Whats wrong with this? breaking chance

[antonio glez]

whats wrong with this? breaking chance please help

imagine you bet every time to a single number at a casino roulette:

you have a chance to win of 1/37 and win 35 expectative of winning: 35 /37

expectative of losing: 36/37 and you lose one, expectative: 36/37 ( the 1/37 house edge is the difference between the winning and the losing)

now you bet at 35 numbers at the same time in such a way that when you win you win one and when you lose you lose 35

chance of winning 35/37 and win one expectative of winning: 35 /37

chance of losing: 2/37 and you lose 35 each time you lose , expectatives of losing: 70/37

the house edge is now 35/37

any light on this please?

 

[antonio glez]

what i see:


if i play to one number i win 1 time out of 37 times 35 euro

i lose 36 times of 37 times 1 euro

if i play to all numbers but two(0 and other)

i win one euro 35 times out of 37

while i lose 35 euros 2 times out of 37 times

 

[Yayness]

Ok, I suppose this is what you mean: There are 37 numbers in total. If you bet on one number (probability of winning is 1/37), you get 35 euros if you win, and you lose 1 euro if you lose.

If you bet on 35 numbers (probability of winning is 35/37), you get 1 euro if you win, and you lose 35 euros if you lose.
----
If you bet on one number, in one out of 37 cases, you'll win 35 euros. In 36 out of 37 cases, you'll lose one euro. So you'll (averagely) after 37 rounds, win 35 euros and lose 36 euros. That means you'll in total lose 1 euro after 37 rounds if you keep betting on one number. So you don't benefit on it.

If you bet on 35 numbers, in 35 out of 37 cases, you'll win 1 euro. In 2 out of 37 cases, you'll lose 35 euros. So you'll (averagely) after 37 rounds, win 35 euros and lose 70 euros. That means you'll in total lose 35 euros if you keep betting on 35 numbers.
You won't benefit on doing either.

 

[antonio glez]

yes but odds say house has an edge of 1/37 and theoretically no strategy can change your winning or losing odds

this startegy increases your chances of losing dramaticaññy

if this is true this would be a non locality of randomness

 

[antonio glez]

actually i don't understand how this non locality is not known

if you bet on 35 numbers your betting against your self¡¡¡¡ is not posible the number one and two turn out at the same time

so the proportion between winning an losing can be changed, at least to lose with huge more eficiency than the 1/37 house edge would say

so certain strategies vary the chances of winning or losing

 

[Yayness]

If you bet on 35 numbers, you win if 1 of them is correct. But you should still not try this, because the few times you lose, you lose a lot more euros than what you gain. If you play for long enough, you'll lose all your money.

 

[antonio glez]

yes that's the whole point:

house edge of 1/37 says you should lose one euro for 37 you play WHATEVER STARTEGY YOU FOLLOW

but with this strategy you lose 35 euro for 37 you play

i think this proves randomness is a bit non local like quantum entanglement

 

[Yayness]

Oh, yeah I know what you mean. That's weird.

Perhaps some people just try it a few times because they believe they'll win anyway since the probability of winning is so high.

 

[antonio glez]

its thought no startegy can change the house edge at roulete of 1/37

i can play a chance game that has an edge against me of 1/37 and reverse the edge to a chance in my favour of 33/37, still seems imposible to have any info of what's coming next

the rules:

i play as the house, that gives me an edge of +1/37, to make my edge -1 /37 negative i give two twokens to my virtual opponent every 37 games, now chance is against me on -1/37, my opponent has an edge over me of 1/37

i decide the startegy of my opponent, theoretically this shouldn't vary the edge but it does

theoretically if chance is local there's no way to change that ofr every 37 games ill lose one token, this is wrong:

the strategy of my opponent will be bet on 35 numbers, all except a random number and zero

every time he wins he wins one euro, every time he loses he loses 35 euro:chance of my opponent winning:

