Thread: Calculating variance question.

Hello all, forgive me my maths is on the level of the four basics of x,+,- and divide.

part -1

I have a question for your hopefully great minds.

52 random variables, that all have a random numbers in 4 groups, of 1-13.

distributed to nine random variables in order, twice rotated, giving the 9 random variables 2 of the 52 random variable each.

then out of the remaining 52 random variable, 5 are distributed to the middle, to none of the nine random variables.


firstly can this be calculated for variance and what would we use for the means?


part-2


The same situation as above, except this time there is 3 of the 52 random variables, distributed to the middle seperate from the 5 random variables.

Can this be calculated for variance?



By now you may of guessed this is texas holdem poker.

I believe that online poker which uses the part 1 system , as more variance than part two system, which is used in live play.

I believe the distribution to be BI-modal or multi modal,

I also believe that a sequence test would show a difference. I believe the 3 extra cards make more outs. So therefore the lesser player is favored keeping their interest.

Part -3

live play poker uses 1 deck of 52 cards shuffled and distributed at your table.

Internet poker as a shuffle server, this server shuffles 1000000s of decks and puts them into a que, the next table that then needs a deck, gets the first deck out of the que. A new deck every hand

I also think that this as effect, as by timing of an hand, you could constantly be a downswing of variance. In another words A player could constanty lose by bad timing, always been on the wrong side of variance.


Is there any maths or way I could work this one out?


I thank you for you help and will try to clarify if needed.

part 1 considering it should be completely random you could use the formula (relative frequency)=(chance of occurance)/(number of occurances).therefore since your cards are dealt first you have a (1/52,1/51) chance of getting a specific card. Now for your 5 cards in the middle your chance of getting the card you want in sequence (1/50,1/49,1/47,1/46,1/45). since you have no idea what cards the other players are holding.
part 2 extra 3 (1/44,1/43,1/42)
or 1/(number of cards-number of known cards)
others
((number of cards dealt)(number of players- your players))/((size of deck)-number of known cards)