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gordonm888
gordonm888
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October 30th, 2020 at 2:41:59 PM permalink
I think everyone knows how to play Yahtzee (please google the rules if you don't). You have 13 rounds of rolling dice to make 13 objectives which are awarded with different number of points. In each round you make an initial roll, hold 0 to 5 dice and roll the remainder, then hold 0-5 dice and and proceed with the final roll. You are trying to maximize the number of points you achieve at the end of all 13 rolls.

The tension in Yahtzee game strategy arises from the need to make decisions about which objective to pursue after seeing the outcome of the 1st and 2nd rolls in any given round. There are so many possible situations in Yahtzee, that I don't think anyone has ever calculated optimum strategy for the entire game. There are lots of websites with general strategy principles/rules but some of the strategy decisions when you have only a few rolls left still seem to be relatively unanalyzed.

So, I am starting this thread because this is the best damn forum to take some bites out of the Yahtzee strategy problem. And, if nothing else, we can pose some interesting problems for the math geeks on this forum to analyze.

To get us started, here is some information I've calculated on what occurs on the initial roll of 5 dice.

Initial Roll Category
Permutations
Probability
Yahtzee (5oak)
6
0.0008
4 of a kind
150
0.0193
Full House
300
0.0386
3 of a kind + 2 Sing.
1200
0.1543
2 Pair + 1 Sing.
1800
0.2315
1 Pair +3 Sing.
3600
0.4630
No Pair
720
0.0926
TOTAL
7776
1
No Pair|Long Straight
240
0.0926
No Pair|Short Straight
240
0.0926
66 Pair|Short Straight
60
0.0077
55 Pair|Short Straight
120
0.0154
44 Pair|Short Straight
180
0.0231
33 Pair|Short Straight
180
0.0231
22 Pair|Short Straight
120
0.0154
11 Pair|Short Straight
60
0.0077
Total Short Straight
960
0.1235


And if you are chasing a single top category such as 6's, here are the probabilities I've calculated for the first roll.

First Roll
Permutations
Probability
Five 6s
1
0.0001
Four 6s
25
0.0032
Three 6s
250
0.0322
Two 6s
1250
0.1608
One 6
3125
0.4019
Zero 6s
3125
0.4019
TOTAL
7776
1


Here are some problems to get us started:

Problem #1: There is only one round left, and you need to fill the FULL HOUSE line. What is the probability of being successful?

Problem #2 There is only one round left, and you need to fill the LONG STRAIGHT line with either 65432 or 54321. Your initial roll, unluckily enough, is a Yahtzee: 11111. Do you hold one dice and roll four dice? Or do you reroll all five dice? What is the strategy for the third roll? What is the expected probability of making a LONG STRAIGHT with this initial roll?

Problem #3 There are three rounds left. You need to fill "THREES", "SIXES" and "FULL HOUSE". You need 27 points between the "THREES" and "SIXES" to make the 35 point bonus for the top categories. The FULL HOUSE category is 25 points.

You first roll is: 66333. What should you do?

Including the 35 point bonus as appropriate, what is the EV in points for the last three rounds if you decide to :
- keep all five dice as a FULL HOUSE?
- keep the two 6s and reroll hoping to roll more 6s
- keep the three 3s and reroll hoping to roll more 3s

(I never said these would be simple problems.)

Any general comments on the Yahtzee Strategy problem are welcome as well!
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
DRich
DRich
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gordonm888
October 30th, 2020 at 3:14:17 PM permalink
Do you have or have you read Olaf's Yahtzee book? He used to be a slot machine designer and BJ strategy guy.

Living longer does not always infer +EV
gordonm888
gordonm888
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October 30th, 2020 at 3:54:48 PM permalink
Quote: DRich

Do you have or have you read Olaf's Yahtzee book? He used to be a slot machine designer and BJ strategy guy.



No, I have never heard of it until this moment. I imagine that he has probably worked out all the expectations for most of the situational decisions that I am proposing.

