game idea but dont know if it is possible

Hi all

I am back :) and need help with a Question . (I apology in advance for my english)

I saw a game offered and it looks simple and not fair to me and I want to make it a zero house edge so the player will have longer fun.

the game is called -find the coin- there are 3 coins and it looks like this

http://gyazo.com/02eae1c4e638246a302c4076e53733bd

the player is asked to click on one coin and then all 3 coins are switched and player will see if he clicked the winning coin. it is a chance 1 out of 3 to win = player will lose 2 out of 3 games.

after clicking a coin it looks like this

http://gyazo.com/a8f1535f6f469a6ec3567b8f11f60819

now lets get to my question:
I want to change the game by leaving the 1 out of 3 option in place (where player will lose 2 out of 3 leaving him with a loss of 1 after 3 games ).
and add another 3 coins option where the outcome would be 2 out of 3 (where the player will win 2 out of 3 games with a win of 1 after 3 games.

so lets say a player now will have the option to play 6 games and 3 of those games he will be offered with the 1 out of 3 win chance and 3 with the 2 out of 3 win chance. but the 6 games should be mixed so the player doesnt know when the 1 out of 3 or the 2 out of 3 version is offered. but to put in simple words, if he would bet all 6 games 1 chip house and player should end up not losing and not winning = zero HE

sure that the 6 games were only an example and what I meant is that it should be an infinite number where both options come mixed 50/50 so on the long run it should be 50/50 and actually zero HE game.

now would this be possible with a RNG (for online game) to mix the 2 options in a a fair way and outcome?

sorry if I confused you but any help and any questions are very much appeciated.

thanks

I think it would be simple to do this, where the second game had 2 coins out of 3 available, and over time those games were weighted and interspersed 50% of the time. But who would offer the game like that? The house can't make any money.

However, that 50% weighting could be manipulated to be 52/48% towards the house, or whatever they had to have to offer it.

Quote: beachbumbabs

I think it would be simple to do this, where the second game had 2 coins out of 3 available, and over time those games were weighted and interspersed 50% of the time. But who would offer the game like that? The house can't make any money.

However, that 50% weighting could be manipulated to be 52/48% towards the house, or whatever they had to have to offer it.

thank You very much for the answer. I understand now that this would be no problem. actually I wanted to add the game to a bitcoin faucet. IMO in a faucet with a zero house game would keep the players longer on the website.

but what do You mean with -where the second game had 2 coins out of 3 available - ? do You mean to change the 2nd game to 2 coins instead of 3? or did I misunderstand You and You just repeated the game 2 out of 3?

if we change the weighting to 52/48% what would be the house edge? 4%? that would be a lot IMO

thanks again and good to know that this would be possible. oh forgot to ask if this weighting would be done in connection with a RNG?

I'm suggesting that it would be much easier to program if you had two games with identical appearance to the player, but 1 game was programmed with 1 coin, and the 2nd game was programmed with 2 coins. This would accomplish exactly what you want to do. You don't have to expose to the player how many coins are available unless you want to, though I think it would be a huge come-on for the players to know that some games had 2 of 3 chances to win.

The RNG seeding would control which game was randomly offered to the player on any given spin, and the long-term percentage of which game appears could be customized to provide whatever house edge you want to offer.

Quote: beachbumbabs

I'm suggesting that it would be much easier to program if you had two games with identical appearance to the player, but 1 game was programmed with 1 coin, and the 2nd game was programmed with 2 coins. This would accomplish exactly what you want to do. You don't have to expose to the player how many coins are available unless you want to, though I think it would be a huge come-on for the players to know that some games had 2 of 3 chances to win.

The RNG seeding would control which game was randomly offered to the player on any given spin, and the long-term percentage of which game appears could be customized to provide whatever house edge you want to offer.

thanks again.

it looks that I still dont understand Your point regarding 1 coin and 2 coin.........
please let me again explain what I had in mind. the player would be informed that after he places his bet chances are 50/50% that each of the 2 versions could come up. oh now I see where probably my mistake is :) lets change the rule (thank You...it looks You enlightened me to get the right rule now)

game 1 is 1 out of 3 the player may chose 1 coin only. that means he has 1 out of 3 chance to get the winning coin
game 2 is also 1 out of 3 but player may chose 2 coins. that means he has 2 out of 3 chance to get the winning coin
now each time a player makes a bet he will be informed when presenting him the 3 coins to chose 1 or 2 coins.

now we should have a zero HE game if weighted 50/50%

The RNG seeding would control which game was randomly offered to the player on any given spin, and the long-term percentage of which game appears could be customized to provide whatever house edge you want to offer.

yes that is exactly what I hoped for when I presented my 1st question.

