Question about statistics

[Bardman]

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

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 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.

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

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[Phildo]

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?

[birdcat90]

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

[Peeps Place]

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%

that's known as Regression to the mean.

[Peeps Place]

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

[CryForMeSky]

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

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 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.

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

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.

[magicmexican]

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

[funtotry]

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!

[Peeps Place]

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

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.

[ndnh]

House edge maters only if you are frequent bettor.

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..

[Peeps Place]

House edge maters only if you are frequent bettor.

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

[ndnh]

House edge maters only if you are frequent bettor.

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

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

[CryForMeSky]

House edge maters only if you are frequent bettor.

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

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.

[Phildo]

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.

[Bardman]

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

[Phildo]

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.

[WEBcreator]

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 

[Phildo]

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.