Primedice.com | Since 2013 | Longest Running Crypto Casino | 113 BTC Jackpot!

[Shogen]

Received the 0.01 bitcointalk giveaway, thanks a lot. I didn't notice the one on FB until now though...

[sukamasoto]

.....
only in the primedice could happen like this  
you also have to try
love Primedice

Nice shot .... Just after 15 try ?

COngrats for your winnings. I'm guessing that you made it from PD giveway , right ?
I'm gonna try my luck too  

Edit : lost them all 

[adaseb]

Just for future gamblers this will stay in history:
That's how I tried winning some (Yeah I know I'm not the most lucky )

No it's impossible even at 1000 try ...... But nice risk .......... i love big risk because big wins behind it

not impossible. if the user is very lucky he could hit it in 1 try. really depends on luck

I would bet against that bet actually for the first few 20-30 rolls or so and win 0.002X20= 0.04BTC with very little risk.

[Monopoly]

I know nothing is impossible .......... but my experience said it's impossible you hit 0.00 at even 1000 try , not even at 10000 try  , not even in 100000 try ... just pick up your seeds and go to dicesite's verifying and set last nonce at 100000 you see everything except 0.00 ....... most of seeds are like that

check mine :

http://dicesites.com/primedice/verifier?ss=4c091d9c6001593bb7c59899b2274f7b858019f098a2925b8b7377434f02de41&cs=JfkSUcQJMC7XI3qMUC60MCAZ5qla7b&ln=100000&n=0&ssh=9A2FAC46EFEB6379DE9EBFD3941B975CAF3E229FD23AB2FBE787A5A28E1B5F58

[rivoke]

Just for future gamblers this will stay in history:
That's how I tried winning some (Yeah I know I'm not the most lucky )

No it's impossible even at 1000 try ...... But nice risk .......... i love big risk because big wins behind it

not impossible. if the user is very lucky he could hit it in 1 try. really depends on luck

I would bet against that bet actually for the first few 20-30 rolls or so and win 0.002X20= 0.04BTC with very little risk.

I'm bet Stunna will get frightened if many people bet with that huge amount

[arallmuus]

I'm bet Stunna will get frightened if many people bet with that huge amount

He will be extremely happy if someone tried to pull a stun like that. There are chances that someone might not hit it in 9900 times roll which means that an easy profit for the house. Besides that unless someone has quite a bankroll to spend than that strategy to roll for max profit of 19.8 BTC isnt a feasible at all though

I know nothing is impossible .......... but my experience said it's impossible you hit 0.00 at even 1000 try , not even at 10000 try  , not even in 100000 try

Hui has proved that it is not impossible to roll for that though what you need is an extra luck. Combination of a good seed might help to roll it below 9900 times but the bigger question would be no one knows that the "magical seed" is

[Monopoly]

I'm bet Stunna will get frightened if many people bet with that huge amount

He will be extremely happy if someone tried to pull a stun like that. There are chances that someone might not hit it in 9900 times roll which means that an easy profit for the house. Besides that unless someone has quite a bankroll to spend than that strategy to roll for max profit of 19.8 BTC isnt a feasible at all though

I know nothing is impossible .......... but my experience said it's impossible you hit 0.00 at even 1000 try , not even at 10000 try  , not even in 100000 try

Hui has proved that it is not impossible to roll for that though what you need is an extra luck. Combination of a good seed might help to roll it below 9900 times but the bigger question would be no one knows that the "magical seed" is

Don't be stupid and don't loss your money for a impossible shot !
Over  99.95 is welcome but over 99.99 is just for a super lucky man , I am talking about 1 man between 10 million men .

[katerniko1]

i would never risk that much on 9900x its just idiotic and you will not hit it 99,9999999%
and why then waste money for it better go all in on 2x and repeat it 10 times
regards.
-Katerniko1

[arallmuus]

-snip-

don't loss your money for a impossible shot !

Impossible is not the word for it, as long as there are still chances for that then it is not impossible. It is highly unlike to happen, YES but not impossible. Someone could be lucky and hit the max payout with it, luck is pretty much different to each individual . Hui has proven that it is not literally impossible to hit it as he/she had a solid record on hitting the 9900x

but over 99.99 is just for a super lucky man , I am talking about 1 man between 10 million men .

