Welcome › Forums › Mathematics › A little [1] [2] [3]problem
Tagged: dice rolls, probability
 This topic has 7 replies, 4 voices, and was last updated 2 years, 5 months ago by Inverted Dice.

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21st March 2016 at 01:24 #1160Girl PowerParticipant
Let’s say it’s the last round and I only have the number 15 left to score. In my first roll, I get [6] [6] [6] [5] [4]. What should I do?
Keep the 3 sixes and hope to get another two sixes in TWO rolls (five sixes gives the result 15), OR reroll ALL five dice and hope to get only ones, twos and threes in TWO rolls?
What are the probabilities in each case?
30th August 2016 at 10:21 #1943Girl PowerParticipantJust realized, that if I choose to reroll all 5 dice, I still have the possibility of achieving both five sixes and combinations of only ones and fives OR only twos and fours. So, the problem should be stated like this:
Is the probability of getting [6] [6] with TWO dice in two rolls higher than the sum of the probabilities of achieving the following combinations with FIVE dice in two rolls?
[6] [6] [6] [6] [6]
[5] [5] [5] [5] [1]
[5] [5] [5] [1] [1]
[5] [5] [1] [1] [1]
[5] [1] [1] [1] [1]
[4] [4] [4] [4] [2]
[4] [4] [4] [2] [2]
[4] [4] [2] [2] [2]
[4] [2] [2] [2] [2]
[3] [3] [3] [2] [1]
[3] [3] [2] [2] [1]
[3] [3] [2] [1] [1]
[3] [2] [2] [2] [1]
[3] [2] [2] [1] [1]
[3] [2] [1] [1] [1]
Is this difficult to calculate? (I don’t know much about probability theory)27th September 2017 at 12:49 #4115JonasParticipantI don’t know I anyone is still interested in the answer, but there is a problem you might not see in your statement.
When you write of “achieving the following combinations with five dices in two rolls”, you omit the part that you have to chose which dices to keep between the two rolls. While it’s possible to compute for each possible result of the first reroll which choice of keeping is the best, and say : “suppose we use that strategy”, there are many of them, and could be painful without a bit of programming.
But then, one may note that programming the calculus is the only reasonable way to tackle that kind of calculus ^_^. I could do it, but I’ll not bother to unless somone says he’s still interested in the matter !
27th September 2017 at 18:55 #4117Girl PowerParticipantStill interested, that would be fun to see!
28th September 2017 at 22:11 #4121JonasParticipantI have been trying to do that in Python, but it hits me that it was going to be really painful to chose the strategy on the fly. Well, I see how it could be done, but still painful. That’s why I tried pen&paper to chose the strategy : for 2turns, it’s a reasonable task.
Well, it turns out that if yo go for the 5reroll, you have *at least* ~9.7% to get 15 (assuming you chose the best strategy, which is keeping what you would keep normally), instead of a mere ~9.3%.“At least” because I only counted the “good” first rerolls (not the “very good” nor the “bad”) : the ones that have exactly 3 or exactly 4 dices on [1],[2] or [3]. The true chances are then above those 9.7% but it shouldn’t be much more.
I’ll finish the programming part later ^^.
29th September 2017 at 11:02 #4126Girl PowerParticipantOK, so to say it in short: The two approaches only differ by approximately ½% and rerolling all 5 dice is slightly better than rerolling two dice. I understand that this type of calculations should be done with a little programming. Anyhow, thanks so far!
5th April 2019 at 03:04 #4745Noel JohnsonParticipantHas anyone solved this?
5th April 2019 at 14:47 #4746Inverted DiceKeymasterYes, the answer to
It’s the last round and I only have the number 15 left to score. In my first roll, I get [6] [6] [6] [5] [4]. What should I do?
is that you should reroll all 5 dice instead of keeping the 3 sixes (according to the answer by Jonas above).
But nobody has calculated the exact probabilities, which would require some programming. 
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