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Author Topic: FIRST Peer-2-Peer betting website. NO HOUSE & NO DEALER 1$ UNLIMITED GAINS  (Read 1789 times)
betroom.eu (OP)
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January 07, 2020, 04:11:40 PM
 #81

1. First off, there is only a 3% fee deducted when the players buy their tickets, therefore this is not a rake.

How is this different than a poker tournament that costs $100 to enter and $97 goes to the prize pool, 3% to the house.

Betroom does NOT have any odds of winning in any of its games and, most importantly

Huh?  If there are 10 players who each bought the same amount of tickets, would they not each have a 10% chance to win?\

Poker tournament - 9 players and a dealer with a LIMITED pot

Betroom's games - UNLIMITED number of players with an UNLIMITED pot
  

     Everything else, relate to my last reply.


Poker tournament, unlimited players with unlimited prize pool, there is no dealer if it's online, they use software...same as you.

Also Betroom's games are FAST - a game starts every 20 seconds allowing a winner every 20 seconds. A feat that NO other website is capable of providing.
Poker games take forever to complete while our games complete at an average rate of 20 seconds.

Now you suddenly shifted from comparing traditional casinos where people bet against the house to a single game "poker". It is indeed close to online poker tournaments but it s far different from it, from gameplay to rate of games played per minute.

Now you are starting to realise the GENIUS in Betroom. That is a good thing.

All I'm saying is that with a 3% fee, each ticket is worth 97% of the price, so the more tickets you buy, the more you lose.  Buying more tickets for one game will increase your chances of winning, but it won't make you more successful.  It doesn't matter how much you spend, 3% of it will go to you and the rest will go to the players.

You're the owner of the site right? 

"Buying more tickets for one game will increase your chances of winning, but it won't make you more successful."

WHAT is that? Increasing the chances of success WILL and DOES make you more successful in all walks of life, the MORE you put in, the MORE you get out of it. There are some concepts in life that you seem to NOT understand. I am unsure if it s your desire to overlook them or pure lack of knowledge, but before making statements, please go on and study the matter in question. Of course it matters how much you spend, if you spend more you CAN become more successful.

No.

Risking $1 to win $100 1% of the time and risking $100 to make $1 99% of the time will be equally 'successful' bets.

The only thing that changes is variance. 

Buying more tickets allows you to win more often, but your profit will be less than if you bought just one ticket.  This is basic stuff here.

You certainly did NOT understand the games i built that combine skill with a random number generator proving classic statistics to be inaccurate the least. I can back up what i m saying by directly challenging him to a game of Rocket Crash where i hold only 2 tickets while you hold 1 ticket and i can prove that i will beat you way better than the ratio of 2:1, proving the point that the higher number of tickets gives a more of an exponential than a proportional edge over competition. This has been tested by ourselves many times over, that s why the ticket prices rise exponentially and not linearly.

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January 07, 2020, 07:08:41 PM
Last edit: January 07, 2020, 10:03:01 PM by TwitchySeal
 #82

I can back up what i m saying by directly challenging him to a game of Rocket Crash where i hold only 2 tickets while you hold 1 ticket and i can prove that i will beat you way better than the ratio of 2:1, proving the point that the higher number of tickets gives a more of an exponential than a proportional edge over competition. This has been tested by ourselves many times over, that s why the ticket prices rise exponentially and not linearly.

It would need to be better than a 3:1 win ratio to prove your point (I spend 1, you spend 1 + 2).

Let's say you bought 10 tickets though and 10 other people each bought 1 ticket.

Let's also just say first ticket costs $1 for simplicity.

You would spend a total of $1,023 on your 10 tickets and the other 10 players would spend $1 each, so the total spent on tickets would be $1,033.

$1,033 total spent
3% fee = $30.99
Prizepool = $1002.01

You would have a very high chance of successfully turning your $1,023 into $1002.01.


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January 07, 2020, 07:52:46 PM
 #83

Looks cool, any promo codes?
betroom.eu (OP)
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January 08, 2020, 08:54:44 AM
 #84

I can back up what i m saying by directly challenging him to a game of Rocket Crash where i hold only 2 tickets while you hold 1 ticket and i can prove that i will beat you way better than the ratio of 2:1, proving the point that the higher number of tickets gives a more of an exponential than a proportional edge over competition. This has been tested by ourselves many times over, that s why the ticket prices rise exponentially and not linearly.

It would need to be better than a 3:1 win ratio to prove your point (I spend 1, you spend 1 + 2).

Let's say you bought 10 tickets though and 10 other people each bought 1 ticket.

Let's also just say first ticket costs $1 for simplicity.

You would spend a total of $1,023 on your 10 tickets and the other 10 players would spend $1 each, so the total spent on tickets would be $1,033.

