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Let’s say I’m playing Yahtzee and my aim is to get any two pair, the actual numbers don’t matter.

As usual, I can save any dice I want and re-roll the rest. Let’s say that my hand contains the following: {1, 3, 3, 5, 6}

I have 3 options:

  1. Re-roll everything and hope to get any two pair.
  2. Save the two 3s and roll the rest. Rolling three dice and hoping to get any pair.
  3. Save the two 3s as well as any other random number, let’s say the 5. Now rolling only two dice hoping to get a 5 or any other new pair.

Which of these strategies maximizes the probability of getting a two pair?

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    Optimal as in just getting 2 pairs or as well in maxing your points?
    – Zibelas
    Apr 25, 2022 at 11:52
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    @Zibelas Just getting two pairs. Ignoring points.
    – hb20007
    Apr 25, 2022 at 11:53
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    Do you wish to maximize the chance of getting two pair, even if it reduces the chance of getting an "even better" result like a full house or Yahtzee? My gut feel is that the difference between (2) and (3) comes down to the chance of getting 3 or more 3s - e.g. if you keep the 5, you have zero chance of getting a Yahtzee. Apr 25, 2022 at 12:36
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    @PhilipKendall That's a good question. For the sake of simplicity, let's just say that I want to maximize the chance of getting two pair even if it reduces the chance of getting a better result.
    – hb20007
    Apr 25, 2022 at 12:58

1 Answer 1

5

This is probably closer to being suited to the mathematics SE, but let's go through it.

Also, funnily enough as far as I can tell there is no scoring option for "Two Pairs" in Yahtzee as officially released by Hasbro. In Yatzy or Yacht, which are common variants released elsewhere, the values of the pairs must be distinct values, but the fifth die may be a copy of one of the pairs (i.e. you can score a Full House as Two Pairs, but not Four of a Kind). I will be working on that assumption here.

Rolling all

If you roll all 5 dice, then there are 6^5 = 7776 possible rolls.

To make Two Pair with an off-value 5th die, there are 6 ways to choose the off-value, and 10 possible choices for the values of the pairs. The off-value die can be any of 5 choices, and then there are 6 ways the two pairs could be placed among the other 4 dice. Hence there are 6x10x5x6 = 1800 ways to roll Two Pair with an off-value.

In addition, to make a Full House you have 6 choices of value for the triple, and 5 choices for the double (a Yatzy can't be scored as Two Pair). There are 10 ways to choose which dice form the double, so there are 300 ways to roll a Full House.

Hence, there are 1800+300=2100 ways to roll Two Pair, and a probability of 2100/7776 = 27.0% that you will do so.

Keeping the pair

If you just keep the pair of 3s, you're rolling 3 dice (of which there are 6^3 = 216 combinations) and hoping for either:

  1. 3 of the same value, not equal to 3.
  2. A pair not equal to 3, and a value not equal to it (but possibly equal to 3).

For (1), there are 5 possible values for the triple, and for each of those there is exactly one way to roll it (since 111 can't be re-arranged to a different roll).

For (2), there are 5 possible values for the pair, and 5 possible values for the off-value (including 3 but not the new pair), and there are 3 dice which could hold the off-value, so there are 75 different rolls.

Hence there are 80 rolls to get Two Pair, and a probability of 80/216 = 37.0%.

Keeping the pair + 1

In this case, you're rolling 2 dice (and there are 6^2 = 36 possible rolls), and to get your Two Pair you must roll a 5 on at least one die, with the other value being anything you want. There are lots of ways to calculate this, but there are 11 ways for this to happen (e.g. 6 ways to roll 5 on the first die, plus 6 ways to roll it on the second, but subtract one because we've double-counted the roll of 55). So the probability is 11/36 = 30.5%

This makes sense, since you're saving the "progress" of having already rolled one of the needed pairs (just as in the second option), but you're also restricting yourself to only allowing one possible value of second pair show up.

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    I found this article helpful in understanding your calculations: datagenetics.com/blog/january42012
    – hb20007
    Apr 26, 2022 at 9:57
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    "there are 5 possible values for the pair, and 5 possible values for the off-value (including 3 but not the new pair" No, once you choose one of them, there are 4 for the other. "This makes sense" No, that doesn't make sense. If you roll an extra die, it can either come up 3 (which is bad), or come up as something that restricts your other rolls just as much as saving one of the dice. For instance, if the die comes up 2, that restricts you as much as if you had saved a 2. So saving a die can't possibly make your chances worse. Apr 27, 2022 at 3:37
  • @Acccumulation I definitely have messed something up in my calculations, I will try to re-do them more thoroughly (I admit I was typing this up rather quickly at the time).
    – ConMan
    Apr 27, 2022 at 23:25
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    Shouldn't "Keeping the pair +1" have slightly higher odds? You could also roll a pair with the 2 dice: 11, 22, 44 or 66, which would all also work to form two pairs together with the already picked 33.
    – zo0x
    May 3, 2022 at 13:50

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