I\'m totally confused thinking about this...non poker game problem
 

Okay i have a little situation I'm curious if any of you hard core math types could come up with a definite answer to, because i'm dumb and the more i think about it the worse my vision gets [img]/images/graemlins/tongue.gif[/img]

let's say I have a game with 10 baskets.

before each game begins 10 random numbers from 0 to 999 are generated, one for each basket.

Then, each of these numbers in the set of 10 is assigned to a specific basket in a random fashion.

up to this point the number selection process is certified random.

so now i have 10 baskets, each basket assigned a random value from 0 to 999 in a random order.

to finish the game, I roll a 10 sided die. if the die lands on 1, I win the value from associated with basket 1, die shows 2, i win basket 2 etc.

simple enough.

now what if the die i rolled was not 100% fair. Let's say 40% of the time it landed showing 1 or 2, or some similar defect in the die.

is this game still random even though the die is not fair? is it possible to calculate the cycle for such a game assuming the die is honest? how is the cycle affected if the die is not honest? how are my odds of winning a specific amount affected if the die is not fair.



Thanks,

bonanz

Re: I\'m totally confused thinking about this...non poker game problem
 

An unfair die does not affect anything.

Even if the die only rolled 6's, because the number in the 6th basket is completely random, your end result is still random.

Re: I\'m totally confused thinking about this...non poker game problem
 

right thats what i thought, but what about the cycle? is it just 1,000 since you're only grabbing one number in the end? the number of baskets and all that doesnt matter right?

I'm feeling kinda dumb now cause it seems easier when you hear someone confirm what you were thinking, but doubting yourself.

Re: I\'m totally confused thinking about this...non poker game problem
 

Your expected value (EV) on this game is 1+2+.....+999/1000
=999(1000)/2000=499.5 per draw.

Re: I\'m totally confused thinking about this...non poker game problem
 

[ QUOTE ]
Your expected value (EV) on this game is 1+2+.....+999/1000
=999(1000)/2000=499.5 per draw.

[/ QUOTE ]

Sweet deal, I'll play for up to 250 a draw [img]/images/graemlins/grin.gif[/img]