Wrong, over 99.99 is impossible . What you mean would be probably be over 99.98 . As per stated, all you need would be luck. Winning the lottery is pretty much "impossible" for some people but there will be always someone that is lucky enough to win it

[Stratobitz]

Just for future gamblers this will stay in history:
That's how I tried winning some (Yeah I know I'm not the most lucky )
-snip-

No it's impossible even at 1000 try ...... But nice risk .......... i love big risk because big wins behind it

Nah nothing is impossible. You can win the x9900 bet in your very first try if you are very lucky.
With 1000 tries, the chance to win at least one of them is 1-((1-0.01%)^1000) = 9.5%, which is not small at all.

Your math assumes that 1 win in 9900 rolls MUST happen. Your factoring your chances of winning on certainty where playing a dice site is anything but. But in truth you could roll 30,000 rolls and not win. Even 100,000. Unlucky streaks happen just like lucky ones do. Your formula assumes past rolls affect future outcomes within a given series. Flip a coin twice. If Heads flips first, tails then has to flip second, when in reality you could flip 20 heads in a row.

Calculating odds of either single wins or chances of a loss streak with in a series is complex math... Theres a number of formulas one can use - and even many of the greatest matheticians often disagree on the most reliable method.

Lots of info on Wikipedia about this. Complex math for sure. But a 10% chance of winning is not anywhere near accurate. The actual odds fall on a curve, between <1% and 100%.

In an absolute random provably fair system I would estimate on 1000 rolls at 9900x your odds of winning with the PD house edge is 2-3% in a single sitting.

Strato

[adaseb]

Just for future gamblers this will stay in history:
That's how I tried winning some (Yeah I know I'm not the most lucky )
-snip-

No it's impossible even at 1000 try ...... But nice risk .......... i love big risk because big wins behind it

Nah nothing is impossible. You can win the x9900 bet in your very first try if you are very lucky.
With 1000 tries, the chance to win at least one of them is 1-((1-0.01%)^1000) = 9.5%, which is not small at all.

Your math assumes that 1 win in 9900 rolls MUST happen. Your factoring your chances of winning on certainty where playing a dice site is anything but. But in truth you could roll 30,000 rolls and not win. Even 100,000. Unlucky streaks happen just like lucky ones do. Your formula assumes past rolls affect future outcomes within a given series. Flip a coin twice. If Heads flips first, tails then has to flip second, when in reality you could flip 20 heads in a row.

Calculating odds of either single wins or chances of a loss streak with in a series is complex math... Theres a number of formulas one can use - and even many of the greatest matheticians often disagree on the most reliable method.

Lots of info on Wikipedia about this. Complex math for sure. But a 10% chance of winning is not anywhere near accurate. The actual odds fall on a curve, between <1% and 100%.

In an absolute random provably fair system I would estimate on 1000 rolls at 9900x your odds of winning with the PD house edge is 2-3% in a single sitting.

Strato

Yes exactly what he wrote. Most people assume that every 2 rolls with 50/50 win, that a winner must happen at least once.

But they don't understand probability and that each roll is completely indepednant of the prior roll hence, a win doesn't always happen. Its just due to luck.

[Monopoly]

Just for future gamblers this will stay in history:
That's how I tried winning some (Yeah I know I'm not the most lucky )
-snip-

No it's impossible even at 1000 try ...... But nice risk .......... i love big risk because big wins behind it

Nah nothing is impossible. You can win the x9900 bet in your very first try if you are very lucky.
With 1000 tries, the chance to win at least one of them is 1-((1-0.01%)^1000) = 9.5%, which is not small at all.

Your math assumes that 1 win in 9900 rolls MUST happen. Your factoring your chances of winning on certainty where playing a dice site is anything but. But in truth you could roll 30,000 rolls and not win. Even 100,000. Unlucky streaks happen just like lucky ones do. Your formula assumes past rolls affect future outcomes within a given series. Flip a coin twice. If Heads flips first, tails then has to flip second, when in reality you could flip 20 heads in a row.

Calculating odds of either single wins or chances of a loss streak with in a series is complex math... Theres a number of formulas one can use - and even many of the greatest matheticians often disagree on the most reliable method.

