GreedyStats 2

In a dice game, six fair dice are rolled. Depending on their faces or combinations of faces, points can be scored. I am trying to find the probability of a successful roll that can yield points. See Appendix A for a table of all the probabilities. Successful Rolls include:

One(s) or five(s) are present

Three, four, five, or six of a kind

Full house (3 pair / four of a kind and a pair)

Long straight (1-6 are present)

Short straight (1-5 are present / 2-6 are present)

Six Dice / Full Roll Here is a Venn-diagram for what I believe contains the correct probabilities. The colored regions are represented by the equation the arrow points to. If I sum all the probabilities together, the successful roll probability becomes 211 in 216 chance or 97.6851851851% chance.

One(s) or Five(s) are Present

Concept A successful roll may have one or more ones or fives. To find this, I calculate the probability of a roll with no ones or fives and find the complement.

Math There is a 4/6 chance of a not rolling a one or five. For all six dice to not roll a one or a five, the probability would be 4/6 to the sixth power.

: (4

6)

6

Then subtract that number from one to finish the equation.

: 1 (4

6)

6

Statistical Reasoning Each die roll is independent of the each other, and each face is always a potential outcome. Under these circumstances, the probability of failure is p^n with p = 4/6 and n = 6. To find the complement, or ones or fives probability, I subtract that from one.

Three, Four, Five, or Six of a Kind

Concept I dont want to find the probability of all three, four, five, or six of a kinds. The three, four, five, or six of a kind with the ones and fives are already calculated in the one(s) or five(s) are present section. Also, the rolls that have a three, four, or five of a kind with the additional dice containing a one or five are already calculated in the one(s) or five(s) are present section. I only want to calculate the probability of the rolls that contain twos, threes, fours, and sixes. To do this, I take the probability of a three of a kind, four of a kind, five of a kind, and six of a kind for one face value, and multiply it by a combinational factor to account for the four possible face values. However, when finding the probability of three of a kind through this method, there is a chance that two three of a kinds occur. This means that each three of a kind is counted twice. This needs to be accounted for and removed.

Math The probability of the successful face is 1/6. For the unsuccessful faces (one, five, or the successful face), the probability is 3/6. For three of a kind, the probability is 1/6 (desired) raised to the third power and 3/6 (undesired) raised to the third power. All is multiplied by a combinational factor.

: 63 (1

6)

3

(3

6)

3

For a four of a kind, the successful face is raised to the fourth, and the unsuccessful faces are raised to the second. All is still multiplied by a combinational factor.

: 64 (1

6)

4

(3

6)

2

For a five of a kind, the successful face is raised to the fifth, and the unsuccessful faces are raised to the first. All is still multiplied by a combinational factor.

: 65 (1

6)

5

(3

6)

1

For a six of a kind, 1/6 only the successful face is raised to the sixth power (but the pattern still fits).

: 66 (1

6)

6

(3

6)

0

I sum all these together and multiply the sum by an additional combinational factor. , , , :

41 ( 63 (1

6)

3

(3

6)

3

+ 64 (1

6)

4

(3

6)

2

+ 65 (1

6)

5

(3

6)

1

+ 66 (1

6)

6

(3

6)

0

)

For the two, simultaneous, three of a kind overlap probability, the unsuccessful faces are only one face with probability 1/6. And then that is multiplied by another combinational factor.

: 42 ( 63 (1

6)

3

(1

6)

3

)

I take this value and subtract it from the overall sum to reach the final probability. , , , :

41 ( 63 (1

6)

3

(3

6)

3

+ 64 (1

6)

4

(3

6)

2

+ 65 (1

6)

5

(3

6)

1

+ 66 (1

6)

6

(3

6)

0

) 42 ( 63 (1

6)

3

(1

6)

3

)