$1,033 total spent
3% fee = $30.99
Prizepool = $1002.01

You would have a very high chance of successfully turning your $1,023 into $1002.01.




Spending doesn t work like that. The formula for tickets' cost is 2 pow (n-1), where n is the number of tickets bought, so if you buy 1 ticket it costs you 1 Betcoin (1 Betcoin = 0.0001 Bitcoins). If i buy 2 tickets it costs me 2 Betcoins.
Now, if we bet against each other EVERY time, i would put down 2, you would put down 1, so 3 in total. If we keep betting continuously like this, Betroom's fee would be equal to the ticket's cost in 33 and 1/3 games. So, in order for me to be profitable i would need to break out of the 2:1 ratio, which holds me captive in an equal risk/return loop, PLUS Betroom's fee per game. To make up for Betroom's fee i would need an advantage of 3% of the 3 tickets, that is 9% of your ticket. So, every game i need a 9% advantage over you to break even, that is, i need an additional win every 11 games, leading to a 2.27:1 ratio - let s round it up to a 2.3:1 ratio. That s ONLY a 15% advantage over the 2:1 ratio, far from your 3:1 ratio, to start making money. Our tests indicate that every 25 games, with a 2:1 ticket ratio, you get a 18.5:6.5 ratio, that is roughly 2.85:1. At this pace you MAKE money. So, if you want to test this with me we can go ahead and do it, if not, do revise your negative review over my website, a review that was not even tested before it was put up.

betroom.eu (OP)
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January 08, 2020, 09:00:10 AM
 #85

Looks cool, any promo codes?

Thank you so much. We don t have any promos, we are a 100% Peer-to-Peer betting website, there s no house, we only take a 3% fee off of ticket sales. Once the platform generates more interest i will add gifts. For the moment, since we are at the very beginning, we do not have promotions.

TwitchySeal
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January 08, 2020, 09:13:32 AM
 #86

I can back up what i m saying by directly challenging him to a game of Rocket Crash where i hold only 2 tickets while you hold 1 ticket and i can prove that i will beat you way better than the ratio of 2:1, proving the point that the higher number of tickets gives a more of an exponential than a proportional edge over competition. This has been tested by ourselves many times over, that s why the ticket prices rise exponentially and not linearly.

It would need to be better than a 3:1 win ratio to prove your point (I spend 1, you spend 1 + 2).

Let's say you bought 10 tickets though and 10 other people each bought 1 ticket.

Let's also just say first ticket costs $1 for simplicity.

You would spend a total of $1,023 on your 10 tickets and the other 10 players would spend $1 each, so the total spent on tickets would be $1,033.

$1,033 total spent
3% fee = $30.99
Prizepool = $1002.01

You would have a very high chance of successfully turning your $1,023 into $1002.01.




Spending doesn t work like that. The formula for tickets' cost is 2 pow (n-1), where n is the number of tickets bought, so if you buy 1 ticket it costs you 1 Betcoin (1 Betcoin = 0.0001 Bitcoins). If i buy 2 tickets it costs me 2 Betcoins.
Now, if we bet against each other EVERY time, i would put down 2, you would put down 1, so 3 in total. If we keep betting continuously like this, Betroom's fee would be equal to the ticket's cost in 33 and 1/3 games. So, in order for me to be profitable i would need to break out of the 2:1 ratio, which holds me captive in an equal risk/return loop, PLUS Betroom's fee per game. To make up for Betroom's fee i would need an advantage of 3% of the 3 tickets, that is 9% of your ticket. So, every game i need a 9% advantage over you to break even, that is, i need an additional win every 11 games, leading to a 2.27:1 ratio - let s round it up to a 2.3:1 ratio. That s ONLY a 15% advantage over the 2:1 ratio, far from your 3:1 ratio, to start making money. Our tests indicate that every 25 games, with a 2:1 ticket ratio, you get a 18.5:6.5 ratio, that is roughly 2.85:1. At this pace you MAKE money. So, if you want to test this with me we can go ahead and do it, if not, do revise your negative review over my website, a review that was not even tested before it was put up.
Direct question:
In the rocket game, if 10 people bought 1 ticket each, and you bought 11 tickets total, how much would you spend for the 11 tickets, and how much would the prize be if you won.  

Same scenario, but you bought 12 tickets instead of 11.  How much would the 12 tickets cost, and what would the prize be.

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CRYPTO EXCLUSIVE
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FAST & SECURE
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betroom.eu (OP)
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January 08, 2020, 09:36:38 AM
 #87

I can back up what i m saying by directly challenging him to a game of Rocket Crash where i hold only 2 tickets while you hold 1 ticket and i can prove that i will beat you way better than the ratio of 2:1, proving the point that the higher number of tickets gives a more of an exponential than a proportional edge over competition. This has been tested by ourselves many times over, that s why the ticket prices rise exponentially and not linearly.