I enjoy working on the mathematics of game strategies. Many of us have worked out the probabilities of many blackjack decisions even though books and the WOO site are available. The same is true for many casino games -many of us do the math ourselves. I was proposing that we do that for Yahtzee; if no one else is interested I'll just go ahead an do it myself unless I lose interest.
Last edited by: gordonm888 on Oct 30, 2020
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
FinsRule
FinsRule
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gordonm888
October 30th, 2020 at 3:56:59 PM permalink
Yahtzee is going to be hard. There’s so many scenarios when you figure in the top half bonus.
charliepatrick
charliepatrick
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gordonm888
October 30th, 2020 at 5:14:40 PM permalink
Quote: gordonm888

...Any general comments on the Yahtzee Strategy problem are welcome as well!

There's an interesting read at https://en.wikipedia.org/wiki/Yahtzee which describes some of the decisions. It looks a tough game to analyse.
gordonm888
gordonm888
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October 30th, 2020 at 5:54:35 PM permalink
Quote: charliepatrick

There's an interesting read at https://en.wikipedia.org/wiki/Yahtzee which describes some of the decisions. It looks a tough game to analyse.



Thanks Charlie. I also found this viewchart presentation from the Netherlands Optimal YAHTZEE strategy . Its slow to get going, but the second half of it has some interesting analysis.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
gordonm888
gordonm888
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charliepatrick
October 30th, 2020 at 6:17:45 PM permalink
Here is some statistical information I found for the outcomes (without the Extra Yahtzee Bonus) when its your last turn and these are the single remaining Yahtzee hand category you are targeting.

LAST TURN Category
Expected Value
Prob. of Zero Pts, %
Aces
2.11
6.49
Twos
4.21
6.49
Threes
6.32
6.49
Fours
8.43
6.49
Fives
10.53
6.49
Sixes
12.64
6.49
3 of a kind
15.19
28.76
4 of a kind
5.61
72.26
Full House
9.15
63.39
Short Straight
18.48
38.40
Long Straight
10.61
73.47
Yahtzee
2.3
95.40
Chance
23.33
0.00


This is from the link I referenced in my previous post.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
unJon
unJon
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October 30th, 2020 at 6:40:16 PM permalink
Amazing thread. Thanks for starting it.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
charliepatrick
charliepatrick
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gordonm888
October 31st, 2020 at 2:45:52 AM permalink
Quote: gordonm888

Here is some statistical information...when its your last turn...

I can see the recursive approach, similar to how to do BJ. Also I noticed that One thru Sixes are always going to be X thru 6X, since it's just a matter of how many of "what you want" you get multiplied by the value.

So I guess at N options to go you look at the {throw,# rolls left} and which dice to keep. Then like BJ look at where you're at and where you might land up and work out the EVs (or expected score). I suspect once you get going (programming wise) it's a matter of cranking a very big handle.

I feel a challenge looming!
gordonm888
gordonm888
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October 31st, 2020 at 11:00:14 AM permalink
Quote: charliepatrick

I can see the recursive approach, similar to how to do BJ. Also I noticed that One thru Sixes are always going to be X thru 6X, since it's just a matter of how many of "what you want" you get multiplied by the value.

So I guess at N options to go you look at the {throw,# rolls left} and which dice to keep. Then like BJ look at where you're at and where you might land up and work out the EVs (or expected score). I suspect once you get going (programming wise) it's a matter of cranking a very big handle.

I feel a challenge looming!



Parts of the problem involving the three rolls in a given round can be addressed by Markhov chains e.g., pursuing 6's, i.e. whenever you have one 6 you will roll four dice and the outcome probabilities for one, two, three, four, five 6's are always the same; when you start a roll with 2 sixes you will roll 4 dice and there will be another set of transiotion probabilities for two, three, four and five dice, etc.

The problem comes when you are rolling for Yahtzee or 4oak and you get say, two 6s on the first roll, but the second roll is three 4's, then you switch to fours. The rolls then switch to looking for 4s.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

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