Statistical Reasoning To find the three, four, five, or six of a kinds probability, I need to consider that each die roll is independent of the each other, and each face is always a potential outcome. Because of this, the probabilities of each die roll needs to be multiplied together. A certain amount of rolls must be of the same die face value (probability p), and the rest of the rolls must not be the same die face value, a one, or a five (probability q). Total rolls (power of p + power of q) must equal six. Also, because order doesnt matter, and there are various combinations of dice that yield the same result, a combinational factor needs to be included. For a three of a kind: p^3 * q^3, combinational factor six dice, choose three to be the same face value. For a four of a kind: p^4 *q^2, combinational factor six dice, choose four to be the same face value. For five of a kind: p^5 * q^1, combinational factor six dice, choose five to be the same face value. For a six of a kind: p^6 * q^0, combinational factor six dice, choose six to be the same face value. p = 1/6 and q = 3/6. There are four potential face values that I care about (not ones or fives) to roll a three, four, five, or six of a kind, and the math is based around calculating one face value. This shows the need and determines the values of the additional combinational factor four faces, choose one of them. This was all done assuming the events were mutually exclusive. When events are mutually exclusive, the events probabilities can be summed together as was done. However, three of a kinds are not mutually exclusive: Two, simultaneous three of a kinds can occur. By doing the above process while including the three of a kinds, the overlap (two, simultaneous three of a kinds) needs to be removed.

To find the probability of the overlap, I need to apply a similar method to finding the three of a kind probability. The difference is that the unsuccessful faces (probability q) must all be the same face, bringing down its value to 1/6. The combinational factor, six dice choose three to be the same face value, is still present. There are still four potential face values that I care about (not ones or fives) to roll a three of a kind, and the math is based around calculating two face values. This shows the need and determines the values of the additional combinational factor four faces, choose two of them. After finding the overlap probability, I can subtract it from the previous three, four, five, of six of a kind probability, thus accounting for the overlap.

Full House

Concept I dont want to find the total probability of all full houses. I only want to calculate the probability of the rolls that contain twos, threes, fours, and sixes. The ones that contain ones and fives are already calculated in the one(s) or five(s) present section. In addition, the full houses that contain a four of a kind and a pair are calculated in the three, four, five, or six of a kind section. To find this limited full house probability, I break down the probabilities of three two of a kinds while multiplying by various combinational factors.

Math Using the same pattern from the three, four, five, or six of a kind section, the two of a kind probability can be found.

: 62 (1

6)

2

(3

6)

4

However, the remaining four unsuccessful face rolls need to be broken down into another two of a kind probability with the unsuccessful faces not including ones, fives, old two of a kind faces, or the new two of a kind faces. This gets multiplied by another combinational factor. This yields the two simultaneous two of a kinds probability.

: 62 (1

6)

2

42 (1

6)

2

(2

6)

2

Because I am calculating a full house, the last unsuccessful faces need to be the same, bringing its probability down to 1/6. All of this is finally multiplied by another combinational factor.

: 43 ( 62 (1

6)

2

42 (1

6)

2

(1

6)

2

)

Statistical Reasoning To find the full houses probability, I need to consider that each die roll is independent of the each other, and each face is always a potential outcome. Because of this, the probabilities of each die roll needs to be multiplied together. I use the same pattern from the three, four, five, or six of a kind section to find a two of a kind probability. For a two of a kind: p^2 * q^4, combinational factor six dice, choose two to be the same face value. p = 1/6 and q = 3/6. I breakdown the last four rolls represented by q^4 into another two of a kind probability with the unsuccessful faces not including ones, fives, old two of a kind faces, or the new two of a kind faces. For two simultaneous two of a kinds: p1^2 * p2^2 *q^2, combinational factors six dice, choose two to be the same face value, and four dice, choose two to be the same face value. p1 = 1/6, p2 = 1/6, and q = 2/6. To find the probability of a full house, the unsuccessful faces (probability q) of the two simultaneous two of a kinds must all be the same face, bringing the down its value to 1/6. The combinational factors are remain the same. There are four potential face values that I care about (not ones or fives) to roll a full house, and the math is based around calculating three face values. This shows the need and determines the values of the additional combinational factor four faces, choose three of them.

Long/Short straight Because the long and both short straights must contain a five, its probability is contained within the one(s) or five(s) present section.

Five Dice Remain Now I want to check the probabilities if there are five dice that remain. With five dice, it is impossible to get a six of a kind, a full house, or a long straight. If I sum all of the probabilities together, the successful roll probability becomes 299 in 324 chance or 92.283950617% chance.

One(s) or Five(s) are Present

Concept (See previous section.) The only difference is that there are five dice instead of six.

