Recent content by bernd

Yeah, we got it that there is no strategy to beat the bank. but nobody cares! -.- aside from that, not sure how else to explain my point. "The best you can do is wait for 15 reds. Then you know the next color is not red, so you put 18/19 of your money on black and 1/19 on zero. Doing better...

while you are right about the maximum 15 times red or black, waiting for actually 15 times red in a row would take way too long since it basically almost never happens (but is still possible, so cant ignore it). so as explained before one might wait for 7 times red and then start betting. bet...

Thinking for a bit, maybe I have higly overestimated the problem. let's say so far base bet was 1 buck and we waited for 5 times red.given that probabilty for red is 18/37 (18 fields are red, 18 black and 1 zero), that probabilty is (18/37)^5. Now, we would instead only wait for 4 times red...

Hey, gotta do some explanation first: I assume you know how roulette works. (if you dont: ball is thrown into a pit and it can either land on red, black or zero, each having a certain likeliness to land there. you can bet on where the ball will land) let's assume unrealistically you have the...

I should probably elaborate where the whole problem is even coming from:In Germany there is a lottery "6 of 49" where, as you might guess, 6 numbers are drawn from a pool of 49 numbers, building the lotto number of that session. people can fill out forms and bet on 6 numbers. and depending on...

exactly that! sorry if my explanations are just.. too bad to understand. I cant even imagine any smart or systematic way to do this, except since brute forcing. Anyone got any good idea?

"A 6-tuple contains 6 elements, not 20." Well, yeah. What I meant: in the sixtuple (1,2,3,4,5,6) the triples (1,2,3)(1,2,4)(1,2,5)(1,2,6)(1,3,4)(1,3,5)(1,3,6)(1,4,5)(1,4,6)(1,5,6) (2,3,4)(2,3,5)(2,3,6)(2,4,5)(2,4,6)(2,5,6)(3,4,5)(3,4,6)(3,5,6)(4,5,6) so 20 triples is the maximum that can be fit...

Hi, I am not sure if this can be called directly a mathematical problem but it kind of is: Say we have the list of all (ascendingly ordered) triple with natural numbers from 1-49 (is about lotto 6 out of 49): T={(a,b,c) | a,b,c natural numbers, 1<=a,b,c<=49, a<b<c} Now we are given a subset of...

Well, while in PeroK's case in order to play a game, I basically have to randomly draws cards following the rules, in my case I have to eliminate not-possible scenarios by finding rules broken by a given end table. one way or another I wont get around checking every single rules for a game. it's...

Well, yeah. since if I add the odds of games where A wins, then that number should be As overall win likeliness. same for B and draw result. Sure, I can go and jsut do 1 million games or such, but I thought to be more precisely just evaluate every possible game and add up the winning odds for...

Well It's still about the baccarat game. Just though that using a similar game in comparison would make me question a bit easier to understand. Here's the rules if anyone cares: https://www.wikihow.com/Play-Baccarat Just ignore anything that wont matter for the online version anyways (like where...

Well, I could copy in the source code of those 3 java files but not sure if it is of much help. If I want to describe an analog issue instead: Say we have play A and B and a 6 sided dice. first A throws the dice 2 times. then B throws the dice 2 times. if A threw a total of 8 or more so far...

Hello, this question isnt really much about calculation but rather which view point is more correct. See, in a gambling game called Baccarat a game is played where a player A ("player") and a player B ("bank") draw cards according to a fixed ruleset from a given card pot. First A and B both...

Basically, yeah. Except in my case here, b is always 49. but otherwise, you got the idea. Haven't really checked your calculation there, just checked the basically recursive formula for S(a,b,n) that you described there. Cant exactly followed how you did those calculations there. but yeah, as...

1. I can follow. 2. I will jsut accept it even though it "seems sus" to me that youc an basically let the outermost border go from 1 to (50-a) insted of a to 49 like before. But I'll just accept it for the moment. after that, I am kind of lost. S(r,m) still depends on n even though you havent...