Please double check my math

i have created a dice game, and if i may ask, will someone please double check my math?

the game is very simple. it uses 1 die. (the die that i have has a devil in place of the usual place for the 6)

the goal of course is to win the bet. there are two ways to win, first, the player must roll a 1 on the first roll. and if failed, must roll a 2 on the second roll, and again if failed must roll a 3 on the third, and 4 on the forth. or the second way is to successfully roll the die 4 times without rolling the devil.

the rules;

on the first roll only, the player loses if he rolls a 5 or the devil (6). and wins if he rolls a 1. (1:1)

on the second and third rolls, the player loses if he rolls the devil. and wins if he rolls the 2, and or of course the 3. (5 is now just a dead number) (1:1)

and on the forth roll the player loses if he rolls the devil, and wins 1:1 if he rolls any number.

i came up with 33.333% dealer edge on the first roll, and 16.666% dealer edge all around.

is this correct??

what might be good upgrade ideas for this game?

Quote: ponyboy

what might be good upgrade ideas for this game?

use two dice? [g]

I hope you are just having fun with trying to invent a game and checking the math, to see if you could do it. Perhaps two of you could play and switch off who is the dealer. In that case whether you have the math right wouldnt matter.

edit: I had some further remarks but I missed that the player also loses on rolling a '5'

Quote: odiousgambit

use two dice? [g]

I hope you are just having fun with trying to invent a game and checking the math, to see if you could do it. Perhaps two of you could play and switch off who is the dealer.
edit: I had some further remarks but I missed that the player also loses on rolling a '5'

two dies would ruin the effect of the game.

yes it is just for fun and a math project. im testing to see if i can create a game that the dealer has the advantage.

yes, the player loses if he rolls a 5, but only on the first roll. in theory, the first roll is the most deadly.

i find that alot of chips can be won to the player, but the house edge hits extremely hard when the odds catch back up to you.

this game is also influenced by a gambling story i once heard, of a guy who went around an bet people that he could roll a 6 in 4 rolls or less. so i switched it around and made it to where the player would win if they successfully rolled the die 4 times without rolling the 6 (or in this particular die...the devil) and added on top of it if you roll the number of the roll you are on you would win as well, and added that a 5 is a lose on the first roll and the the 5 is just a dead number place holder for the other rolls. making it 2 ways to win and 2 ways to lose. and if my math is correct, the dealer has the advantage in the long run.

i was also thinking that if the player rolls the 5 on the first roll they lose, but if they roll the 5 at any other time the game is a push. and a 2:1 payout for rolling the 4 on the forth roll.

the theory behind the game is that the player cant roll the die 4 times on average without rolling the devil. and that the dealer has the greatest chance to win at the beginning of the game, and the player has the greatest chance to win at the end of the game. making it much like blackjack where the player plays first and risks losing first ect.

I get a 1.85% dealer edge overall, if I understand the rules.

First roll: a 1 is a win (1/6), a 5 or a 6 is a loss (1/3), and a 2, 3, or 4 goes to the second roll (probability 1/2 of having a second roll).

Second roll: a 2 is a win (1/2 x 1/6), a 6 is a loss (1/2 x 1/6), and anything else goes to the third roll (probability 1/2 x 2/3 = 1/3 of having a third roll).

Third roll: a 3 is a win (1/3 x 1/6), a 6 is a loss (1/3 x 1/6), and anything else goes to the fourth roll (probability 1/3 x 2/3 = 2/9 of having a fourth roll).

Fourth roll: a 6 is a loss (2/9 x 1/6), and anything else is a win (2/9 x 5/6).

Sum of the win probabilities = 1/6 + 1/12 + 1/18 + 5/27 = 53/108.
Sum of the loss probabilities = 1/3 + 1/12 + 1/18 + 1/27 = 55/108.
Expect to lose 2 for every 108 bet.

Quote: ThatDonGuy

I get a 1.85% dealer edge overall, if I understand the rules.

First roll: a 1 is a win (1/6), a 5 or a 6 is a loss (1/3), and a 2, 3, or 4 goes to the second roll (probability 1/2 of having a second roll).

Second roll: a 2 is a win (1/2 x 1/6), a 6 is a loss (1/2 x 1/6), and anything else goes to the third roll (probability 1/2 x 2/3 = 1/3 of having a third roll).

Third roll: a 3 is a win (1/3 x 1/6), a 6 is a loss (1/3 x 1/6), and anything else goes to the fourth roll (probability 1/3 x 2/3 = 2/9 of having a fourth roll).

Fourth roll: a 6 is a loss (2/9 x 1/6), and anything else is a win (2/9 x 5/6).

Sum of the win probabilities = 1/6 + 1/12 + 1/18 + 5/27 = 53/108.
Sum of the loss probabilities = 1/3 + 1/12 + 1/18 + 1/27 = 55/108.
Expect to lose 2 for every 108 bet.

right on thank you! your math makes total sence! =] im already at work to increase that 1.85%!!!

Quote: ponyboy

right on thank you! your math makes total sence! =] im already at work to increase that 1.85%!!!

the 5's are now a lose on the first roll, and a push on the second, third, and forth rolls. and a 2:1 win of 5 on the fifth roll amd a 1:1 win for any other number for successfully rolling 5 times without a "6" comming up.

idk the extra advantage this would create, but being able to "push" and end a game would help alot as it not not paid, and it restarts the odds again from the beginning of the game and sets up for another 1/3 chance of a lose. as well as force a 5th roll ;]