It would need to be better than a 3:1 win ratio to prove your point (I spend 1, you spend 1 + 2).

Let's say you bought 10 tickets though and 10 other people each bought 1 ticket.

Let's also just say first ticket costs $1 for simplicity.

You would spend a total of $1,023 on your 10 tickets and the other 10 players would spend $1 each, so the total spent on tickets would be $1,033.

$1,033 total spent
3% fee = $30.99
Prizepool = $1002.01

You would have a very high chance of successfully turning your $1,023 into $1002.01.




Spending doesn t work like that. The formula for tickets' cost is 2 pow (n-1), where n is the number of tickets bought, so if you buy 1 ticket it costs you 1 Betcoin (1 Betcoin = 0.0001 Bitcoins). If i buy 2 tickets it costs me 2 Betcoins.
Now, if we bet against each other EVERY time, i would put down 2, you would put down 1, so 3 in total. If we keep betting continuously like this, Betroom's fee would be equal to the ticket's cost in 33 and 1/3 games. So, in order for me to be profitable i would need to break out of the 2:1 ratio, which holds me captive in an equal risk/return loop, PLUS Betroom's fee per game. To make up for Betroom's fee i would need an advantage of 3% of the 3 tickets, that is 9% of your ticket. So, every game i need a 9% advantage over you to break even, that is, i need an additional win every 11 games, leading to a 2.27:1 ratio - let s round it up to a 2.3:1 ratio. That s ONLY a 15% advantage over the 2:1 ratio, far from your 3:1 ratio, to start making money. Our tests indicate that every 25 games, with a 2:1 ticket ratio, you get a 18.5:6.5 ratio, that is roughly 2.85:1. At this pace you MAKE money. So, if you want to test this with me we can go ahead and do it, if not, do revise your negative review over my website, a review that was not even tested before it was put up.
Direct question:
In the rocket game, if 10 people bought 1 ticket each, and you bought 11 tickets total, how much would you spend for the 11 tickets, and how much would the prize be if you won.  

Same scenario, but you bought 12 tickets instead of 11.  How much would the 12 tickets cost, and what would the prize be.

10 people bought 1 ticket each means 10 Betcoins
1 person buying 11 tickets that is 1024 Betcoins
total pot = 1034 Betcoins

10 people bought 1 ticket each means 10 Betcoins
1 person buying 12 tickets that is 2048 Betcoins
total pot = 2058 Betcoins

It s the same calculation as before, but the ratio does not get in a 2:1 since the tickets' ratio is 11/10 and 12/10 respectively. If you want to get into a 2:1 ratio with the rest, and get an almost 50% advantage, you need the BIG bucks to do it, otherwise play along with the rest. This is what i tried to achieve here from the onset.

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January 08, 2020, 09:44:18 AM
 #88

I can change the game such that each additional ticket will cost the same as the previous one, therefore having a linear increase in ticket prices with quantity but it would get into a bidding game, which i can do but it depends if people want that as well. It s up to them, not to me. It wouldn t be fair if you ask me since if 1 person wins 1 pot then he will win all upcoming pots in the same circumstances. I tried hard to avoid this, that s why i came up with an exponential increase in ticket prices, so that it gets exponentially harder to control a game as more people join it.

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January 08, 2020, 09:44:26 AM
 #89

10 people bought 1 ticket each means 10 Betcoins
1 person buying 11 tickets that is 1024 Betcoins
total pot = 1034 Betcoins

10 people bought 1 ticket each means 10 Betcoins
1 person buying 12 tickets that is 2048 Betcoins
total pot = 2058 Betcoins

What about the 3% fee?  Wouldn't the total pot be 1034 - 31.02 fee = 1002.98 pot in the first example and 2058 - 61.79 fee =1996.27 pot in the second example?

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January 08, 2020, 09:48:05 AM
 #90

10 people bought 1 ticket each means 10 Betcoins
1 person buying 11 tickets that is 1024 Betcoins
total pot = 1034 Betcoins

10 people bought 1 ticket each means 10 Betcoins
1 person buying 12 tickets that is 2048 Betcoins
total pot = 2058 Betcoins

What about the 3% fee?  Wouldn't the total pot be 1034 - 31.02 fee = 1002.98 pot in the first example and 2058 - 61.79 fee =1996.27 pot in the second example?


Yes i missed that, that s correct. The calculations still hold, i just forgot the fee, i m doing a lot of things right now.

