About This Tool

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This Liar’s Dice Probability Calculator helps players analyze probabilities to make better decisions. Liar’s Dice is a game of skill, psychology, and probability. While intuition plays a big role, understanding the math behind the game can give you a sharper edge.

Benefits of Using This Tool

Make more accurate calls on whether to challenge a claim.
Improve your bluffing strategies by estimating plausible claims.
Learn how probability impacts decision-making in hidden information games.

Try it out, refine your gameplay, and take your Liar’s Dice strategy to the next level!

How to Use It

Enter Your Dice – Input your dice as comma-separated numbers (e.g., 1,2,3).
Enter Total Dice in Play – The total number of dice still in the game.
Enter the Previous Claim – The quantity and face value of the last claim.
Enter the Claimant’s Dice Count – How many dice the claimant has.
Select if 1s are Wild – If enabled, 1s count as any number.
Click Calculate – The tool will display probabilities for the claim and next moves.

How the Math Works

Estimating Unknown Dice

Since Liar’s Dice involves hidden dice, we only have partial information about the total distribution.

Each die outside of the ones you see has an equal probability of showing any number (1 through 6).
If wild ones are enabled, the probability of rolling the claimed face increases to 2 out of 6.

Counting Known Dice

The tool first counts how many of the claimed face value you personally have.

If "wild ones" are enabled, the count includes 1s as well as the claimed face.

Binomial Probability Calculation

The probability of rolling a specific number on a single die is:

p = \frac{1}{6}, \quad \text{or} \quad p = \frac{2}{6} \text{ (if wild ones are enabled)}

We use the binomial distribution to model the probability of rolling at least $ Y $ of the claimed number among $ X $ unknown dice:

P = \sum_{i=Y}^{X} \binom{X}{i} p^i (1 - p)^{X - i}

Where:

$ X $ = Number of unknown dice
$ Y $ = Additional dice needed to meet the claim
$ p $ = Probability of rolling the claimed face
$ \binom{X}{i} $ = Binomial coefficient (number of ways to choose $ i $ successes from $ X $ trials)

Adjusting for Claimant’s Dice

If you enter a suspected number of dice the claimant has, the tool assumes they contribute to the count and recalculates the probability accordingly.

Step-by-Step Adjustments:

Estimate the Claimant’s Contribution: If the claimant has $ C $ dice, their expected contribution for a given face is:

E = C \cdot p

With E being reduced to the lowest whole number.

Adjust the Remaining Dice Pool: If the current face matches the previous claim, we assume the claimant contributes $ E $ dice.
Modify the Remaining Dice Count: Subtract the claimant’s estimated contribution from the total unseen dice before computing the probability.

Example Calculation

Suppose there are 12 total dice, and you have 1, 3, 3, 5. The last claim was 6 threes, and the claimant has 4 dice. Wild 1s are enabled.

Step-by-Step Calculation:

Not Including Claimant's Contribution:

Your Contribution: You have two 3s, plus any 1s (which are wild).
Remaining Dice Pool: $ 12 - 4 = 8 $ unknown dice.
Remaining Threes Needed: $ 6 - (2 + 1) = 3 $.
Binomial Probability Calculation: The probability of rolling a three on an unknown die is: $p = \frac{2}{6} = \frac{1}{3}$
The probability of rolling at least 3 threes among the 16 unknown dice follows a binomial distribution: $P = \sum_{i=3}^{8} \binom{8}{i} \left(\frac{1}{3}\right)^i \left(\frac{2}{3}\right)^{8 - i}$

Final Probability Result:

After computing the binomial sum, we find:

P \approx 0.5318 \quad (53.18\% \text{ chance the claim is valid})

So, given the current game state, there is a 53.18% probability that there are at least 6 threes on the table, using only the known information

Including Claimant's Contribution:

Your Contribution: You have two 3s, plus any 1s (which are wild).
Claimant’s Expected Threes: $E = 4 \times \frac{2}{6} = 1.33 \approx 1$
Remaining Dice Pool: $ 12 - 4 - 1 = 7 $ unknown dice.
Remaining Threes Needed: $ 6 - (2 + 1 + 1) = 2 $ .
Binomial Probability Calculation: The probability of rolling a three on an unknown die is: $p = \frac{2}{6} = \frac{1}{3}$
The probability of rolling at least 2 threes among the 15 unknown dice follows a binomial distribution: $P = \sum_{i=2}^{7} \binom{7}{i} \left(\frac{1}{3}\right)^i \left(\frac{2}{3}\right)^{7 - i}$

Final Probability Result:

After computing the binomial sum, we find:

P \approx 0.7366\quad(73.66\% \text{ chance the claim is valid})

So, given the current game state, there is a 73.66% probability that there are at least 6 threes on the table, assuming the claimant has at least 1 one or three on the table.