Bitcoin Forum

Well lets say i go to a dice site that has no house edge and i start betting on 50%

Now the chances of loosing once are 50% and the chances of loosing twice are 25% and so on. So im betting 1 satoshi per bet and i lose 10 times in a row does that mean that the next bet has more chances of winning than the first bet that i made before the 10th loss?

no it doesnt

gambling is about luck you could always hit a losing streak no matter what your calculation is, thats the beauty of owning a gambling sites i guess lol  ::) ::)

No, the next bet chances are still 50%. You will either win or loss that bet. I've heard some crazy losing streak on dice site like Primedice, so even though the chances of loss streak are 1 in a trillion, it might still happen.

You say the next bet chances are 50% but then you say they are 1 in a trillion

he is saying that it is all luck, you know with 1% of losing chance you can even score up to 5x lose streak if you are pretty badluck  :-\ :-\

I mean in a single bet. Even though you have 10 losses before, the chances for the next bet is still 50%.

How can you calculate the chances of 30 lost bets then? If they are always 50%

No but human nature will keep you betting until you win.  That's why casinos show the results of the last spins on Roulette, even though they are utterly pointless.
You have to think logically about it, what about the last spin/roll would affect the next spin/roll... nothing

2³⁰ gives you the chances of 30 consecutive losses. So it is 1073741824:1 about a billion to one against. 

So it is better to wait to lose 10 times and bet after or it doesnt matter

It doesn't matter, in fact, it doesn't affect anything. The "lucky numbers" doesn't affected by your previous games.

I still dont get it, if it means you are unlucky means that there are more chances of winning the next bet.

It absolutely does not matter - the chances of 10 losses in a row are 1024, the chances of 11 losses are 2048.  The chance difference for each roll, in advance of the first roll, is ever increasing, but after the 10th consecutive losing roll the chance of the next is still 2¹. i.e. 50:50

it doesnt matter, it is just a calculation of the probabilty numbers which in fact may or may not happens because there is always a factor that is call luck

Ofc may not happen but if you get the possibility of betting at 80% or 50%, you would chose 80% even tho there is the luck factor because its a better probability

The chances are always reset everytime you must make a bet, the previous one does not add up to you luck, just to the statistics.

That is the fallacy of the three door/goat showhost riddle. You have to reset all your chances every time and not add them up for the next gamble you take.

indeed thats why i call it luck factor, because with 98% chance to win, you can get a losing streak, so there is never a guarantee about that

If the odds are truly 50/50 past results don't impact future probabilities. Even if you hit a long streak 1 way or the other, the results will approach 50/50 eventually, even though each individual flip will be 50/50. Let's use flipping a coin for an example.

You flip the coin 10 times and get 7 heads and 3 tails. So you have 70% heads

If the next 10 are 50/50 you have 12 heads and 8 tails: 60% heads

If the next 10 are 50/50 you have 17 heads and 13 tails: 56 2/3% heads

If the next 10 are 50/50 you have 22 heads and 18 tails: 55% heads

...

If the next 100 are 50/50 you have 72 heads and 68 tails: 51.4% heads

...

If the next 1000 are 50/50 you have 572 heads and 568 tails: 50.1% heads.


So the odds for each individual flip are still 50/50, but as we flip more and more coins the overall results head back towards 50%

It is very unlikely to lose 30 times in a row flipping a coin. But, if you've already lost 29 in a row, you have already done the hard part, you've lost 29 in a row. The coin doesn't know that you've lost the first 29. After you've already lost 29 in a row, your odds of losing the 30th in a row are 50/50.

House edge maters only if you are frequent bettor.

yeah.. i think i get it, still a little bit confused tho

why are you confused? I got a little lost in the top of the thread too so i started going off on my own.

Did my example with flipping the coin 1,000 times help?

well actually in gambling does not affected by statistic

on 10x pay out, if we rolls for 10x then theres no guaranteed for happening..

even on 2x there will be still a chance for lose

that's known as Regression to the mean.

Since coin flips are independent events, each flip is 50/50.

Nothing to be confused about, every flip is an independent event.  You can say "if I flip this coin 4 times, there is a 6.25% chance that it'll be heads every time" but that doesn't mean that if you flip three heads in a row the next flip will be 93.75% to be tails...it's still 50/50.

http://en.wikipedia.org/wiki/Gambler's_fallacy

Anything is possible. Its even possible for a casino to open and not win a SINGLE bet there. Keep in mind that the longer you gamble the more you will lose. Don't gamble more than you can lose!

Even though roulette isn't beatable, a professional gambler may be playing roulette for fun with his friends. If Black came up 10 times in a row, an amateur would play Red the next time.

A professional would play Black thinking that there may be something wrong with the wheel. This isn't true for all.

Not really it matters for every single roll.
With 1% edge, you have 1% extra chance of losing than with 0% house edge.

For instance, you make a below 49.5 roll and hit 49.9..

I think what he is saying that you have a better chance of winning with less trials. You can play with this to see why http://stattrek.com/online-calculator/binomial.aspx (http://stattrek.com/online-calculator/binomial.aspx)

What I was saying was that house edge matters even if you made only a single bet.

This is true, but I think the point was that the more you bet, the more you get raked by the house edge.  If your bankroll was 100 and your goal was 200, and you were betting @ 2x, you'd have a higher % of success betting once at 100 than you would betting 1 over and over.

House edge always matters, it's just a different way of expressing the expected value of a bet.

If the house edge is 1%, you expect to lose 1% of your bets overall.

Now, if you are playing a 50% chance of winning dice game that pays out 1.98 x your bet, you can't lose 1% of your bet in 1 roll, but the house edge is still 1, and it still matters because if you do win you only get 1.98 your bet instead of the 2x you should get.

But even if the payout is 2x and 50% seems like strategies would still fail with not much difference, its what i read around here

If the payout is 2x and the odds of winning is 50% you expect to break even.

you are right but also note that there are 50% chance to lose too, either you break even or make your lost a lot more, and so the probabiity of you hiting 30x lose streak is always there  ::) ::)

Yes, but you have the same probability of a 30x win streak. And if the odds of winning are 50% and the payout is 2x you will win the same amount on your 30x win streak as your 30x losing streak, so you expect to break even.

If you are going to be doing any sort of regular/serious betting please everyone look up and understand the term expected value. It's basic statistics and will at least give you some of the information so you can be aware of what's happening. I'm not saying to never take a -ev bet, but at least look up the basics so you know the odds and can understand what's happening and how the casinos make their money.