Street poker.

At school my friends and me liked to play our own games with dices.
One of the games we invented was to throw with 6 cubical dices, each numbered from 1 to 6.
You had to throw 6 different pokers (6 ones, 6 twos etc.) and a street. You were free to choose or change the sequence after any throw.
We never agreed about the best strategy.
How many throws does one need (average), assuming one follows the best strategy?

Fjord

Originally posted by fjord
At school my friends and me liked to play our own games with dices.
One of the games we invented was to throw with 6 cubical dices, each numbered from 1 to 6.
You had to throw 6 different pokers (6 ones, 6 twos etc.) and a street. You were free to choose or change the sequence after any throw.
We never agreed about the best strategy.
How many throws does one need (average), assuming one follows the best strategy?

Fjord

To get this clear - Is this how it works?:

a)You throw all 6 dice (this counts as 1 throw).
b)You then note any poker or street that has appeared.
c)You then rethrow as many of the dice as you want (this is your second throw), and you carry on repeating b and c until you have got all the special combinations.

Originally posted by iamatiger
To get this clear - Is this how it works?:

a)You throw all 6 dice (this counts as 1 throw).
b)You then note any poker or street that has appeared.
c)You then rethrow as many of the dice as you want (this is your second throw), and you carry on repeating b and c until you have got all the special combinations.

That is the way we played it.
Mind you, you are also allowed to pick up dices you left on the table during earlier throws.

E.g.
First throw: 1, 1, 2, 3, 4, 4.
You leave on the table: 1, 1,
Second throw: (1, 1) 3, 3, 3, 6
Now you may leave on the table: 3, 3, 3.

Or another variation for the second throw: 1, 1, 2, 3, 4, 5,
Now you may choose to pick up just one 1 and try to make a street.

Fjord

Originally posted by fjord
That is the way we played it.
Mind you, you are also allowed to pick up dices you left on the table during earlier throws.

E.g.
First throw: 1, 1, 2, 3, 4, 4.
You leave on the table: 1, 1,
Second throw: (1, 1) 3, 3, 3, 6
Now you may leave on the table: 3, 3, 3.

Or another variation for the second throw: 1, 1, 2, 3, 4, 5,
Now you may choose to pick up just one 1 and try to make a street.

Fjord

This is going to take some heavy analysis just to work out the best strategy. And maybe a computer program to work out the expected number of throws. Please be patient!

Here's the first part of the analysis:
If you throw 6 dice, the possible results are:

out of 46656 possible throws (in order of probability):

two pairs............16200 = 4 to a straight
Pair only............10800 = 5 to a straight
Three of a kind...7200 = 4 to a straight
full house...........7200 = 2 to a straight
four of a kind.....1800 = 3 to a straight
3 pairs...............1800 = 3 to a straight
straight.............720
pair and four......450 = 2 to a straight
2 threes.............300 = 2 to a straight
five of a kind......180 = 2 to a straight
six of a kind.......6 = 1 to a straight

Now the next things I (or someone else!) need to work out are:
When should you keep a partial straight (6 different)?
When should you keep a partial flush (6 of a kind)?

Originally posted by iamatiger
This is going to take some heavy analysis just to work out the best strategy. And maybe a computer program to work out the expected number of throws. Please be patient!

Here's the first part of the analysis:
If you throw 6 dice, the possible results are:

out of 46656 possible throws (in order of probability):

two pairs............16200 = 4 to a ...[text shortened]... you keep a partial straight (6 different)?
When should you keep a partial flush (6 of a kind)?

I will be patient ;-)
Just one question for the moment.
The probabilities you mention are right as long as you assume you will throw every time with six dice (sorry for the wrong plural I used in the previous posts; at school I learnt that 'mice' and 'lice' were the only irregular plurals of this kind).
What happens when your first throw is e.g. 1, 1, 2, 2, 3, and 4?
You will keep 1 and 1.
The second throw is (1, 1), 2, 2, 2 and 2.
Now you will keep 2, 2, 2, and 2.
It seems to me that this will change the number of probabilities you found. But maybe I'm wrong.

Fjord

Originally posted by fjord
I will be patient ;-)
Just one question for the moment.
The probabilities you mention are right as long as you assume you will throw every time with six dice (sorry for the wrong plural I used in the previous posts; at school I learnt that ...[text shortened]... umber of probabilities you found. But maybe I'm wrong.

Fjord

Yes, one of the the next things I will calculate is your "probablity of improving" from various configurations of kept dice - but it is complex because it depends on what you have already got - for instance if you already got your 6 twos, the twos in your above example would be no use to you. But before I can this I think that I need to decide, if you should keep 1,1 or 1,2,3,4,5 from a dice throw of 1,1,2,3,4,5.

Originally posted by iamatiger
Yes, one of the the next things I will calculate is your "probablity of improving" from various configurations of kept dice - but it is complex because it depends on what you have already got - for instance if you already got your 6 twos, the twos in your above example would be no use to you. But before I can this I think that I need to decide, if you should keep 1,1 or 1,2,3,4,5 from a dice throw of 1,1,2,3,4,5.

Let me give it a try.

According to your data the ratio of probabilities between throwing a flush of ones and a straight is 1 : 720. (1)
When you have got 1,1 on the table the probability of a flush of ones will be (after 46656 throws): 36
When you have got 1,2,3,4,5 on the table the probability of a straight will be (after 46656 throws): 7776.
So the ratio of probabilities is here: 36 : 7776 = 1 : 216. (2)

When we compare situation (1) and (2) we see that the probabilities for making a straight in relation to the probabilities of a flush of ones are diminishing from 720 to 216.
This means we should go for the flush of ones after a throw of 1,1,2,3,4,5.

Fjord

Originally posted by fjord
Let me give it a try.

According to your data the ratio of probabilities between throwing a flush of ones and a straight is 1 : 720. (1)
When you have got 1,1 on the table the probability of a flush of ones will be (after 46656 throws): 36
When you have got 1,2,3,4,5 on the table the probability of a straight will be (after 46656 throws): 7776.
So the rat ...[text shortened]... to 216.
This means we should go for the flush of ones after a throw of 1,1,2,3,4,5.

Fjord

You calculation looks fine up to "this means..." - but I don't understand why it means that! Another way of looking at it is if you keep the partial straight you have multiplied your chance of getting a straight on the next go by 10.8 times, but if you keep the partial flush you have only multiplied the chance of getting a flush on the next throw by 6 times.

What I am inclined to think is that you might not want to go for a straight because 1 1 2 3 4 5 would be much worse for you once you have your ones, and you might want to save the option of going for a straight until then. But I'm not sure of this.

Maybe I need to work out the maths for both options and hopefully it may then become apparent which one is working best.