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January 08, 2020, 09:51:41 AM
 #91

10 people bought 1 ticket each means 10 Betcoins
1 person buying 11 tickets that is 1024 Betcoins
total pot = 1034 Betcoins

10 people bought 1 ticket each means 10 Betcoins
1 person buying 12 tickets that is 2048 Betcoins
total pot = 2058 Betcoins

What about the 3% fee?  Wouldn't the total pot be 1034 - 31.02 fee = 1002.98 pot in the first example and 2058 - 61.79 fee =1996.27 pot in the second example?


Yes i missed that, that s correct. The calculations still hold, i just forgot the fee, i m doing a lot of things right now.

So in both of these scenarios, the player who bought the most tickets would lose the most no matter what.  And if they bought another ticket, they would lose even more.  

Is this true, or am I missing something?

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January 08, 2020, 10:15:04 AM
 #92

10 people bought 1 ticket each means 10 Betcoins
1 person buying 11 tickets that is 1024 Betcoins
total pot = 1034 Betcoins

10 people bought 1 ticket each means 10 Betcoins
1 person buying 12 tickets that is 2048 Betcoins
total pot = 2058 Betcoins

What about the 3% fee?  Wouldn't the total pot be 1034 - 31.02 fee = 1002.98 pot in the first example and 2058 - 61.79 fee =1996.27 pot in the second example?


Yes i missed that, that s correct. The calculations still hold, i just forgot the fee, i m doing a lot of things right now.

So in both of these scenarios, the player who bought the most tickets would lose the most no matter what.  And if they bought another ticket, they would lose even more, right?

You have to commit to a larger number of tickets when the pot (excluding your investment) reaches at least 3% of your total investment, given that at every 11 games you must have a 1 game advantage in order to break even. Since your investment is exponentially larger and your win/lose ratio decreases, you would lose money if you were to have a big investment. There are 2 viable solutions to make gains, buy all possible combinations OR get in with a maximum of 4 tickets to keep your win ratio above the rest (given that the rest is a low number of players). As i said before, i discourage a bidding game  up to a certain level, the level being a handful of players where the bid is relatively small.

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January 08, 2020, 10:22:22 AM
 #93

10 people bought 1 ticket each means 10 Betcoins
1 person buying 11 tickets that is 1024 Betcoins
total pot = 1034 Betcoins

10 people bought 1 ticket each means 10 Betcoins
1 person buying 12 tickets that is 2048 Betcoins
total pot = 2058 Betcoins

What about the 3% fee?  Wouldn't the total pot be 1034 - 31.02 fee = 1002.98 pot in the first example and 2058 - 61.79 fee =1996.27 pot in the second example?


Yes i missed that, that s correct. The calculations still hold, i just forgot the fee, i m doing a lot of things right now.


In both of these scenarios, the player who bought the most tickets would lose the most no matter what.  And if they bought another ticket, they would lose even more.   <== is this true or false, just want to be clear

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January 08, 2020, 10:27:50 AM
 #94

10 people bought 1 ticket each means 10 Betcoins
1 person buying 11 tickets that is 1024 Betcoins
total pot = 1034 Betcoins

10 people bought 1 ticket each means 10 Betcoins
1 person buying 12 tickets that is 2048 Betcoins
total pot = 2058 Betcoins

What about the 3% fee?  Wouldn't the total pot be 1034 - 31.02 fee = 1002.98 pot in the first example and 2058 - 61.79 fee =1996.27 pot in the second example?


Yes i missed that, that s correct. The calculations still hold, i just forgot the fee, i m doing a lot of things right now.

So in both of these scenarios, the player who bought the most tickets would lose the most no matter what.  And if they bought another ticket, they would lose even more, right?

You have to commit to a larger number of tickets when the pot (excluding your investment) reaches at least 3% of your total investment, given that at every 11 games you must have a 1 game advantage in order to break even. Since your investment is exponentially larger and your win/lose ratio decreases, you would lose money if you were to have a big investment. There are 2 viable solutions to make gains, buy all possible combinations OR get in with a maximum of 4 tickets to keep your win ratio above the rest (given that the rest is a low number of players). As i said before, i discourage a bidding game  up to a certain level, the level being a handful of players where the bid is relatively small.

In both of these scenarios, the player who bought the most tickets would lose the most no matter what.  And if they bought another ticket, they would lose even more.   <== is this true or false, just want to be clear


what scenarios are you talking about? I said that if a player buys all possible combinations he wins every time, if he buys less than or equal to 4 tickets and the player number is lower than 4 he also gains every time. The rest of the cases he loses. What else?

Oh, your 2 scenarios, yes, he loses.

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January 08, 2020, 10:30:19 AM
 #95

As i said, i discourage a bidding game up to a certain point, but if the community so wishes, i will adapt to it.

