# Dice probability for Yahtzee large straight

I had a discussion with a friend of mine about what is the best course of action in the following Yahtzee scenario:

The first throw with 5 dice results in the following values:

1, 3, 4, 5, X, with X = 3, 4 or 5


Based on probability, what will be the best decision to make to end up with a large straight (1-2-3-4-5 or 2-3-4-5-6):

1. Try to throw a 2 with the X die in the remaining two throws
2. Take the 1 and the X dice to throw a 1 and a 2 or a 2 and a 6 in the remaining two throws.

I claim the first option has the best probability to succeed while my friend claims it is the second option.

I tried calculating the probabilities, but it has been a while since I followed my probability classes, so I am not so sure. For the first option I believe the probability is 1/6 + (5/6 * 1/6) = 11/36. But what about the second option? What is the correct way to calculate this probability?

• The goal is to get five sequential dice, that is what I meant with a large straight. In Yahtzee you also have the small straight, which is four sequential dice. Nov 21, 2011 at 20:49
• OK, it appears that rules are different in different countries here. Sorry about that. The game I play has a small straight 1-2-3-4-5 and a large straight 2-3-4-5-6.
– NRH
Nov 21, 2011 at 20:58
• Ok, so to make things clear, in my scenario both 1-2-3-4-5 and 2-3-4-5-6 are considered a success. I'll edit it into my question. Nov 21, 2011 at 21:00
• I showed how to obtain an exact answer to questions like this in the course of addressing a very similar question about how to end up with a Yahtzee: stats.stackexchange.com/questions/154519. The code given there would need to be modified to handle the question about large straights, because its data structures do not distinguish straights from any other combination of five unique values.
– whuber
Dec 13, 2016 at 19:08

Lets take it one throw at the time. Then we have the two questions

• What is the probability of getting a 2 in one throw with one die? (clearly $1/6$)
• What is the probability of getting {1,2} or {2,6} in one throw with two dice?

The latter is slightly more complicated, but certainly not impossible. There are 36 possible outcomes, 2 of those result in {1, 2} and 2 result in {2, 6}. Thus the probability is $4/36 = 1/9$ of getting {1,2} or {2,6}.

One way to compute the probability correctly of getting a 2 in one of two throws is to look at the complement, which is to not get a 2 in both of the two throws. Because the latter is an intersection of two events that are independent (not getting a 2 in the first throw and not getting a two in the second throw), the probability is the product of the probabilities of the single events. Thus, the probability of not getting a 2 in both of the two throws is $(5/6) \times (5/6) = 25/36$, and the probability of getting a 2 in one of the two throws is $1 - 25/36 = 11/36 \simeq 0.3056$.

The same argument can be used for the second strategy. If we throw two dice two times the probability of getting {1,2} or {2,6} in one of the throws is $$1 - (8/9) \times (8/9) = 1 - 64/81 = 17/81 \simeq 0.2099.$$

Thus if the goal is to get a straight it is better to throw the single die twice to get a 2 than to throw two dice twice to get {1,2} or {2,6}. This is actually clear from the first computation of the probabilities 1/6 and 1/9, respectively, of being successful in just one throw. If you play a game twice and the probability of success in one game is $p$, the probability of success in one of the two throws is, using the same argument as above,

$$1 - (1-p)^2 = 1 - 1 - p^2 + 2p = p(2-p)$$

and this is an increasing function of $p$.

An alternative way to derive the formula above, which I believe is the argument the OP uses, is to decompose the event of success in one of the throws into the disjoint union of the event of success in the first throw and the event of not success in the first throw combined with success in the second throw. The former event has probability $p$ and independence gives that the latter event has probability $(1-p)p$. Since the probability of disjoint events is the sum of the probabilities of the single events we get the same probability

$$p + (1-p)p = p(2-p)$$

of success in one of the throws.

