number 111 counts as two times 11 while th browser searcher counts it just once

but you could play as the browser searches when you win you dont play next game but keep doubling the previous