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January 08, 2020, 11:36:02 AM
 #96

10 people bought 1 ticket each means 10 Betcoins
1 person buying 11 tickets that is 1024 Betcoins
total pot = 1034 Betcoins

10 people bought 1 ticket each means 10 Betcoins
1 person buying 12 tickets that is 2048 Betcoins
total pot = 2058 Betcoins

What about the 3% fee?  Wouldn't the total pot be 1034 - 31.02 fee = 1002.98 pot in the first example and 2058 - 61.79 fee =1996.27 pot in the second example?


Yes i missed that, that s correct. The calculations still hold, i just forgot the fee, i m doing a lot of things right now.


In both of these scenarios, the player who bought the most tickets would lose the most no matter what.  And if they bought another ticket, they would lose even more.   <== is this true or false, just want to be clear


Regarding your statement : "A lot of red flags with betroom.eu

The owner/developer is either being intentionally deceitful or doesn't really understand the basics of how casinos work."

I showed you that you are totally wrong. I exposed my point of view, i said that bidding is not on the table (only up to a certain point) AND buying more tickets GIVES you an advantage in the scenarios i have showed you. But, since all this information is publicly available (ticket prices, game stats, general stats, etc) for people to make their own assumptions and objectively prove that i m right, i do not see how i am "deceitful or doesn't really understand the basics of how casinos work.". You are looking for the scammers very hard i see and when you can t find one, you make one up. Well, given the circumstances and the proof i have put forth, you deceive people into thinking you have found something when in reality you are just lying to them. 

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January 08, 2020, 04:36:10 PM
 #97

10 people bought 1 ticket each means 10 Betcoins
1 person buying 11 tickets that is 1024 Betcoins
total pot = 1034 Betcoins

10 people bought 1 ticket each means 10 Betcoins
1 person buying 12 tickets that is 2048 Betcoins
total pot = 2058 Betcoins

What about the 3% fee?  Wouldn't the total pot be 1034 - 31.02 fee = 1002.98 pot in the first example and 2058 - 61.79 fee =1996.27 pot in the second example?


Yes i missed that, that s correct. The calculations still hold, i just forgot the fee, i m doing a lot of things right now.


In both of these scenarios, the player who bought the most tickets would lose the most no matter what.  And if they bought another ticket, they would lose even more.   <== is this true or false, just want to be clear


Regarding your statement : "A lot of red flags with betroom.eu

The owner/developer is either being intentionally deceitful or doesn't really understand the basics of how casinos work."

I showed you that you are totally wrong. I exposed my point of view, i said that bidding is not on the table (only up to a certain point) AND buying more tickets GIVES you an advantage in the scenarios i have showed you. But, since all this information is publicly available (ticket prices, game stats, general stats, etc) for people to make their own assumptions and objectively prove that i m right, i do not see how i am "deceitful or doesn't really understand the basics of how casinos work.". You are looking for the scammers very hard i see and when you can t find one, you make one up. Well, given the circumstances and the proof i have put forth, you deceive people into thinking you have found something when in reality you are just lying to them.  

I don't think you're a scammer, I just think you're misleading potential players and then trying to defend the misleading statements with a bunch of deflection and convoluted arguments.

You're also not very transparent at all.  The rules are not clear, nothing is provably fair, and, at least in the rocket game, players just have to trust that you aren't simply beating them yourself because you know exactly when the rocket will explode.

You take a 3% fee from each bet, that's fine.  But whether it's a fee or a house edge doesn't matter, the players will on average lose %3 of what they bet.

You claim the more you bet the more 'success' you have.  Yet, your rocket games offer scenarios where the player has literally a 0% chance to profit.  The more they bet the more they lose.  

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▀▀▀████████▀▀▀
█████████████LEADING CRYPTO SPORTSBOOK & CASINO█████████████
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CRYPTO EXCLUSIVE
CLUBHOUSE
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PAYMENTS
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January 08, 2020, 08:57:20 PM
 #98

10 people bought 1 ticket each means 10 Betcoins
1 person buying 11 tickets that is 1024 Betcoins
total pot = 1034 Betcoins

10 people bought 1 ticket each means 10 Betcoins
1 person buying 12 tickets that is 2048 Betcoins
total pot = 2058 Betcoins

What about the 3% fee?  Wouldn't the total pot be 1034 - 31.02 fee = 1002.98 pot in the first example and 2058 - 61.79 fee =1996.27 pot in the second example?


Yes i missed that, that s correct. The calculations still hold, i just forgot the fee, i m doing a lot of things right now.


In both of these scenarios, the player who bought the most tickets would lose the most no matter what.  And if they bought another ticket, they would lose even more.   <== is this true or false, just want to be clear


Regarding your statement : "A lot of red flags with betroom.eu

The owner/developer is either being intentionally deceitful or doesn't really understand the basics of how casinos work."

