Topic: Is Chances of Winning in Gambling Higher? (Read 613 times)

Thank you everyone for your comments .

I have updated the topic with an edit. Please have a look at it.

Now, I'm locking the topic.

You are making a false assumption in your calculations.

You assumption is that you cannot lose more than 3 times in a row at the start, and that you always reach 1001 bitcoins again within a few bets.

The odds may be low, but they are catastrophic to your winning chance. It is highly likely that you will NEVER reach 1001 bitcoins again if you drop to say 990 right at the start.

You have use the expected value when calculating this. With odds of 50% of winning, your Expected value E[X] would be:

E[X] = -1B * 0.5 + -1B * 0.5 = 0B, so your net outcome would be exactly what you start with.

Are you referring on playing dice site? 0 house edge doesn't exist nowadays and we cant really assume that 1000 bitcoin bankroll on playing gambling. Regarding on the calculation I could say that it would be useless because you cant be sure on the next random result. You cant state if its a win or lose because they are independent and doest rely on each other.

Okay, I think I get what you're describing here.  You seem interested in the math behind this stuff so you might appreciate this.  The danger with your permutation analysis lies within what is termed the "random walk."  You can easily end up in a situation where it comes highly improbable that you'll ever come back toward the break even mark, let alone a positive profit.  And, because your strategy dictates a "quit" on a win of one unit, the risk outweighs the reward significantly.  You might find it enlightening to study the math behind the "random walk."  Start here: https://en.wikipedia.org/wiki/Random_walk

I don't know how you came up with those numbers but if you don't have a large enough sample size, nothing matters.  And what gambling game is this?  If you can win 70% of the time then you would be a millionaire.  That's something that just doesn't happen a lot and if it does, it's all due to variance.

i think the chance of winning on gambling is 50% if you really play on 50% chance rate and keep betting with fixed amount, But if you are doing a random amount of bets, then i think your calculations is quiet wrong. Your computation is just for the probability of winning without considering the settings of chance rate. But the game result is always varies for every chance rate setup.

What about the losing cases(L-Win W-Loss) that could lead you to less than 1000 bitcoins have you ever added those parts? You should put a tl;dr just stop betting if you already made a profit.

What if you keep losing 1000 times ? Since each event is simply independent of the previous one, it can happen. And even let us say you don't, eventually the luck will sway to one side until bad luck bankrupts one of you. You should try simulating this and seeing for yourself.

Why did you stop there?

What if you lose 10 in a row right off the bat?

If there is no EV, it should balance out in the end, if you are able to stop as soon as you get a profit, good for you, but you haven't done anything spectacular here.

Do the impossible task of finding a 0 house edge game and bet the same amount every time until you reach profit is great, except or all the times you hit a losing streak.

I dont think those assumption or patterns would really work anytime even what kind of gambling game you would play. 50% chance doesnt always give you a LWW and you could be assure that pattern would go out because most of the times those patterns arent useful at all because anytime you will suffer LLL even on 50% chance that would really depend on your luck and no pattern should be followed.Better to play on random.

Okay, so let's consider an heads tail bet between you and your friend.

First thing, chances of you winning a bet is 50%
Second thing, you both have $1000 initially in your pockets.
Third, you and your friend bet $1 in each bet. Winner takes $2.
Fourth, you'll leave as soon as your pocket has $1001


So the match starts,
If you win first bet, your pocket has $1001 , you leave . Chances of winning fist my was 50%.

Now chances of losing first match was 50% . In case first match is a loss, you'll continue to play till your pocket has more than 1000 dollar.

I'm talking about a two player Roulette with zero house edge (hypothetical).

Yes, there's no gaurantee of winning fist game.
But see, for a fair game with zero house edge, chances of winning and losing are both 50 %.

If I win first game, I immediately withdraw all my balance. - chances of winning fist match is 50%

In case I lose first match (chances 50%) , I'll continue to play. An event LWW could occur which has chances 12.5% of happening.

I have considered that you can bet each time exactly 1 Bitcoin.

Consider a case , I lose first game (represented by L), so my balance becomes 999 Bitcoins. Chances of losing first match were 50 %. Now I again bet 1 Bitcoin and I win (Balance becomes 1000) . Now again I bet 1 Bitcoin , I win (balance becomes 1001) . Now I will withdraw all my balance .
So the above event is represented by LWW . Chances of happening of above event is 50% * 50% * 50% = 12.5%


Yes, there's no website with zero house edge. But I'm taking an assumption that house edge is zero. If we consider house edge too , the calculated chances would be around 65% .