Lots of info on Wikipedia about this. Complex math for sure. But a 10% chance of winning is not anywhere near accurate. The actual odds fall on a curve, between <1% and 100%.

In an absolute random provably fair system I would estimate on 1000 rolls at 9900x your odds of winning with the PD house edge is 2-3% in a single sitting.

Strato

Yes exactly what he wrote. Most people assume that every 2 rolls with 50/50 win, that a winner must happen at least once.

But they don't understand probability and that each roll is completely indepednant of the prior roll hence, a win doesn't always happen. Its just due to luck.

Yes i wanted say something like that but for terrible English writing i couldn't .....

[FanEagle]

i would never risk that much on 9900x its just idiotic and you will not hit it 99,9999999%
and why then waste money for it better go all in on 2x and repeat it 10 times
regards.
-Katerniko1

I think going allin 10 times straight in an action like that is certainly a suicide,at least in that picture I show the most improbable of the happening.

[Buziss]

Just for future gamblers this will stay in history:
That's how I tried winning some (Yeah I know I'm not the most lucky )
-snip-

No it's impossible even at 1000 try ...... But nice risk .......... i love big risk because big wins behind it

Nah nothing is impossible. You can win the x9900 bet in your very first try if you are very lucky.
With 1000 tries, the chance to win at least one of them is 1-((1-0.01%)^1000) = 9.5%, which is not small at all.

Your math assumes that 1 win in 9900 rolls MUST happen. Your factoring your chances of winning on certainty where playing a dice site is anything but. But in truth you could roll 30,000 rolls and not win. Even 100,000. Unlucky streaks happen just like lucky ones do. Your formula assumes past rolls affect future outcomes within a given series. Flip a coin twice. If Heads flips first, tails then has to flip second, when in reality you could flip 20 heads in a row.

Calculating odds of either single wins or chances of a loss streak with in a series is complex math... Theres a number of formulas one can use - and even many of the greatest matheticians often disagree on the most reliable method.

Lots of info on Wikipedia about this. Complex math for sure. But a 10% chance of winning is not anywhere near accurate. The actual odds fall on a curve, between <1% and 100%.

In an absolute random provably fair system I would estimate on 1000 rolls at 9900x your odds of winning with the PD house edge is 2-3% in a single sitting.

Strato

Wrong. The calculation is only based on that the bets are independent. The chance for not winning a 0.01% bet is 99.99%, and the chance for not winning such bet 1000 times = 99.99%^1000 because this is how you calculate the probability of independent events.

In fact, the chance of not getting a win in 9900 rolls is (99.99%)^9900 = 37.16%, which is far far away from zero.

[british]

just doing 1 max bet is the best way to do it other thatn that all fail.
You can never overcome casinos profit in the long run and that is why they always profit.

[Brewins]

I lost the last giveaway

but it is not that much anyway, but good to know that PD is back to giveaways

[mattiadeabtc]

I lost the last giveaway

but it is not that much anyway, but good to know that PD is back to giveaways

Fortunately primedice offer giveaway often,  stay tuned for the next

[madonnino]

I almost did, but thanks to pd i was able to eliminate any loss of purchasing power after the dump!
Yeah! some days it goes well!

[Stratobitz]

Just for future gamblers this will stay in history:
That's how I tried winning some (Yeah I know I'm not the most lucky )
-snip-

No it's impossible even at 1000 try ...... But nice risk .......... i love big risk because big wins behind it

Nah nothing is impossible. You can win the x9900 bet in your very first try if you are very lucky.
With 1000 tries, the chance to win at least one of them is 1-((1-0.01%)^1000) = 9.5%, which is not small at all.

Your math assumes that 1 win in 9900 rolls MUST happen. Your factoring your chances of winning on certainty where playing a dice site is anything but. But in truth you could roll 30,000 rolls and not win. Even 100,000. Unlucky streaks happen just like lucky ones do. Your formula assumes past rolls affect future outcomes within a given series. Flip a coin twice. If Heads flips first, tails then has to flip second, when in reality you could flip 20 heads in a row.

Calculating odds of either single wins or chances of a loss streak with in a series is complex math... Theres a number of formulas one can use - and even many of the greatest matheticians often disagree on the most reliable method.