Quote: seven

thanks again.

it looks that I still dont understand Your point regarding 1 coin and 2 coin.........
please let me again explain what I had in mind. the player would be informed that after he places his bet chances are 50/50% that each of the 2 versions could come up. oh now I see where probably my mistake is :) lets change the rule (thank You...it looks You enlightened me to get the right rule now)

game 1 is 1 out of 3 the player may chose 1 coin only. that means he has 1 out of 3 chance to get the winning coin
game 2 is also 1 out of 3 but player may chose 2 coins. that means he has 2 out of 3 chance to get the winning coin
now each time a player makes a bet he will be informed when presenting him the 3 coins to chose 1 or 2 coins.

now we should have a zero HE game if weighted 50/50%

The RNG seeding would control which game was randomly offered to the player on any given spin, and the long-term percentage of which game appears could be customized to provide whatever house edge you want to offer.

yes that is exactly what I hoped for when I presented my 1st question.

You and I are talking about 2 different ways to accomplish the same thing; either would work at your discretion. I'm saying the player only ever gets 1 pick, but game 1 has only 1 winning spot of 3, game 2 has 2 winning spots of 3. You're saying the RNG could randomly offer 1 or 2 picks on the same game. Either could be set to whatever HE you wanted to offer; the frequency of game 1 appearing vs. game 2 appearing could be controlled the same way the frequency of offering 1 pick vs. 2 picks.

Quote: beachbumbabs

Quote: seven

thanks again.

it looks that I still dont understand Your point regarding 1 coin and 2 coin.........
please let me again explain what I had in mind. the player would be informed that after he places his bet chances are 50/50% that each of the 2 versions could come up. oh now I see where probably my mistake is :) lets change the rule (thank You...it looks You enlightened me to get the right rule now)

game 1 is 1 out of 3 the player may chose 1 coin only. that means he has 1 out of 3 chance to get the winning coin
game 2 is also 1 out of 3 but player may chose 2 coins. that means he has 2 out of 3 chance to get the winning coin
now each time a player makes a bet he will be informed when presenting him the 3 coins to chose 1 or 2 coins.

now we should have a zero HE game if weighted 50/50%

The RNG seeding would control which game was randomly offered to the player on any given spin, and the long-term percentage of which game appears could be customized to provide whatever house edge you want to offer.

yes that is exactly what I hoped for when I presented my 1st question.

You and I are talking about 2 different ways to accomplish the same thing; either would work at your discretion. I'm saying the player only ever gets 1 pick, but game 1 has only 1 winning spot of 3, game 2 has 2 winning spots of 3. You're saying the RNG could randomly offer 1 or 2 picks on the same game. Either could be set to whatever HE you wanted to offer; the frequency of game 1 appearing vs. game 2 appearing could be controlled the same way the frequency of offering 1 pick vs. 2 picks.

:) ok well understood now.........I am glad that I can offer this kind of game and lets see what the user will say.

great help ....thanks

Hi again

my partner started to confuse me, so I came back with another question. hopefully one can help me to find the answer. we know now that 50/50% weighted will result in zero HE. what should be the weighting to get a 1% HE?

my partner said that 50.5%/49.5% weighting is under the line not a 1% HE. he is convinced that it is 0.33% HE and if I want a 1% HE I need to weight 51.5%/48.5%

he explained it like this:
lets take 1800 random games of the 2 versions weighted 50.5%/49.5%.
version 1 with 1 out of 3 win chance will be shown to the player 50.5% = 909 times
version 2 with 2 out of 3 win chance will be shown to the player 49.5% = 891 times

now version 1 should win 303 times and lose 606 times = 303-
now version 2 should win 594 times and lose 297 times = 297+
all together the player will lose 6 and 6 out of 1800 = 0.33% (HE)

now he is claiming that to get the 1% HE the weighting should be 51.5/48.5%

I understood his explanation but I am confused now :( please help me to understand what I missed here?

thanks

Let,
P = the probability of getting game 1
1-P = the probability of getting game 2

In other words, P is the weighting you are trying to find. Also,

Probability player wins game 1 = 1/3
Probability player wins game 2 = 2/3

Therefore, the player’s expected return from the game is…
(1/3)P + (2/3)(1-P) – (2/3)P – (1/3)(1-P)
= -(2/3)P + 1/3

Now, a house edge of 1% is the same as the player’s expected return of -1%. So…
-(2/3)P + 1/3 = -1/100
P = (1/100 + 1/3) * 3/2
P = 103/200
or 51.5%

You can find P for whatever HE you would like.