No matter how many times you roll a dice, the chance is always 1:2 or 50% without house edge.  A sample question would be this.  If you toss a coin 9 times and the result is H, T, T, T, T, T, T, T,  what is the chance that the coin will land in H on the 9th toss? (H=head, T=tail) .  Please  realize that the result in gambling depends entirely on its internal system.

There is no math for the chance of winning in gambling no matter how good you are in a gambling game whether the house edge has big or small percentage. And I appreciate on how you did calculate or you wherever you got this calculation for the luck/chance of winning. But still whether you do this or not, chance of winning is always on how you are going to risk your money, on how lucky you are and on how broad your knowledge with the game you are playing.

Yes chances of winning in gambling is higher if you know how to play the game and you are knowledgeable with bankroll management. But without full knowledge of you cannot have higher chances of winning and most probably more loses is posible for you when you gamble, i dont think thats everyones opinion but i am certain with mine.

Sports betting is always required skills if you really targeting to make a good profit but if you are just gamble in sports betting without them and relaying on luck i think 50% that you can be lose.. but if you have knowledge how to choose a team 80% that you can make a good profit when choosing the best team in the game..  not same in other game in gambling like dice that actually base only for luck.

Honestly if that method do by others there is different result because the game is almost random and all result are not the same..
Like other said it is base on luck so if you really want for best profit better to go in trading and make your gambling activity as your past time or play with fun.. because they are made that game is just for entertainment purposes..

Yes, I was wondering the same thing.  The math is off on the face of it---even with the zero percent edge assumption.  It doesn't look like random walk is being considered, nor are mutually exclusive events being taken into account.  If we had a little more information as to the reasoning here, it would be much easier to provide some beneficial feedback.

If you are gambling on bet sites and are a sport betting gambler then it will not all depend on your luck but it will also require some skills in the sports and it will also require your knowledge about the players and the sports on which you are betting your money.

Not exactly sure what you mean here. Care to add more details ? Like whats your base bet, whats your subsequent bet after your first loss ? First of all none of the gambling websites will give you 0 house edge, because that would mean your or them have an equal chance of winning. And even if they did, from what I have seen, it doesn't mean you will have an edge by changing your betting pattern.

Hah! Chances in gambling totally depend on your luck at the respective moment. We cannot predict it, "strategies" won't help us either. It's all about them making your mind believe they work, but in the long run (or even in the short one) you usually end up losing more than you planned to.

What a gambling platform you used for take this calculation?
What you think gamble have a big % win chance? 75% win chance is so high result I think.
With your calculation W & L with % that's no guarantees if in the first bet you can win, means 50% chance win in the first bet is also big calculation.

I just did some calculations. According to them, chances of winning in gambling comes out to be around 75%. Please correct me if I'm wrong somewhere


My assumptions-

I'm considering a case, with zero house edge.
It would be X2 if you win. And you get 0 of you lose.
Each game would charge 1 Bitcoin. Which means, 2 Bitcoin if I win, 0 if I lose.
I have 1000 Bitcoin ( many Bitcoins) in the balance of company X (Gambling Website)
Chances of winning and losing are equally likely - 50%
I would withdraw all my balance as soon as my balance becomes 1001 Bitcoins
I'm betting in two player Roulette. (I'm not talking about any sports bet)


So here it goes,

Here L  represents Lost game and W represents  win.  I'm writing the chances of event occurring in front of it. So, LWW means that I lose first game, and win second and third game giving me a net profit of 1 Bitcoin.

Cases which could lead to 1001 Bitcoins,

W -50%
LWW -12.5%
LWLWW -3.125%
LLWWW - 3.125%
LLLWWWW - 0.78125%
LLWLWWW- 0.78125%
LLWWLWW- 0.78125%
LWLWLWW- 0.78125%
LWLLWWW- 0.78125%

You can further continue this.

This sums up to be 72.65 %

Which means you have higher chances of winning in gambling if you follow these assumptions.




Edit 1: I have considered that you can bet each time exactly 1 Bitcoin.

Consider a case , I lose first game (represented by L), so my balance becomes 999 Bitcoins. Chances of losing first match were 50 %. Now I again bet 1 Bitcoin and I win (Balance becomes 1000) . Now again I bet 1 Bitcoin , I win (balance becomes 1001) . Now I will withdraw all my balance .
So the above event is represented by LWW . Chances of happening of above event is 50% * 50% * 50% = 12.5%




Edit 2 : now I've understood it. Took some help from my friend and after reading your comments .
I was right that probability  that ending up with net profit of 1 Bitcoin (1001 balance) is about 75%.
But there are chances that I can lose up everything.
As someone pointed out here, it is due to binomial variance.