I showed you that you are totally wrong. I exposed my point of view, i said that bidding is not on the table (only up to a certain point) AND buying more tickets GIVES you an advantage in the scenarios i have showed you. But, since all this information is publicly available (ticket prices, game stats, general stats, etc) for people to make their own assumptions and objectively prove that i m right, i do not see how i am "deceitful or doesn't really understand the basics of how casinos work.". You are looking for the scammers very hard i see and when you can t find one, you make one up. Well, given the circumstances and the proof i have put forth, you deceive people into thinking you have found something when in reality you are just lying to them.  

I don't think you're a scammer, I just think you're misleading potential players and then trying to defend the misleading statements with a bunch of deflection and convoluted arguments.

You're also not very transparent at all.  The rules are not clear, nothing is provably fair, and, at least in the rocket game, players just have to trust that you aren't simply beating them yourself because you know exactly when the rocket will explode.

You take a 3% fee from each bet, that's fine.  But whether it's a fee or a house edge doesn't matter, the players will on average lose %3 of what they bet.

You claim the more you bet the more 'success' you have.  Yet, your rocket games offer scenarios where the player has literally a 0% chance to profit.  The more they bet the more they lose.  


I have ZERO interest in favouring a winner since i am only taking a 3% fee. Again, i am at the opposing end of traditional casinos that take 97% instead of 3% and that i showed in one of my first messages in the posts i wrote you.

Nothing is convoluted, i explained above if wee bet against each other i WILL have an edge and i WILL make money even adding Betroom's 3% fee, i will make GURANTEED money as a player.

If there are more players involved, as i said, that money is not guaranteed, but you have the all the ingredients you need to make the assumptions yourself, i don t need to tell people how to do their due diligence. There are scenarios that make them money, as i showed you. If people so desire i can create a system so that players who bet more have guaranteed success, like the following system: every ticket costs 1% less than the previous ticket. This way if you buy 20 tickets it costs you roughly 17 Betcoins and if you play against 10 other players that buy 1 ticket each it costs them 10 Betcoins in total. Your risk/reward ratio is 17/10 and your winners/losers ratio is 2:1. If you account for Betroom's 3% (6% out of your profits) it leads to a winning system for the player who buys more tickets (33 betcoins profit for the 2:1 player to be exact every 33 matches played). I purposely put the tickets' prices higher in order to discourage them to buy more tickets. I couldn t care less who is the winner, i want EVERYONE to win and return. I don t want to create a money making system for the high bidders, i want to discourage that but, again, if people want to win in every circumstance where they buy more tickets, then i am transparent and open to that as well.

I am not a scammer, i can even show you the source code since i am the programmer myself, i never bet against my players and i can show that as well. I have nothing to hide and if you blame me of something you need to bring facts, otherwise just shut up. You might look intelligent to the average person but i m an engineer and a damn good one, numbers and facts are my specialty and i can prove all my statements ANYTIME with solid proof.

Provably fair is a nonsense as you already know, there is a random number generator which any programmer has access to, but since my system is built around players playing against other players, why would i hit myself in the head when it s really not necessary? Why should i risk my business and taint my name in doing so? If there is a way to show people that, i am going to ABSOLUTELY do it. But if you haven t got solid proof to accuse me of something, i advise you to stay in your lane and mind your business because you look stupid coming up with imaginary claims. Here you didn t find your scammer, you just found a site that s totally different than the rest and a person open to back up anything he says, anytime.



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January 08, 2020, 09:35:17 PM
 #99

I have ZERO interest in favouring a winner since i am only taking a 3% fee.
The fact you're taking a fee has nothing to do with whether or not you would play on your site and use information that only you have access to to gain an unfair advantage over players.  You could easily do this undetected and make more money.


Again, i am at the opposing end of traditional casinos that take 97% instead of 3% and that i showed in one of my first messages in the posts i wrote you.
The traditional casino doesn't take 97%.  If they offer a game with a 3% house edge, they will profit 3% of the total wagered, just like you.


Provably fair is a nonsense as you already know
You're uninformed about provably fair.  Do some research.

I am not a scammer, i can even show you the source code since i am the programmer myself,
Cool, post it.

i never bet against my players and i can show that as well.
Cool, prove it.  Make it so that you couldn't exploit the players even if you wanted to also...of course this would require a provably fair system which you think is nonsense.


  ▄▄███████▄███████▄▄▄
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January 09, 2020, 08:21:54 AM
 #100

I have ZERO interest in favouring a winner since i am only taking a 3% fee.
The fact you're taking a fee has nothing to do with whether or not you would play on your site and use information that only you have access to to gain an unfair advantage over players.  You could easily do this undetected and make more money.


Again, i am at the opposing end of traditional casinos that take 97% instead of 3% and that i showed in one of my first messages in the posts i wrote you.
The traditional casino doesn't take 97%.  If they offer a game with a 3% house edge, they will profit 3% of the total wagered, just like you.


