Probability of rolling a straight in Yahtzee
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I am trying to compute a very specific probability given a game of Yahtzee.
Here are the rules:
- There are 5 fair, 5-sided dice with equal probability of rolling each number.
- You have 3 rolls to obtain a straight [1,2,3,4,5].
- Per Yahtzee rules, you can choose which dice to reroll, in order to obtain a straight. (ie, if on roll 1 you get [1,2,3,3,4], you can keep [1,2,3,4] and reroll 1 dice)
- Your first roll must have a 3 of a kind (ie: [1,2,3,3,3] or [1,1,3,3,3]
- First rolls with 4 of a kind, 5 of a kind, or a perfect straight DO NOT count (ie: [3,3,3,3,4])
I am trying to write this in terms of mathematic notation. My intuition is telling me that this is a conditional probability, where
$P(C)$ = probability that we get a 3 of a kind on roll 1
$P(A)$ = probability that we roll a straight in 3 rolls
$P(A mid C)$ = probability that we roll a straight in 3 rolls, given we get a 3 of a kind on roll 1
$$P(A mid C) = frac{P(A {cap} C)}{P(C)}$$
I am worried my intuition is incorrect. I ran simulations (verified correct, 20000 trials) that say the probability of getting 3 of a kind on the first roll is roughly 20.06%, and the probability of getting a straight, given you get 3 of a kind on the first roll, is roughly 4.34%.
However, plugging in these values to my conditional probability results in:
$$P(A mid C) = frac{0.0434}{0.2006} = 0.2163$$
I am confused - isn't $P(A mid C)$ the probability of getting a straight in 3 rolls, given you roll a 3 of a kind on the first try?
Any guidance would be very helpful - I wrote code simulations first, and now need to write a general verification of my result (0.0434) by hand.
Thanks!
probability dice conditional-probability
edited Dec 4 '18 at 9:15
Tki Deneb
32710
asked Dec 4 '18 at 8:12
Megan ByersMegan Byers
2616
$endgroup$
add a comment |
$begingroup$
I am trying to compute a very specific probability given a game of Yahtzee.
Here are the rules:
- There are 5 fair, 5-sided dice with equal probability of rolling each number.
- You have 3 rolls to obtain a straight [1,2,3,4,5].
- Per Yahtzee rules, you can choose which dice to reroll, in order to obtain a straight. (ie, if on roll 1 you get [1,2,3,3,4], you can keep [1,2,3,4] and reroll 1 dice)
- Your first roll must have a 3 of a kind (ie: [1,2,3,3,3] or [1,1,3,3,3]
- First rolls with 4 of a kind, 5 of a kind, or a perfect straight DO NOT count (ie: [3,3,3,3,4])
I am trying to write this in terms of mathematic notation. My intuition is telling me that this is a conditional probability, where
$P(C)$ = probability that we get a 3 of a kind on roll 1
$P(A)$ = probability that we roll a straight in 3 rolls
$P(A mid C)$ = probability that we roll a straight in 3 rolls, given we get a 3 of a kind on roll 1
$$P(A mid C) = frac{P(A {cap} C)}{P(C)}$$
I am worried my intuition is incorrect. I ran simulations (verified correct, 20000 trials) that say the probability of getting 3 of a kind on the first roll is roughly 20.06%, and the probability of getting a straight, given you get 3 of a kind on the first roll, is roughly 4.34%.
However, plugging in these values to my conditional probability results in:
$$P(A mid C) = frac{0.0434}{0.2006} = 0.2163$$
I am confused - isn't $P(A mid C)$ the probability of getting a straight in 3 rolls, given you roll a 3 of a kind on the first try?
Any guidance would be very helpful - I wrote code simulations first, and now need to write a general verification of my result (0.0434) by hand.