Edit: This was added later to clarify what happens with the following mixed strategy: You throw two dice, then either you get {1,2} or {2,6}, or you get one of 1,2 or 6 but not "the other", and throw the remaining single die, or you get none of 1,2 and 6 and you throw two dice. If we write down a 6 by 6 table with all possible combinations of the first throw of two dice we can count that there are 4 possibilities that gives the first situation, 23 that gives the second and 9 that gives the third. Given that we are in the second situation we have success if we get the last die right (which has probability 1/6) and given the third situation we have success if the final throw with two dice is successful (which has probability 1/9). By this decomposition into disjoint events according to the first throw of two dice we get the probability of success $$1/9 + (23/36) \times (1/6) + (1/4) \times (1/9) \simeq 0.2454.$$

• Does your second computation account for the case where you throw the first value of one of the pairs ({1,2}, {2,6}) in the first throw with two dice, and you throw the second value of the pair with the single die in the third throw? Because that is also allowed in the Yahtzee rules we use. Nov 21, 2011 at 20:59
• BTW, you are right with your statement about how I reasoned to get the 11/36 probability for the first option. However, this line of reasoning gets a bit more complex for the second option. Nov 21, 2011 at 21:09
• @Jan-Henk, no, that is a mixed strategy that I did not consider. The answer considers only the "pure" strategy of either throwing two dice or one die both times. The rationale was that the mixed strategy can not be better than the best pure strategy. Specifically, if you get one of 1, 2 or 6 in the first throw of two dice you are not in a better position than you started out with, but have only one throw left. If it is a good idea at this point to change strategy and only throw one die, it would have been a better idea to throw only one die from the beginning.
– NRH
Nov 21, 2011 at 21:31
• Ok, thx for your answers. You confirmed my own line of reasoning, only you are far better at expressing it all :) Nov 21, 2011 at 21:35
• @Jan-Henk, I have added the actual computation of the probability for a mixed strategy. Now it is crystal clear that the mixed strategy falls in-between the two pure strategies.
– NRH
Nov 21, 2011 at 22:04

The method your friend is suggesting is one that I use sometimes when I am playing Yahtzee. Your friend is wrong to assert that this method has a higher probability of giving a high straight, but it does have the benefit of giving you a substantial probability of getting a low straight if you fail to get the high straight (which is impossible with the first method). It therefore represents a good low-risk strategy in cases where you do not already have a low straight, since it gives a "fall back position" if the high straight does not accrue. Below I show the probabilties of these outcomes.

Method 1: In this method you reroll the X die on each roll in an attempt to get a 2 to complete the high (large) straight. For each roll you have 6 possible outcomes and 1 of these leads to a high straight. So the probability of getting a high straight with this method is:

\begin{align} \mathbb{P}(\text{High Straight}|\text{Method 1}) &= 1- (1-\tfrac{1}{6})^2 \\[6pt] &= \tfrac{11}{36} \\[6pt] &= 0.3055556, \\[6pt] \mathbb{P}(\text{Low Straight}|\text{Method 1}) &= 0. \\[6pt] \end{align}

Bear in mind that with this method there is no good "fall back" option; if you do not get the high straight then you also do not have a low straight, and the best you will have is a pair.

Method 2: In this method you reroll the 1 and X dice in the first roll. In this case I will assume that if you do not get a high straight on the second roll then you keep a 2 if you rolled this, but reroll all other numbers. On the second roll you have 36 possible outcomes and 4 of these leads to a high straight (the outcomes 1,2, 2,1, 2,6 and 6,2). If you do not complete the high straight on this roll then there are another 7 outcomes where you get a 2 that you will keep for the third roll (the outcomes 2,2, 2,3, 3,2, 2,4, 4,2, 2,5 and 5,2) and 25 outcomes where you don't get a 2. So the probability of getting a high straight or a low straight with this method is:

\begin{align} \mathbb{P}(\text{High Straight}|\text{Method 2}) &= \tfrac{4}{36} + \tfrac{7}{36} \cdot \tfrac{2}{6} + \tfrac{25}{36} \cdot \tfrac{4}{36} \\[6pt] &= \tfrac{144}{1296} + \tfrac{84}{1296} + \tfrac{100}{1296} \\[6pt] &= \tfrac{328}{1296} \\[6pt] &= \tfrac{41}{162} \\[6pt] &= 0.2530864, \\[6pt] \mathbb{P}(\text{Low Straight}|\text{Method 2}) &= \tfrac{7}{36} \cdot \tfrac{4}{36} + \tfrac{25}{36} \cdot \tfrac{16}{36} \\[6pt] &= \tfrac{28}{1296} + \tfrac{400}{1296} \\[6pt] &= \tfrac{428}{1296} \\[6pt] &= \tfrac{107}{324} \\[6pt] &= 0.3302469. \\[6pt] \end{align}

As you can see, this method has a lower probability of giving you a high straight, but it gives you a substantial probability of the "fall back" position of a low straight. Thus, in cases where you do not alreaedy have a low straight, this method may be a better option.