Provably fair is a nonsense as you already know
You're uninformed about provably fair.  Do some research.

I am not a scammer, i can even show you the source code since i am the programmer myself,
Cool, post it.

i never bet against my players and i can show that as well.
Cool, prove it.  Make it so that you couldn't exploit the players even if you wanted to also...of course this would require a provably fair system which you think is nonsense.





I will take care of all your nonsense one step at a time.

1. I m posting the code (backend) and since it s a lot, i will have 1 post for each game. Here's Keno Frenzy


Keno Frenzy

This is the code that generates the numbers, broadcasts the results and controls the auto-pilot (automated betting system for players)

var ev
var kenoBet = require ('../../models/Keno Frenzy/keno.model')
var keno = []
var result
var kenoValue
var kenoWinner = require('./winner')
var objects = require ('../../objects')
var duplicates = require ('../../duplicates')
var autoPilot = require('../../auto-pilot')
var time = 1

module.exports = function (socket, kenoCB, breakCB, event, kenoPotSchema, User, kenoSchema) {

    function getRandomIntInclusive(min, max) {
        min = Math.ceil(min);
        max = Math.floor(max);
        return Math.floor(Math.random() * (max - min + 1)) + min; //The maximum is inclusive and the minimum is inclusive
      }

(function () {
      var caller = arguments.callee
    var interval1 =  setInterval(() => {
        kenoValue = getRandomIntInclusive(1, 80)
        result = keno.find((val) => {return val === kenoValue})
        while( result !== undefined ){
            kenoValue = getRandomIntInclusive(1, 80)
            result = keno.find((val) => {return val === kenoValue})
        }
       
        keno.push(kenoValue)
        socket.of('/api/keno-frenzy').emit('keno-plot', {kenoValue: kenoValue, draw: keno, event: ev})
        kenoCB(kenoValue)
       
        if (time === 1) {
            clearInterval(interval1)
   
                    kenoSchema.find({event: ev}, (e1,r1) => {
                            r1.forEach((val) => {
                                  const result = keno.concat(val.chosenNumbers)
                                  kenoSchema.findByIdAndUpdate(val._id, {matched: duplicates(result), $set: {draw: keno}}, {new: true}, (e2, r2) => {
                                        socket.of('/api/keno-frenzy').emit('table', {avatar: r2.avatar, username: r2.username, matched: r2.matched})
                                  })
                            })
                            keno.length = 0
                            var pause = 20
                            kenoWinner(socket, ev, kenoPotSchema, User, kenoSchema, () => {
                                objects.kenoPlayers = 0
                                ev = Date.now()
                                event(ev)
                                breakCB(pause)
                                socket.of('/api/keno-frenzy').emit('break', {break: pause})
                                kenoPotSchema.find({}, (e, rf) => {
                                    var pot = 0
                                    rf.forEach((obj) => {
                                        pot = pot + obj.pot
                                    })
                                kenoPotSchema.remove({}, () => {

                                kenoPotSchema.create({
                                    pot: 0 + pot,
                                    timeOfInsertionms: Date.now(),
                                    event: ev
                                }).then((finalPot) => {
                                autoPilot.preKeno(pause, ev, socket)
                                var int = setInterval(() => {
                                kenoBet.countDocuments({event: ev}, (err, count) => {
                                    if (count >= 1) {
                                        clearInterval(int)
                                        socket.of('/api/keno-frenzy').emit('remaining', {remaining: 0})

                                var interval2 = setInterval(() => {
                         
                                     socket.of('/api/keno-frenzy').emit('break', {break: pause})
                                     breakCB(pause)
                                     if (pause === 1) {

                                         clearInterval(interval2)
                                         time = 10
                                         kenoBet.countDocuments({event: ev}, (err, count) => {
                                             objects.kenoPlayers = count !== undefined ? count : 0
                                            socket.of('/api/online').emit('players', {rocket: objects.rocketPlayers, keno: objects.kenoPlayers, stock: objects.stockPlayers})
                                         })
                                         caller()
                                     }
                                     pause--
                         
                                 }, 1000)

            }
            else {
                socket.of('/api/keno-frenzy').emit('remaining', {remaining: 10 - count})
            }
        })
                                                                        }, 1000)
                          })
                   
                })
    })
            })})
        }

            time--
       

      }, 1000)
    })()
   
    }




This is the code called inside the previous block to split the pot between the winner(s).




