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Subject: Dice bidding game concept rss

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Mister Neutron
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I’m working on a dice roll with bidding game concept.

The simple mechanic is bidding on a dice roll using a D8 8-sided die.

Players bid before rolling the die. They bid that they will roll a certain number or higher.

Here is a simple bid scoring rule:

Bid 2 - a roll of 2 or more scores 2
Bid 3 - a roll of 3 or more scores 3
Bid 4 - a roll of 4 or more scores 4
Bid 5 - a roll of 5 or more scores 5
Bid 6 - a roll of 6 or more scores 6
Bid 7 - a roll of 7 or more scores 7
Bid 8 - a roll of 8 scores 8 or optionally a higher amount as a bonus

Simple odds chart:

Bid of 2 - 7 to 1 odds of winning
Bid of 3 - 3 to 1 odds of winning
Bid of 4 - 5 to 3 odds of winning
Bid of 5 - 1 in 2 odds of winning or 50/50
Bid of 6 - 3 in 8 odds of winning
Bid of 7 - 1 in 4 odds of winning
Bid of 8 - 1 in 8 odds of winning


At first it seems reasonably balanced, with players winning less on bids that are more likely to succeed and more when taking more risk, but is the payoff for bidding on 7 versus 5 being only 2 points more enough to be worth it? Is there an overall strategy of bidding 4 on every turn, 5 on every turn, or 6 on every turn that would make any other strategy unviable and break the game?

I’m sure there is a better odds based way of scoring but I don’t have the math skills to calculate it.

Would it be too complicated for the average player to form a strategy? Obviously the farther behind a player is the more they will need to risk near the end of the game, this could be fun if the game is balanced right.

Another mechanic of the game is that the player’s bids are placed face down so the players do not see what the others are bidding.

I’m also considering a bonus if the roll is the same number as the bid.

I am also considering using 2 D8 dice. Using 2 normal 6-sided dice would be more common, but this game mechanic is part of a game where I need to use a D8 die or dice.

Are there any other games that use a dice bidding mechanic?

I would appreciate any advice.
 
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Jeremy Lennert
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The average amount that you'll win is the payoff times the probability of winning. For your suggested payout table, that's:

bid 2: 2 points * 7/8 chance = 14/8 = 1.75
bid 3: 3 points * 6/8 chance = 18/8 = 2.25
bid 4: 4 points * 5/8 chance = 20/8 = 2.5
bid 5: 5 points * 4/8 chance = 20/8 = 2.5
bid 6: 6 points * 3/8 chance = 18/8 = 2.25
bid 7: 7 points * 2/8 chance = 14/8 = 1.75
bid 8: 8 points * 1/8 chance = 8/8 = 1

On average, you get the largest payouts with a "bid" of 4 or 5. (In general, when you're trying to maximize the product of 2 numbers without changing their sum, you want them to be as equal as possible.)
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Mister Neutron
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Thank you for the numbers Jeremy.

This brings up an interesting point. Should the average payoff be the same for each bid or should the average payoff increase by a certain amount for each higher bid?

Players win nothing if their bid is lower than the roll.
 
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BT Carpenter
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Look to Istanbul's Tea Room to see this mechanic in action.

Players state a number 3 or higher, roll 2d6 and gain their *bid* in coins if the number rolled equals or exceeds their bid. Failures gain 2 coins.

Two things make this interesting:

1/ Players may need more or less money to do other things they want to do, which will influence their spoken bid.

2/ Another power in the game allows manipulation of the die roll. (Specifically, either alter one of the two dice to 4, or reroll both dice, accepting only the second result.

By itself, there isn't a whole lot of 'game' in the rules you've described. It's more a mathematical exercise of expected value. I think you're further hindered by the linear nature of the distribution (1dX) as opposed to a 2dX or 3dX bell type curve.

It is more useful as a *mechanic* as part of a larger game, where there is varying risk/reward that can influence the amount "bid" or the results themselves.
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