Why not just level the payouts for a 0 house edge so you can just stick to/offer one game and not have to get complex about it?

i.e. Player wins 1/3 times and loses 2/3 times, thus when the player wins they should win 2 to 1 since they're 2 times more likely to lose. Thus if a player loses twice, then wins once (1 out of 3) they, and the house, remain even.

I assume you're asking this for a non casino environment, as this would never, ever, make it near a casino environment. Interesting choice of bitcoins for the buttons btw =p.

Quote: PeeMcGee

Let,
P = the probability of getting game 1
1-P = the probability of getting game 2

In other words, P is the weighting you are trying to find. Also,

Probability player wins game 1 = 1/3
Probability player wins game 2 = 2/3

Therefore, the player’s expected return from the game is…
(1/3)P + (2/3)(1-P) – (2/3)P – (1/3)(1-P)
= -(2/3)P + 1/3

Now, a house edge of 1% is the same as the player’s expected return of -1%. So…
-(2/3)P + 1/3 = -1/100
P = (1/100 + 1/3) * 3/2
P = 103/200
or 51.5%

You can find P for whatever HE you would like.

thank You very much for taking the time and the explanation. I am not the mathematical formula guy but if I understood You right You are confirming my partners result that if we want to have 1% HE we need to weight 51.5/48.5%
P = 51.5%
-1P = 48.5%

Quote: Romes

Why not just level the payouts for a 0 house edge so you can just stick to/offer one game and not have to get complex about it?

i.e. Player wins 1/3 times and loses 2/3 times, thus when the player wins they should win 2 to 1 since they're 2 times more likely to lose. Thus if a player loses twice, then wins once (1 out of 3) they, and the house, remain even.

I assume you're asking this for a non casino environment, as this would never, ever, make it near a casino environment. Interesting choice of bitcoins for the buttons btw =p.

I am not sure if I understood Your point, but it is not for an land based casino environment, it is for an online gaming site or a btc faucet. by the way the bitcoins button screenshot was taken from a btc faucet.

thanks

Quote: seven

Quote: PeeMcGee

Let,
P = the probability of getting game 1
1-P = the probability of getting game 2

In other words, P is the weighting you are trying to find. Also,

Probability player wins game 1 = 1/3
Probability player wins game 2 = 2/3

Therefore, the player’s expected return from the game is…
(1/3)P + (2/3)(1-P) – (2/3)P – (1/3)(1-P)
= -(2/3)P + 1/3

Now, a house edge of 1% is the same as the player’s expected return of -1%. So…
-(2/3)P + 1/3 = -1/100
P = (1/100 + 1/3) * 3/2
P = 103/200
or 51.5%

You can find P for whatever HE you would like.

thank You very much for taking the time and the explanation. I am not the mathematical formula guy but if I understood You right You are confirming my partners result that if we want to have 1% HE we need to weight 51.5/48.5%
P = 51.5%
-1P = 48.5%

Yea, and maybe to offer a little more clarification on the math...

The player wins:
1/3 of the time when they get game 1
OR
2/3 of the time when they get game 2.

The probability of getting game 1 is P. Therefore, to get game 1 AND win is (1/3)*P.
Likewise, the probability of getting game 2 (which is 1-P) AND win is (2/3)*(1-P).

Therefore, the total probability of the player winning is the sum or: (1/3)P + (2/3)(1-P). call this eq.1

Do the same for losing. The player loses:
2/3 of the time when they get game 1
OR
1/3 of the time when they get game 2.

Therefore, the total probability of the player losing is: (2/3)P + (1/3)(1-P). call this eq.2

The House Edge is the difference between the player losing and the player winning. Or…
HE = P(Player Lose) – P(Player Win)
HE = eq.2 - eq.1
HE = (2/3)P + (1/3)(1-P) - (1/3)P - (2/3)(1-P)
HE = (2/3)P – 1/3

And again, plug in any HE you want and solve for P (or vice versa).

@PeeMcGee

thank You again for taking the time to simplify the formula and explanation. it was extremely helpful and very much appreciated.

BTW a family member told me to think about adding the game idea to facebook. as I am not familiar with it I will start
to research if it is worth the pain:)
in case I find out that it does make sense I will have another question for a formula guy like You :)

How can I do a simulation of the weighting of 51.5/48.5%? just to get a picture if a martingale would work
against the game.

thanks