Math (See previous section.) The only difference is that there are five dice instead of six.

: (4

6)

5

: 1 (4

6)

5

Statistical Reasoning (See previous section.) The only difference is that there are five dice instead of six (n = 5).

Three, Four, or Five of a Kind

Concept (See previous section.) Other than the inability to roll a six of a kind, the only difference is that it is impossible to roll two simultaneous three of a kinds. This means that the three, four, and five of a kinds are mutually exclusive.

Math (See previous section.) The differences are that there is no six of a kind section, and there is no overlap.

: 53 (1

6)

3

(3

6)

2

: 54 (1

6)

4

(3

6)

1

: 55 (1

6)

5

(3

6)

0

, , : 41 ( 53 (1

6)

3

(3

6)

2

+ 54 (1

6)

4

(3

6)

1

+ 55 (1

6)

5

(3

6)

0

)

Statistical Reasoning (See previous section.) Other than the inability to roll a six of a kind, the difference is that it is impossible to roll two simultaneous three of a kinds. This means that the three, four, and five of a kinds are mutually exclusive. Also, the power of p + power of q must equal five. In addition, the intermediate combinational factors change. For a three of a kind: p^3 * q^2, combinational factor five dice, choose three to be the same face value. For a four of a kind: p^4 *q^1, combinational factor five dice, choose four to be the same face value. For five of a kind: p^5 * q^0, combinational factor five dice, choose five to be the same face value. p = 1/6 and q = 3/6.

Short straight Because the short straights must contain a five, its probability is contained within the one(s) or five(s) present section.

Four Dice Remain Now I want to check the probabilities if there are four dice that remain. With four dice, it is now impossible to get a five of a kind or a short straight. If I sum all of the probabilities together, the successful roll probability becomes 91 in 108 chance or 84.259259259% chance.

One(s) or Five(s) are Present

Concept (See previous section.) The only difference is that there are four dice instead of five.

Math (See previous section.) The only difference is that there are four dice instead of five.

: (4

6)

4

: 1 (4

6)

4

Statistical Reasoning (See previous section.) The only difference is that there are four dice instead of five (n = 4).

Three or Four of a Kind

Concept (See previous section.) The only difference is that it is impossible to roll a five of a kind.

Math (See previous section.) The only difference is that there is no five of a kind section.

: 43 (1

6)

3

(3

6)

1

: 44 (1

6)

4

(3

6)

0

: 41 ( 43 (1

6)

3

(3

6)

1

+ 44 (1

6)

4

(3

6)

0

)

Statistical Reasoning (See previous section.) The difference is that it is impossible to roll a five of a kind. Also, the power of p + power of q must equal four. In addition, the intermediate combinational factors change. For a three of a kind: p^3 * q^1, combinational factor four dice, choose three to be the same face value. For a four of a kind: p^4 *q^0, combinational factor four dice, choose four to be the same face value. p = 1/6 and q = 3/6.

Three Dice Remain Now I want to check the probabilities if there are three dice that remain. With three dice, it is now impossible to get a four of a kind. However, the rules state that if there is a three, four, or five of a kind rolled with dice remaining, then that three, four, or five of a kinds face behaves as if it were a one or a five (though no points are awarded). If I sum all of the probabilities together, the successful roll probability becomes 13 in 18 chance or 72.22222% chance.

If a three of a kind was previously rolled and its face was not a one or a five, then the successful roll probability becomes 8 in 9 chance or 88.88888% chance.

One(s), Five(s), or Three of a Kind Face(s) are Present

Concept (See previous section.) The difference is that there are three dice instead of four. Also, when accounting for the potential extra, three of a kind face, the successful dice probability changes.

Math (See previous section.) The difference is that there are three dice instead of four. In addition, if a three of a kind (not of a one or five) was previously rolled, there is a 3/6 chance of not rolling a one, five, or the three of a kinds face.

: (4

6)

3

, , : (3

6)

3

: 1 (4

6)

3

, , : 1 (3

6)

3

Statistical Reasoning (See previous section.) The difference is that there are three dice instead of four (n = 3). Also, if a three of a kind (not of a one or five) was previously rolled, p = 3/6.

