Playtest this game please.

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Dice game requiring 5 4-sided dice.

Method of play:
1. Choose how many of the 5 dice you want to roll. Max 4.
2. Roll those dice and sum them.
3. Roll the remaining dice and sum the total.
4. If the total of the 1st roll is evenly divisible by the sum of the second, you win.

Numerator / Denominator
4d4 / 1d4 = Method A
3d4 / 2d4 = Method B
2d4 / 3d4 = Method C
1d4 / 4d4 = Method D

Race to 15 game.
Method A = 2 pts for a win
Method B = 5 pts for a win
Method C = 12 pts for a win
Method D = Automatic win.
Can be played by multiple people, first to reach 15 pts wins.

Gambling against the house
Method A = pays 15:8
Method B = pays 4:1
Method C = pays 9:1
Method D = pays 1000:1
I'm looking for input on if the point values and pay out values seem fair.
Thanks in advance.

Kitty Sinatra

I assume you've calculated the actual odds for the dealer game. I haven't done so, so I'm curious what you figured it being for Method A that you chose not to make the payout odds a much nicer looking 2:1
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My calculations for A are 52.05% chance of a win. And any input on what would be a favourable odds posting for that method would be greatly appreciated. I haven't been inside a casino in years so, I don't know how they post odds these days.

Kitty Sinatra

Well, 15:8 odds would be annoying as all smurf in a casino. Paying out a win involves some awkward math:

For example: You bet $5 and win. How much do you win? And then what do I win if I leave what I just won on the table and let it ride ad won again. What's 15/8 of 15/8 of $5? Do you want you game to include this level of math?

[strike]But there's good news. With odds of 52% for the house, a casino could offer 2:1 and will make money. It probably wouldn't be a game they'd push but they would also be able to point to it advertise it as a "fair" game to show they aren't cheating the gamblers.[/strike]

Doh! That was the odds for the player winning. That hurts, man. It's gonna make the game ugly.
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