var topEarners = require ('../../models/Top 3/top-earners.model')

module.exports = function (socket, event, kenoPotSchema, User, kenoSchema, cb) {

    kenoSchema.aggregate([
        {$match: {$and:[{event: event}]} },
        {$group: {
            _id: "$matched",
            winners: {$push:'$username'}
        }},
        {$sort:{_id:-1}},
        {$limit:1} 
  ]).then((res) => {

        if (res.length > 0 && event) {
            kenoPotSchema.findOne({event: event}, (e, r) => {

                var partialPot = r.pot / res[0].winners.length
                res[0].winners.forEach((winner) => {
                    User.findOneAndUpdate({username: winner}, {$inc:{accountValue: partialPot}}, {new:true}, (e3, r3) => {
                        topEarners.create({
                            username: r3.username,
                            avatar: r3.avatar,
                            winnings: partialPot,
                            timeOfInsertionms: Date.now()                           
                        }).then(() => {})
                        socket.of('/api/keno-frenzy').emit('table', {avatar: r3.avatar, event: event, username: winner, matched: res[0]._id, result: 'winner'})
                    })
                })
                kenoPotSchema.findByIdAndDelete(r._id, () => {
                    cb()
                })
        })
        }
        else {
            cb()
             }

    })

}


Here is the code for the auto-pilot called in the same first block:



exports.preKeno = function (kenoPause, kenoEvent, socket) {
    kenoArr.forEach((player) => {
        User.findOne({username: player.username}, (e1, resp) => {
        if(player.noOfGames !== 0 && resp.accountValue >= Math.pow(2, player.tickets - 1)) {
            if (player.noOfGames !== -1) {
             const index = kenoArr.indexOf(player)
             kenoArr.splice(index, 1, {
                username: player.username,
                avatar: player.avatar,
                chosenNumbers: player.chosenNumbers,
                tickets: player.tickets,
                noOfGames: player.noOfGames - 1
             })
             }
            //  User.findOne({username: player.username}, (eee1, resss) => {
              if (kenoPause > 1 && resp.accountValue >= Math.pow(2, player.tickets - 1) && player.tickets >= 1 && player.chosenNumbers.length === 10 + (player.tickets-1)) {
                    kenoSchema.findOne({username: player.username, event: kenoEvent}, (err, res) => {
                          if (!res || res === null || res === undefined) {
                                kenoSchema.create({
                                      username: player.username,
                                      avatar: player.avatar,
                                      ticketsPrice: Math.pow(2, player.tickets - 1),
                                      tickets: player.tickets,
                                      availableTickets: player.tickets,
                                      chosenNumbers: player.chosenNumbers,
                                      timeOfInsertion: new Date(Date.now()),
                                      timeOfInsertionms: Date.now(),
                                      event: kenoEvent
                                }).then((val) => {
                                      topTickets.create({
                                            username: val.username,
                                            avatar: val.avatar,
                                            tickets: val.tickets,
                                            timeOfInsertionms: Date.now()                           
                                        }).then(() => {})
                                      //  player.soc.emit('buy-ticket', {tickets: val.tickets, event: val.event})
                                            socket.of('/api/keno-frenzy').emit('table', {avatar: val.avatar, username: val.username, event: val.event})
                                            User.findOneAndUpdate({username: val.username}, {$inc: {accountValue: -val.ticketsPrice}}, (e, r555) => {

                                                  revenue.findOneAndUpdate({game: 'Keno Frenzy'}, {$inc:{amount: 0.03 * val.ticketsPrice}}, {new: true}, (ex, rx) => {
                                                        if (rx && rx !== undefined && rx !== null) {
                                                              kenoPotSchema.findOneAndUpdate({event: kenoEvent}, {$inc:{pot: 0.97 * val.ticketsPrice}}, {new: true}, (e2, r2) => {
                                                                    socket.of('/api/keno-frenzy').emit('pot', {pot: r2.pot, event: r2.event})
                                                              })
                                                        }
                                                        else {
                                                              revenue.create({
                                                                    game: 'Keno Frenzy',
                                                                    amount: 0.03 * val.ticketsPrice                                                             
                                                              }).then(() => {
                                                                    kenoPotSchema.findOneAndUpdate({event: kenoEvent}, {$inc:{pot: 0.97 * val.ticketsPrice}}, {new: true}, (e2, r2) => {
                                                                          socket.of('/api/keno-frenzy').emit('pot', {pot: r2.pot, event: r2.event})
                                                                    })
                                                              })
                                                        }
                                                  })


                                            })
                                       
                                       
                                })
                          }
                          else{

                          }
                    })
               
              }
          //  })
     }
     else {
        autoPilot.findOneAndUpdate({username: player.username}, {
            keno: false,
            avatar: player.avatar,
            $set:{'kenoSettings.$.chosenNumbers': [],
                  'kenoSettings.$.tickets': 0,
                  'kenoSettings.$.noOfGames': 0}
       }, {new: true}, (e, r) => {
            const index = kenoArr.indexOf(player)
            kenoArr.splice(index, 1)
       })
     }
    })
    })
}



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