please help w/ 'In Yahtzee, five dice are thrown. Show...'

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yaszine

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The problem is the following:

In the game of Yahtzee (or Yacht), five dice are thrown. Show that the probability of throwing a large straight (5 numbers in a row, the order does not matter) is 5/162.

Alternatively, you may solve the following, harder, problem: Show that the probability of throwing a small straight (4 numbers in a row) is 10/81. Do not count small straights which are also large straights.

I did the first part as follows:
The 5 numbers in a row can be either 65432 or 54321; thus the chances to get them will be 2(5*4*3*2*1)/6^5. But I don't really know how to represent my answer in a professional way.

For the second one, I have no clue how to do it.

P.S. We throw the dice only once, not like in the real game where we throw them twice (I think).

Thank you for your very needed help!
 
number of ways of getting {1,2,3,4,5} when order is not important is the number of ways you can rearrange those elements which is 5! Same goes for {2,3,4,5,6}.

And so the number of ways you can get a straight is 5!+5!.

The total number of possible dice outcomes is 6^5

Therefore the probability of a straight is \(\displaystyle \frac{5!+5!}{6^5}=\frac{5}{162}\)
 
thanks !! :d it looks very nice !!
have u got any idea on how i can start my second part of the problem ?!! :oops:
 
Doesn't "alternatively" mean "do this one OR that one, but both are not needed"?

Eliz.
 
The question is only asking you to do the first question. If you are ambitious you may do the second instead.
 
good then ! thanks ! i don't think i'm that imbitious ! i don't even know why i took this classe !! i only care about passing it ! but still the answer for the second question, if any knows how to do it, would be very wanted!! we never know if he'd put it in our midterm which is couple days ahead!!
thanks again for ur help!!
sincerly,
yassine!
 
possible small straights are:

1) 1,2,3,4,X (X cannot be 5)
2) 2,3,4,5,Z (Z cannot be 6 or 1)
3) 3,4,5,6,Y (Y cannot be 2)

Case 1) When X is an element of {1,2,3,4} then we have a tie. And so the number of ways we can arrange {1,2,3,4,X} is:
\(\displaystyle 4{5 \choose 2}3!\)
In the case where X=6, there are 5! different ways of rearranging {1,2,3,4,6}

And so the number of ways to get the first straight 1) is:

\(\displaystyle 4{5 \choose 2}3!+5!\)

If you do all the others similarily you will get the correct answer
 
i'm sorry ! but could you please explain how u got this line :
Case 1) When X is an element of {1,2,3,4} then we have a tie. And so the number of ways we can arrange {1,2,3,4,X} is...!!

thanks so much for your help!!
 
For example, when X=1, we have {1,2,3,4,1} We need to count the number of ways of rearranging this:

for X=2 we have {1,2,3,4,2} and so on.

The number of ways of rearranging {1,2,3,4,1} is the number of ways of picking 2 of the spots from 5 (and assigning 1 to them) multiplied by the number of ways to rearrange the remaining 3 spots.

The number of ways of choosing 2 from 5 is \(\displaystyle {5 \choose 2}\) and the number of ways of rearranging the remaining 3 spots is 3!

therefore the number of ways of rearranging {1,2,3,4,1} is \(\displaystyle 3!{5 \choose 2}\)

The calculations for X=1,2,3,4 are all similar and so the total number of ways in the cases when X=1 or X=2 or X=3 or X=4 is:

\(\displaystyle 4 \cdot 3!{5 \choose 2}\)

But there is also the case where X=6. in this case we need to count the number of ways of rearranging {1,2,3,4,6} this is easy its just 5!

The total number of ways we can get the first small straight is then \(\displaystyle 4 \cdot 3! {5 \choose 2}+5!\)

Does this make it more clear?
 
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