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In Maxi Yahtzee (Yahtzee with 6 dices): How to calculate the probability of getting two trios in one round?. Idem for three duos.

Rules: The game consists of a number of rounds. In each round, a player gets three rolls of the dice, although they can choose to end their turn after one or two rolls. After the first roll the player can save any dice they want and re-roll the other dice. This procedure is repeated after the second roll. The player has complete choice as to which dice to roll. They can re-roll a die for the third roll that was not rolled on the second roll.

Note: 333444 is a double trio. But 333333 it is not.

334455 is a triple duo. But 333355 and 333333 they are not.

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  • $\begingroup$ For the benefit of those who don't play the game, perhaps you could spell out the rules? $\endgroup$ – lulu Apr 24 '16 at 0:59
  • $\begingroup$ In particular: Can the pairs or triples have the same numbers? I.e., do 444444 and 444433 count as three pairs? $\endgroup$ – joriki Apr 24 '16 at 7:17
  • $\begingroup$ For your problem, can you save any dice and re-roll the other dice, or are they three independent rolls of $6$ dice ? $\endgroup$ – true blue anil Apr 24 '16 at 9:32
  • $\begingroup$ Do I understand correctly that you're asking two separate questions, one for two triples, one for three pairs, and in each case you want to know the probability of getting that result, assuming a strategy that maximises only that particular probability? I.e. in one scenario the player is only interested in two triples, not in three pairs or any other configuration, and plays optimally to achieve that; and in a second scenario analogously, trying only to get three pairs? $\endgroup$ – joriki Apr 27 '16 at 13:55
  • $\begingroup$ @joriki, excuse the delay in replying. You are right. That's exactly what I ask. $\endgroup$ – Matías Benzo Apr 28 '16 at 19:00
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In the case of two triples, the Transition matrix $P_{33}$ is given by $$ P_{33} = \left( \begin{array}{ccccccc} \frac{5}{324} & \frac{25}{108} & \frac{1531}{7776} & \frac{125}{324} & \frac{425}{2592} & \frac{25}{3888}\\ 0 & \frac{5}{54} & \frac{127}{432} & \frac{35}{108} & \frac{355}{1296} & \frac{5}{324}\\ 0 & 0 & \frac{17}{27} & 0 & \frac{25}{72} & \frac{5}{216}\\ 0 & 0 & 0 & \frac{4}{9} & \frac{1}{2} & \frac{1}{18}\\ 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6}\\ 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right)$$ The $p_{i,j}$ element represents the probability of moving from state i to j, according to the following table

\begin{array}{ccccccc} State\\ 1: & 0 \:pair, 0 \:triple\\ 2: & 1 \:pair, 0 \:triple\\ 3: & 0 \:pair, 1 \:triple\\ 4: & 2 \:pairs\\ 5: & 1 \:pair, 1 \:triple\\ 6: & 2 \:triples \end{array}

Then the probability of getting two triples in 3 or less rolls is given by the sixth element of

$$\left( \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 \end{array} \right)\cdot P_{33}^{3}$$

that is to say $$\frac{260649145}{1632586752} \approx 0.1597$$

In the case of three pairs, the Transition matrix $P_{222}$ is given by $$ P_{222} = \left( \begin{array}{ccccccc} \frac{5}{324} & \frac{3331}{7776} & \frac{4025}{7776} & \frac{25}{648}\\ 0 & \frac{167}{432} & \frac{245}{432} & \frac{5}{108}\\ 0 & 0 & \frac{8}{9} & \frac{1}{9}\\ 0 & 0 & 0 & 1 \end{array} \right) $$ The $p_{i,j}$ element represents the probability of moving from state i to j, according to the following table

$$ \begin{array}{ccccccc} State\\ 1: & 0 \:pair\\ 2: & 1 \:pair\\ 3: & 2 \:pairs\\ 4: & 3 \:pairs \end{array} $$

Then the probability of getting three pairs in 3 or less rolls is given by the fourth element of

$$ \left( \begin{array}{ccccccc} 1 & 0 & 0 & 0 \end{array} \right)\cdot P_{222}^{3} $$

that is to say

$$\frac{221494355}{1088391168} \approx 0.2035$$

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