Three of a Kind

Concept (See previous section.) The difference is that it is impossible to roll a four of a kind. Also, if a three of a kind was rolled (not of a one or five), then I dont want to find the probability of all three of a kinds. The three of a kinds with the ones, fives, and three of a kinds face are already calculated in the one(s), five(s), or three of a kind faces are present section. I only want to calculate the probability of the rolls that contain values that arent ones, fives, or the value of the three of a kinds face.

Math (See previous section.) The difference is that there is no four of a kind section. Also, if a three of a kind was rolled (not of a one or five), the probability of the unsuccessful faces (one, five, three of a kind value, or the successful face) is 2/6. (This number is actually unimportant. It has no statistical use. It only exists for continuing the patterns.) Also, this causes combinational factors changes.

: 33 (1

6)

3

(3

6)

0

( ) :

33 (1

6)

3

(2

6)

0

: 41 ( 33 (1

6)

3

(3

6)

0

)

( ) : 31 ( 33 (1

6)

3

(2

6)

0

)

Statistical Reasoning (See previous section.) The difference is that it is impossible to roll a four of a kind. Also, the power of p + power of q must equal three. In addition, the intermediate combinational factors change. Also, if a three of a kind was rolled (not of a one or five), the probabilities and the outer combinational factor changes. There is one less possible face to be considered unsuccessful (probability q), and thus becomes 2/6. For a three of a kind: p^3 * q^0, combinational factor three dice, choose three to be the same face value. p = 1/6 and q = 3/6. For a three of a kind with a previous (not one or five) three of a kind: p^3 * q^0, combinational factor three dice, choose three to be the same face value. p = 1/6 and q = 2/6 If a three of a kind was rolled (not of a one or five), there becomes only three potential face values that I care about (not ones, fives, or the value of the three of a kinds face) to roll a three of a kind, and the math is based around calculating one face value. This continues to show the need and determines the new values of the additional combinational factor three faces, choose one of them.

Two Dice Remain Now I want to check the probabilities if there are two dice that remain. With two dice, it is now impossible to get a three of a kind. If I sum all of the probabilities together, the successful roll probability becomes 5 in 9 chance or 55.55555% chance.

If a three or four of a kind was previously rolled and its face was not a one or a five, then the successful roll probability becomes 3 in 4 chance or 75% chance.

One(s), Five(s), Three, or Four of a Kind Face(s) are Present

Concept (See previous section.) The difference is that there are two dice instead of three. Also, when accounting for the four of a kind face, it acts no differently than if a previous three of a kind was rolled.

Math (See previous section.) The difference is that there are two dice instead of three.

: (4

6)

2

, , , : (3

6)

2

: 1 (4

6)

2

, , , : 1 (3

6)

2

Statistical Reasoning (See previous section.) The difference is that there are two dice instead of three (n = 2).

One Die Remains Now I want to check the probabilities if there is one die that remains. If I sum all of the probabilities together, the successful roll probability becomes 1 in 3 chance or 33.33333% chance. If a three, four, or five of a kind was previously rolled and its face was not a one or a five, then the successful roll probability becomes 1 in 2 chance or 50% chance.

One(s), Five(s), Three, Four, or Five of a Kind Face(s) are Present

Concept (See previous section.) The difference is that there is one die instead of two dice. Also, when accounting for the five of a kind face, it acts no differently than if a previous three of a kind was rolled.

Math (See previous section.) The difference is that there is one die instead of two dice.

: (4

6)

1

, , , : (3

6)

1

: 1 (4

6)

1

, , , : 1 (3

6)

1

Statistical Reasoning (See previous section.) The difference is that there is one die instead of two (n = 1).

Appendix A Tables of Probabilities Six Dice / Full Roll Successful Roll Failing Roll 97.6851851851% 2.314814814%

Five Dice Remain Successful Roll Failing Roll 92.283950617% 7.716049382%

Four Dice Remain

Successful Roll Failing Roll 84.259259259% 15.740740740%

Three Dice Remain Successful Roll Failing Roll 72.22222% 27.77777% 88.88888%* 11.11111%*

Two Dice Remain Successful Roll Failing Roll 55.55555% 44.44444% 75%* 25%*

One Die Remains Successful Roll Failing Roll 33.33333% 66.6666% 50%* 50%*

* A previous roll contains a three, four, or five of a kind (not of a one or a five).