# Math Help!

SSBBJunky
Posts: 209
Joined: 2009-02-06
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Math Help!

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?

''Black Holes result from God dividing the universe by zero.''

butterbattle
Posts: 3945
Joined: 2008-09-12
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I'm too lazy to analyze the

I'm too lazy to analyze the formulas in detail, but it seems to work.

Our revels now are ended. These our actors, | As I foretold you, were all spirits, and | Are melted into air, into thin air; | And, like the baseless fabric of this vision, | The cloud-capped towers, the gorgeous palaces, | The solemn temples, the great globe itself, - Yea, all which it inherit, shall dissolve, | And, like this insubstantial pageant faded, | Leave not a rack behind. We are such stuff | As dreams are made on, and our little life | Is rounded with a sleep. - Shakespeare

Cpt_pineapple
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Joined: 2007-04-12
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For n players:  The first

For n players:

The first one has a chance of 1/n of guessing correctly.

So they have a [n-1]/n chance of losing leaving the next person to guess, who has a [1/[n-1] chance of guessing.

So we have the probablity of the second person winning [The probability of the first person losing and them guessing correctly]

[n-1]/n*[1/[n-1]]

which is

1/n

and so on