5 dice, 3 rolls, 1 box I.

Fives dice, three rolls, one box - part 1

 
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We have five dice again, but three rolls and you have left only the Ones box empty, so only the ones are taken into account. After rolling you can save the dice if they are 1 (one) and re-roll the rest of the dice. You can score in total 0, 1, 2,3,4,5 depending on the number of Ones. We don’t take into consideration the extra Yahtzee bonus at this point. What is the outcome of the game if we roll many times and calculate the average of the scores?

I will not provide you the formula here (it can be calculated recursively). The probabilistic value is 2.1.

For boxes from Ones to Sixes, the probabilistic values are as follows (without the extra Yahtzee bonus):

 

Prob. Value

Max. value/score

Ones

2.10

5

Twos

4.21

10

Threes

6.31

15

Fours

8.42

20

Fives

10.53

25

Sixes

12.63

30

The minimum scores are zero.

What is the probability after three rolls to get exactly one, two, three, four or five of a kind?

 

probability

0 x number

0,064

1 x number

0,236

2 x number

0,343

3 x number

0,250

4 x number

0,091

5 x number

0,010

The biggest probability is to score 2 of a kind (with 34,3% chance).

!!! Please note that there is still about 0.064 probability (6,4%) after three rolls that you don’t score any Ones or a given number with five dice!! This means roughly once in 16 every cases.

One would think that if we had only the upper section to fill up (Ones, Twos, … Sixes), the probabilistic value of them would be ~ 42. However, this is not true in Yahtzee! Why is the probabilistic value bigger than 42? Read the next section!

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