Lots of info on Wikipedia about this. Complex math for sure. But a 10% chance of winning is not anywhere near accurate. The actual odds fall on a curve, between <1% and 100%.

In an absolute random provably fair system I would estimate on 1000 rolls at 9900x your odds of winning with the PD house edge is 2-3% in a single sitting.

Strato

Wrong. The calculation is only based on that the bets are independent. The chance for not winning a 0.01% bet is 99.99%, and the chance for not winning such bet 1000 times = 99.99%^1000 because this is how you calculate the probability of independent events.

In fact, the chance of not getting a win in 9900 rolls is (99.99%)^9900 = 37.16%, which is far far away from zero.

Using the power function to calculate the odds of getting rolling a win or loss with a specific series of rolls is highly flawed, because the potential series is infinite, and each roll restarts the series from scratch. Meaning, after 5000 rolls of losing, your odds of rolling a 9900x win in the NEXT 9900 rolls is exactly the same as it was when you started 5000 rolls ago. The fact that you've lost 5000 rolls has no bearing on the next 9900. 

Your formula assumes the odds are absolute which is simply not correct, whether it be on PrimeDice or in a physical casino.

To expand, statistical odds represented as a percentage must add up to 100%, while at the same time, will never reach the true round number of 100% --- Being 99.99999... and so on is as close as one could get. Unless of course you're flipping a double headed coin

Lets use your example to demonstrate...

On 9900x you've calculated/assumed the actual odds of not rolling a win rolling 9900 rolls is (99.99%)^9900 = 37.16%

The flaw here is if your odds of 'not' winning is 37.16% then mathematically speaking your odds of winning must be 62.84%.

Betting 0.00001000 per roll would result in a win of 0.09899000. Rolling 9900 times, a player would wager 0.09900000 - he would win 63 out of every 100 times he played this series. (62.84% Rounded).

So every 100 games the player would win 63 and lose 37...? Thats guaranteed profit.

You can use discrete probability to calculate odds for only a single roll... To calculate the odds of winning or losing in a series, continuous probability must be used and a number of statistical paradoxes must be considered.

 Strato




 

[Buziss]

Using the power function to calculate the odds of getting rolling a win or loss with a specific series of rolls is highly flawed, because the potential series is infinite, and each roll restarts the series from scratch. Meaning, after 5000 rolls of losing, your odds of rolling a 9900x win in the NEXT 9900 rolls is exactly the same as it was when you started 5000 rolls ago. The fact that you've lost 5000 rolls has no bearing on the next 9900. 

Your formula assumes the odds are absolute which is simply not correct, whether it be on PrimeDice or in a physical casino.

To expand, statistical odds represented as a percentage must add up to 100%, while at the same time, will never reach the true round number of 100% --- Being 99.99999... and so on is as close as one could get. Unless of course you're flipping a double headed coin

Lets use your example to demonstrate...

On 9900x you've calculated/assumed the actual odds of not rolling a win rolling 9900 rolls is (99.99%)^9900 = 37.16%

The flaw here is if your odds of 'not' winning is 37.16% then mathematically speaking your odds of winning must be 62.84%.

Betting 0.00001000 per roll would result in a win of 0.09899000. Rolling 9900 times, a player would wager 0.09900000 - he would win 63 out of every 100 times he played this series. (62.84% Rounded).

So every 100 games the player would win 63 and lose 37...? Thats guaranteed profit.

You can use discrete probability to calculate odds for only a single roll... To calculate the odds of winning or losing in a series, continuous probability must be used and a number of statistical paradoxes must be considered.

 Strato

It looks to me that you are a bit confused about conditional probability. You are right that the probability won't change after getting 5000 loss, and indeed that is because each bet is independent to each other, and that is exactly the reason why ^ is used in calculating the unconditional probability.


Also, "you have 62.84% chance to make it" doesn't mean you will win 62.84% of the time for sure, the number of expected occurrence and realized occurrence could be hugely different, especially when the sample size is small.
Take a simple event as example, you have 16.6667% chance to roll a '6' with a fair dice. Does that mean you will be guaranteed to get it once every 6 rolls? The answer is no, but the probability tells you how likely an event is going to occur.