Hey guys,

 I was playing Rummikub earlier today with my parents; for those who don't know what game that is, the point is that we did something(take our 'numbers') before each taking out a number to decide who begins(it's supposed to be done before). So I thought of this; I picked a number from 1-3 and asked my father to pick one from 1-3. If he guessed it right, he would start. If not, I would pick a number from 1-2 and then ask my mother to pick one as well. If she got it right, she would start. If not, I would start.

 After short calculations, you can find out that we each have an equal chance at starting the game. Now, being the math-geek that I am, I wanted to prove this little method algebraically to show that it would work with any number of people.

 Now, please bear with my mathematical dribble as I am only a junior in high school, but here goes:

x = # of players, which would also be the amount of numbers I'd start with.

y = # of players that have gone before the player of which the probability is being calculated.

Pⁿ is player number n   ...   and _(W;L) is win; loss chance.

 If each player has en equal chance of winning, then their chance of winning is 1/x  and their chance of losing is

(1-1/x).

 First, we know that P¹_(W;L) = (1/x ; 1-1/x)

 Now, for players other than the first one, I came up with this to calculate the chance of a player losing:

     y/x + [(1-y/x)(1-1/(x-y))] 

 -The chances of a player losing will be the sum of the chances of the players before him winning plus the chance of him losing in his turn(not as an independent event).

 For P² we know that ''the sum of the chances of the players before him winning'' is y/x and will continue to be y/x for other P's if we continue to get 1/x as the P_(W).

 The chance of a player losing in his turn(not as an independent event) will be the product of the chances of the players before him not picking the right number(allowing his turn to be played) multiplied by the chance of him picking a wrong number.

 ''The chance of any player picking a wrong number'' is 1-1/(x-y) since the chance of him picking the right number is 1/(x-y) because the amount of numbers in the ''pool'' goes down by 1 each time you go to the next player.

 Now, for ''the product of the chances of the players before him not picking the right number''; with P² we get:

 (1-1/x) since there is only P¹

 So, the total chance of P² picking the wrong number would be (1-1/x)(1-1/(x-1)) --- This becomes 1-2/x; and when we try to get it for P³, we would multiply (1-2x) by (1-1/(x-2)) --- This in turn becomes 1-3/x; and we get this part of the formula:

 y/x + [(1-y/x)(1-1/(x-y))] 

 So, by solving this we should get 1-1/x :

 y/x + 1 - y/x - 1/(x-y) + y/(x²-xy)

 y/x + 1 - y/x - x/(x²-xy) + y/(x²-xy)

 1 + (-x+y)/(x²-xy)

 1 + (-1/x)

 1-1/x   yay!

 Now, The chances of a player winning will be the product of the chances of the players before him not picking the right number(allowing his turn to be played) multiplied by the chance of him picking the right number.

 So,

 (1-y/x)(1/(x-y)) --- which when solved should equal 1/x

 (1/(x-y)) - (y/(x²-xy))

 (x/(x²-xy)) - (y/(x²-xy))

 (x-y)/(x²-xy)

 1/x   yay!

 And for the last player: y = x-1 --- So,

 P_(W) = (1-(x-1)/x)(1/(x-(x-1)))

             ((x-(x-1))/x)(1/1)

             (1/x)

              1/x   yay!

Did any of this make sense?

Was there any problem in my reasoning?

Did I succeed?