1/9+(23/36)×(1/6)+(1/4)×(1/9)≃0.2454.

Please correct me if I'm wrong, but there is a flaw in your calculation when you only get one of the required numbers to complete the straight. By just looking at the (23/36)×(1/6) portion, if you roll (2,2)(2,3)(2,4)(2,5) you actually have a 1/3 chance of completing the straight on the subsequent roll.

So wouldn't the answer be : 1/9+(16/36)×(1/6)+(7/36)x(1/3)+(1/4)×(1/9) = 10/36

As I am trying to create a bot for a yatzee game I came across this and thought I'll give my 2 cents.

I usually take the 1 and x and keep 3,4,5.

You have 2 throws to accomplish the large straight.

In the first throw you need either 1,2 or 2,6. With 6X6 there are 36 combinations of wich 4 are the once you exactly need. There is a small chance that you'll get it exactly (4/36)=(1/9) to get: (1,2),(2,1)(2,6)(6,2)

There is however a good change you'll get a 2, any 2 will do, the chance is (11/36). These are better odds than taking the x and try for a 2 (1/6). If you have a 2 then the odds for the 2nd throw double as you now have one dice and need either a 1 or a 6 (2/6).

There is also a good chance to get a 6 or a 1: (22/36), if you have no 2 but have the 6 or 1 then you're second turn has the same odds as initially taking the x and fail in the first try.

Not sure what the calculations are but I get a slightly bigger chance only for not keeping the 1 and throw both 1 and X in the first roll. Here are the calculations:

Throw both 2,6 or 1,2 in either first or second throw

1/9 or 1/9 = 2/9

Throw a 2 in the first and either 1 or 6 in the second. Remove (2,6),(6,2),(1,2) and (2,1) in first throw: that case is mentioned above.

7/36 and 2/6 => 7/36 and 1/3 => (7+1)/(36*3) = 8/108

Throw a 1 or 6 in the first and 2 or 6 in the last. Remove (2,6),(6,2),(1,2), (2,1) because they are used in previous scenarios

12/36 and 1/6 => (12+1)/(216) = 13/216

Chance of success: 2/9 + 8/108 + 17/216 => 48/216 + 16/216 + 13/216

77/216 is about 0.356481 This is probably wrong though because I am not sure how to remove combinations used in previous scenarios from further scenarios second roll.

Keep the 2 go for a 1 1/6 or 1/6 = 2/6 => 1/3

These are the only options you have so about 0.333333

you must have the 2 but otherwise can use either a 1 or 6. if you roll two dice on the 2nd roll you have a 1/3 chance of getting a 2 instead of a 1/6 chance. on the third roll:
A) If you get the 2 on the 2nd roll you have a 1/3 chance of rolling either a 1 or 6. B) If you get a 1 or 6 on the 2nd roll you have a 1/6 chance of rolling a 2. C) However if you do not get a 1,2 or 6 on your 2nd roll you now only have a 1/9 chance of getting both a 2 and also a 1 or 6.

worse case scenario rolling 2 dice adding 1/3 and 1/9 = 4/9 (compared to adding 1/6 and 1/6 = 1/3)

Via brute force, trying these strategies 100,000 times each, it seems that if you keep the 1345 and roll twice for the 2, you win 30% of the time. But if you keep the 345 and roll for either the 12 or the 26 in two rolls, you only win about 19% of the time.

I'm not a probabilities guy, but that's the result I'm seeing.

• Oops, had a problem in my code. Fixed that, and now the 345 scenario wins 26% of the time, which is more in line with NRH's calculations.
– Dave
Dec 13, 2016 at 17:52