Mind help: Are there any real math freaks out there? - Help.com

Well, it can give you thinking time, if you depend on it.

Well, actually the problem is not me being kept from my sleep, I get really fascinated by maths, and solve many useless problems just for the sake of increasing my “systematical” ability. I’ll post the math problem’s self as soon as I formulate it (in a couple of min.’s).

Math freak? I wouldn’t label myself that way but Jeffrey Rosenswieg once did.

You know the gamblers fallacy: that if you toss a coin 3 times and H shows up on the first 2 the probability of Tails showing up later on the third, raises. This is actually false.
Here’s the demonstration:

For a toss we have 2 prob.’s (H,T)
Here are the prob.’s for the 3 tosses
(n=2^3=8) :
toss 1 Toss 2 Toss 3
1H H H
2H H T
3H T T
4H T H
5T T T
6T T H
7T H H
8T H T

So if H show’s up on the first 2, we have
Toss 1 Toss 2 Toss 3
H H T
H H H
Still a %50 chance.

But the gambler fallacy disregards the order/combination variable and calculates as following
Toss Toss Toss
H H T
H T H
T H H
H H H

So it believes the probability of T occuring on the third is (%75) favored against H (%25)

This was the background, hope you get the idea.

What if we try to apply this fallacy to Texas hold’em Poker (with 13*4 different probabilities and min 5+2 revealed cards each turn, and plus the
real probability changes with decreasing amounts of cards in the turn, and resetting that variable at the beginning of each turn)

The problem is: According to this fallacy how many turns it will take without seeing any two aces together to have the %30 probability of Having 2 J’s on your hand. (assume 4 players; disregard the effects of the unkown cards (The other players hold) (disregard any kind of order variables - as it is the gamblers fallacy)

The maths in that problem really picked me… Let’s see if you can help.

Correction: “without seeing any two Js together”

i dont really know i love math but i dont do it all the time i just enjoy it.yea so yea. so i acctually dont know.

Ok. I figured out until a point.
* = there are 4 players, so you’ll be unaware of 3 players cards = 6 => a deck will have 46 cards before you pick your cards and 5 for the table.
a = the entire prob. of 2 cards you have in your hand
b = the entire prob. of 5 cards you’ll see on the table
c = the entire prob. of pair of J’s.
d = the entire prob of 2′d comb. with a J in it
a1= The prob. of you having a pair of J’s in your hand (c/a)
a2= The prob. of you having a comb. with a 1 J. at your hand
b1= The prob. of the 5 cards on the table having J’s (nJ>0, nJ 5)
b2= The prob. of the 5 cards on the table having J’s (nj>1, nJ 4)
e1= The prob of having a J at your hand and 1-3 J’s at the table (a2* b2)

f= calculate the entire prob.’s with n>1 J’s at the 7 cards you see (a1+b1+e1)
f1=calculate the entire prob’s with those 7 cards (a*b)
g=divise it by the entire prob.’s you may have with those 7 cards. (f/f1)

After this, it’s pretty much like the coin flip I guess. but can’t systemize the progression of prob.’s at each hand without a pair of J’s. I intiutinally think the progression goes like this, but can’t really figure out=

Prob. of not having at least a single pair of j’s at the:
1st hand
x=(1-g)
2 hand
x2= (1-g)^2

x3= (1-g)^3
…..
x. = %70

Is this correct anyway.. Don’t know..

correction: “intuitionally think..”