You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
yahtzee large straight
- Thread starter bobbixler
- Start date
D
Deleted member 4993
Guest
What is the probability of a large straight in one turn (3 throws) at Yahtzee? Assume you are going for a large straight.
First define large straight.
Next how many ways can we get large straight in Yahtzee?
Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
Yahtzee rules
In Yahtzee one turn consists of up to 3 rolls of the dice. You roll 1-5 dice at once (your choice) on each of the 3 rolls. A large straight consists of either rolling a 12345 or 23456 after your 1-3 rolls. Order of the dice is not important. What is the probability of doing this assuming that is your goal?
In Yahtzee one turn consists of up to 3 rolls of the dice. You roll 1-5 dice at once (your choice) on each of the 3 rolls. A large straight consists of either rolling a 12345 or 23456 after your 1-3 rolls. Order of the dice is not important. What is the probability of doing this assuming that is your goal?
Hello, bobbixler!
This problem is more elaborate that I had anticipated.
I can give you a start on the procedure . . . Let \(\displaystyle LS\) = "Large Straight".
First, consider only this LS: "1-2-3-4-5".
We could get it in one roll: .\(\displaystyle P(\text{1 roll}) \,=\,\left(\frac{1}{6}\right)^5\)
We could get it in two rolls.
There are 4 cases to consider.
4 numbers are in the LS: .\(\displaystyle {5\choose4}(\frac{1}{6})^4(\frac{5}{6})\)
We want the 5th: .\(\displaystyle \left(\frac{1}{6}\right)\)
. . \(\displaystyle P(\text{LS with 4 numbers}) \:=\:{5\choose4}(\frac{1}{6})^5(\frac{5}{6})\)
3 numbers are in the LS: .\(\displaystyle {5\choose3}(\frac{1}{6})^3(\frac{5}{6})^2\)
We want the 4th and 5th: .\(\displaystyle (\frac{1}{6})^2\)
. . \(\displaystyle P(\text{LS with 3 numbers}) \:=\:{5\choose3}(\frac{1}{6})^5(\frac{5}{6})^2\)
2 numbers are in the LS: .\(\displaystyle {5\choose2}(\frac{1}{6})^2(\frac{5}{6})^3\)
We want the 3rd, 4th and 5th: .\(\displaystyle (\frac{1}{6})^3\)
. . \(\displaystyle P(\text{LS with 2 numbers}) \:=\:{5\choose2}(\frac{1}{6})^5(\frac{5}{6})^3 \)
1 number is in the LS: .\(\displaystyle {5\choose1}(\frac{1}{6})(\frac{5}{6})^4\)
We want the 2nd, 3rd, 4th and 5th: .\(\displaystyle (\frac{1}[6})^4\)
. . \(\displaystyle P(\text{LS with 1 number}) \:=\:{5\choose1}(\frac{1}{6})^5(\frac{5}{6})^4\)
At this point, my brain turned to tapioca . . .
What does it mean "One number is in the Large Straight"?
If we rolled 3-6-6-6-6, we'd keep the 3 and reroll the 6's.
But it we rolled 1-1-2-2-6, which die is the "one"?
Wouldn't we keep a 1 and a 2 and reroll the other 3 dice?
And is there another case? ."No numbers are in the LS."
. . A player can reject all five dice and reroll all of them.
But would he?
What is the probability of "No numbers are in the LS" ?
. . Since the player will reject the roll, the probability is \(\displaystyle 1.\)
But he wouldn't reroll if he got a LS on his first roll.
And if he got 1-2-3-4-6, I'm sure he'd reroll the 6 only.
I need a nap . . .
This problem is more elaborate that I had anticipated.
I can give you a start on the procedure . . . Let \(\displaystyle LS\) = "Large Straight".
What is the probability of a Large Straight in one turn (3 throws) at Yahtzee?
Assume you are going for a Large Straight.
First, consider only this LS: "1-2-3-4-5".
We could get it in one roll: .\(\displaystyle P(\text{1 roll}) \,=\,\left(\frac{1}{6}\right)^5\)
We could get it in two rolls.
There are 4 cases to consider.
4 numbers are in the LS: .\(\displaystyle {5\choose4}(\frac{1}{6})^4(\frac{5}{6})\)
We want the 5th: .\(\displaystyle \left(\frac{1}{6}\right)\)
. . \(\displaystyle P(\text{LS with 4 numbers}) \:=\:{5\choose4}(\frac{1}{6})^5(\frac{5}{6})\)
3 numbers are in the LS: .\(\displaystyle {5\choose3}(\frac{1}{6})^3(\frac{5}{6})^2\)
We want the 4th and 5th: .\(\displaystyle (\frac{1}{6})^2\)
. . \(\displaystyle P(\text{LS with 3 numbers}) \:=\:{5\choose3}(\frac{1}{6})^5(\frac{5}{6})^2\)
2 numbers are in the LS: .\(\displaystyle {5\choose2}(\frac{1}{6})^2(\frac{5}{6})^3\)
We want the 3rd, 4th and 5th: .\(\displaystyle (\frac{1}{6})^3\)
. . \(\displaystyle P(\text{LS with 2 numbers}) \:=\:{5\choose2}(\frac{1}{6})^5(\frac{5}{6})^3 \)
1 number is in the LS: .\(\displaystyle {5\choose1}(\frac{1}{6})(\frac{5}{6})^4\)
We want the 2nd, 3rd, 4th and 5th: .\(\displaystyle (\frac{1}[6})^4\)
. . \(\displaystyle P(\text{LS with 1 number}) \:=\:{5\choose1}(\frac{1}{6})^5(\frac{5}{6})^4\)
At this point, my brain turned to tapioca . . .
What does it mean "One number is in the Large Straight"?
If we rolled 3-6-6-6-6, we'd keep the 3 and reroll the 6's.
But it we rolled 1-1-2-2-6, which die is the "one"?
Wouldn't we keep a 1 and a 2 and reroll the other 3 dice?
And is there another case? ."No numbers are in the LS."
. . A player can reject all five dice and reroll all of them.
But would he?
What is the probability of "No numbers are in the LS" ?
. . Since the player will reject the roll, the probability is \(\displaystyle 1.\)
But he wouldn't reroll if he got a LS on his first roll.
And if he got 1-2-3-4-6, I'm sure he'd reroll the 6 only.
I need a nap . . .
program solution ???
Hi, I'm woking on this problem...... may I ask how you implement a program to solve this?
Did you just simply randomly throw the dice? and count the number of large straight?
Thanks!!!!!!!!!!!
As no one has given an answer I went ahead and wrote the code to solve the problem. The answer is about .249.
Hi, I'm woking on this problem...... may I ask how you implement a program to solve this?
Did you just simply randomly throw the dice? and count the number of large straight?
Thanks!!!!!!!!!!!