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Yachtzee involving Mathematical Probabilities...

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Fenix
#1
Oct5-04, 08:57 PM
P: 9
First off, has anyone played this game before?

I haven't, but it would be of great help if someone who knows the game, could answer some questions. (I don't need anyone to solve anything. I solved all the computations, I just want confirmation on the rules, or if my reasoning is flawed.)

If you don't know about the game, but would still like to help, read on.

It's not a requirement, but I guess it would help.

Why do I need to know about this game?

Well, it's about a question about Yachtzee that can be misinterpreted in many ways.

Let me state how you start off Yachtzee to those who do not have a clue what the game is about.

First, you just simultaneously roll 5 dice, and then a bunch of numbers appear in the face value of all 5 dice.

What I am asking is, is order of the dice numbers important?

Ie. If you rolled a 1,2,3,3,4,5, could it also be interpreted as, 1,3,3,2,4,5, or 5,4,3,3,2,1, or 4,3,3,2,1,5?

If the previous sets are all the same with one another, then order is of no importance, but if they are important, then each previous set mentioned in the example is not considered the same with the other aforementioned set.

I think you get the point now.

I'm supposed to calculate the permutations (the possible outcomes) of all the outcomes.

If order IS important, the total outcome, or sample space would be 6^5 = (6x6x6x6x6) = 7776 outcomes.

But if order IS NOT important, then the total outcome would be 10!/(5!)(5!) = 252 total outcomes. (Note: 5! = 5x4x3x2x1)

So is order important in this game, or not?

Also, I got one last thing to say,

The probability of getting two pairs (order important) is 25/108 = 0.23148 = 23.15%.

Yet, the probability of getting two pairs (order not important) is 15/63 = 0.23809 = 23.81%.

The percentage of both probabilities is very close. Why is this? Wouldn't something as making the dice results order important or unimportant causes the two values to be different?

So which result is correct, and why? Also, why are the percentage of both probabilities of getting two pairs of order important and order unimportant so close?

What's your opinion on this matter?

PS: If the answers for the sample space, or probabilities mentioned in this post were wrong, I apologize, and I will try to remedy the problem immediately.

Thank you, Everyone.
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Hurkyl
#2
Oct5-04, 09:45 PM
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P: 16,099
The game doesn't care about order - 12345 is as much of a large straight as 51423.


The thing to realize is that, if you ignore order, you do not have a uniform distribution. The probability of a "good" outcome is not the number of good outcomes vs the total number of possible outcomes.

For example, the (unordered) result 12345 occurs with probability 1.5% whereas the (unordered) result 11111 occurs with probability .013%.


When computing probabilities, you should take order into account somewhere in the analysis, so you can take advantage of the fact the ordered outcomes are uniformly distributed. A typical approach might be to determine the # of outcomes ignoring order, then use that to get the # of outcomes not ignoring order.
Peter_F
#3
Oct7-04, 06:07 PM
P: 3
Quote Quote by R043482/Fenix
I've received other responses that talks about how when order is ignored
responses? :P

Quote Quote by R043482/Fenix
I've used your input and some others on the http://www.physicsforums.com
Some others? :P

Quote Quote by R043482/Fenix
I hope you weren't one of the people there who responded to the same question I asked
You mean that you hope I wasn't the only person to respond to the same question :P


I wondered why you were still asking me questions when you'd apparently got lots of help here. Hehe.


And no, I'm not Hurkyl :P
I used to visit these forums a bit, though I barely posted. I only came back here because you mentioned them.


EDIT> I just realised that I'm logged on with a new account which only had 1 post before this one.
I guess I forgot my password sometime and reregistered :P
I'm Asimir, in case you didn't realise.

Fenix
#4
Oct8-04, 11:00 PM
P: 9
Yachtzee involving Mathematical Probabilities...

Duh, it was obvious.

And you believe this is the only other forum, I've asked?

Well, you're right.

But I've asked many people in real life, mostly my friends, and other students about the problem. Not only that, but I've even tried an old mathematics teacher who once taught me in highschool, and a math professor about it.

So what do you have to say to that, Asimir?
Peter_F
#5
Oct9-04, 12:33 PM
P: 3
My response to that is
Quote Quote by R043482
I've used your input and some others on the http://www.physicsforums.com
:P

Also that we should stop spamming now :P


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