35/37 to win one token

2/37 to lose 35 tokens

therefore every 37 games he wins 35 euro and loses 35*2 and every 37 games he also gets two tokens from me

so following no startegy theoretical chances say i should lose one token evry 37

following this startegy every 37 games i win -35+70-2= 33

so if you consider chance extrictly local every 37 games i lose one token, accounting for the trully nonlocality of chance if i account for that in my startegy i reverse the edge to one in my favour of winning 33 token for every 37 games

still with this method is imposible to know what chance decides but i know enough about chance, the nonlocality as to with a negative edge have a huge profit :)

in one hour playing this game i won playing token by token 600 token, 33 tokens fopr every 37 games :)

 

[HallsofIvy]

imagine you bet every time to a single number at a casino roulette:

you have a chance to win of 1/37 and win 35 expectative of winning: 35 /37

expectative of losing: 36/37 and you lose one, expectative: 36/37 ( the 1/37 house edge is the difference between the winning and the losing)

now you bet at 35 numbers at the same time in such a way that when you win you win one and when you lose you lose 35

chance of winning 35/37 and win one expectative of winning: 35 /37

I am not sure what you are saying here. If you bet 35 numbers at the same time, there is, yes, a chance of winning 35/37 but you do NOT win 35/37. You will 35- 34= 1 because you will lose on the other 34 numbers.

chance of losing: 2/37 and you lose 35 each time you lose , expectatives of losing: 70/37

the house edge is now 35/37

any light on this please?

 

[antonio glez]

yes that's it

ive checked this on practice with a toy roulette:

i bet on a single number, sometimes i win a fair amount of money other i lose not too much

i bet on 35 numbers:

i lose really bad

so roulette is like kind of black jack you can play well or bad

if you don't believe me and you know how to make simple problems simulate this:

bob and alice alternate to be the house

when either one wins or loses 100 the house swithc

the winning losings that pass 100 are acumulated next game

so this should be a 50-50 chance game however the startegy

now check this strategy

alice bets 35 tokens to red each time

bob bets to 35 numbers each time

if you could simulate this with a program it would be great

remember the trick is that bob and alice alternate to be the house

 

[antonio glez]

this is a simpler example on how in a 50/chance game startegy influences dramatically:

imagine bob and alice play a 50% game, luck

bob is going to be the house till alice win or loses 1 million dollars moment that alice will switch to be the house and so on for certain time

they both play betting at red or black with one difference:

alice plays one million dollar each time while bob plays one dollar each time

its obvious that when alice plays and bob is the house she has a chance close to 0.5 to win

while when alice is the house and bob plays bob is going to win nearly never

so they will keep alternating to be the house and bob will lose almost always while alice just a an almost 50%of times

so in a game of luck of 50% risking the minimum posible means losing almost always in a game of luck as is life

 

[willem2]

this is a simpler example on how in a 50/chance game startegy influences dramatically:

imagine bob and alice play a 50% game, luck

bob is going to be the house till alice win or loses 1 million dollars moment that alice will switch to be the house and so on for certain time

they both play betting at red or black with one difference:

alice plays one million dollar each time while bob plays one dollar each time

its obvious that when alice plays and bob is the house she has a chance close to 0.5 to win

while when alice is the house and bob plays bob is going to win nearly never

so they will keep alternating to be the house and bob will lose almost always while alice just a an almost 50%of times

so in a game of luck of 50% risking the minimum posible means losing almost always in a game of luck as is life

For each dollar that they bet they will lose 1/37 dollar. The reason that Alice's expectation is -$1000000/37 = -27027.28, and Bob's expectation is very close to 0, is that Bob will make much more than a million bets, in fact 37 million bets on the average, so he'll loose 37 times as much on a complete game.

 

[antonio glez]

exactly that's why there are skills even at a game as roulete

gambling 10.000 euro at red or black only once can be a good inversion

expect to have fun in a casino gambling litle a lot gives you an almost zero chance of having fun :)

edit:

unless you have skills at playing black jack in which case youll be banned