Thanks!
probability dice conditional-probability
edited Dec 4 '18 at 9:15
Tki Deneb
32710
asked Dec 4 '18 at 8:12
Megan ByersMegan Byers
2616
$endgroup$
1
$begingroup$
I think you are confusing $P(A cap C)$ and $P(A | C)$ in the calculation after your simulation. The numerator ($P(A cap C)$) shouldn't be $0.0434$, but the number of times you got $A$ and $C$, divided by the total amount of simulations. The result should then be $0.0434$. For the mathematical calculation of this probability, you should be more precise on how we choose which dice to reroll after each round.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:19
$begingroup$
@TkiDeneb Thank you - this helped clear some stuff up. Is it right to say that this is question is asking a conditional probability? Do you think that testing for the probability of A (get a straight) given C (get 3 of a kind on first roll) is the right verbal terminology?
$endgroup$
– Megan Byers
Dec 4 '18 at 19:11
1
$begingroup$
That depends on what you want. If you are looking for the probability to get $3$ of a kind in the first roll and then a straight in any attempt, you want $P(A cap C)$. (Imagine placing yourself before the first roll.) If you are looking for the probability to get a straight after you have already gotten $3$ of a kind, then you want $P(A mid C)$. (Imagine placing yourself after a first roll that had $3$ of a kind.)
$endgroup$
– Tki Deneb
Dec 4 '18 at 19:56
add a comment |
$begingroup$
I am trying to compute a very specific probability given a game of Yahtzee.
Here are the rules:
- There are 5 fair, 5-sided dice with equal probability of rolling each number.
- You have 3 rolls to obtain a straight [1,2,3,4,5].
- Per Yahtzee rules, you can choose which dice to reroll, in order to obtain a straight. (ie, if on roll 1 you get [1,2,3,3,4], you can keep [1,2,3,4] and reroll 1 dice)
- Your first roll must have a 3 of a kind (ie: [1,2,3,3,3] or [1,1,3,3,3]
- First rolls with 4 of a kind, 5 of a kind, or a perfect straight DO NOT count (ie: [3,3,3,3,4])
I am trying to write this in terms of mathematic notation. My intuition is telling me that this is a conditional probability, where
$P(C)$ = probability that we get a 3 of a kind on roll 1
$P(A)$ = probability that we roll a straight in 3 rolls
$P(A mid C)$ = probability that we roll a straight in 3 rolls, given we get a 3 of a kind on roll 1
$$P(A mid C) = frac{P(A {cap} C)}{P(C)}$$
I am worried my intuition is incorrect. I ran simulations (verified correct, 20000 trials) that say the probability of getting 3 of a kind on the first roll is roughly 20.06%, and the probability of getting a straight, given you get 3 of a kind on the first roll, is roughly 4.34%.
However, plugging in these values to my conditional probability results in:
$$P(A mid C) = frac{0.0434}{0.2006} = 0.2163$$
I am confused - isn't $P(A mid C)$ the probability of getting a straight in 3 rolls, given you roll a 3 of a kind on the first try?
Any guidance would be very helpful - I wrote code simulations first, and now need to write a general verification of my result (0.0434) by hand.
Thanks!
probability dice conditional-probability
edited Dec 4 '18 at 9:15
Tki Deneb
32710
asked Dec 4 '18 at 8:12
Megan ByersMegan Byers
2616
$endgroup$
I am trying to compute a very specific probability given a game of Yahtzee.
Here are the rules:
- There are 5 fair, 5-sided dice with equal probability of rolling each number.
- You have 3 rolls to obtain a straight [1,2,3,4,5].
- Per Yahtzee rules, you can choose which dice to reroll, in order to obtain a straight. (ie, if on roll 1 you get [1,2,3,3,4], you can keep [1,2,3,4] and reroll 1 dice)
- Your first roll must have a 3 of a kind (ie: [1,2,3,3,3] or [1,1,3,3,3]
- First rolls with 4 of a kind, 5 of a kind, or a perfect straight DO NOT count (ie: [3,3,3,3,4])
I am trying to write this in terms of mathematic notation. My intuition is telling me that this is a conditional probability, where
$P(C)$ = probability that we get a 3 of a kind on roll 1
$P(A)$ = probability that we roll a straight in 3 rolls
$P(A mid C)$ = probability that we roll a straight in 3 rolls, given we get a 3 of a kind on roll 1
$$P(A mid C) = frac{P(A {cap} C)}{P(C)}$$
I am worried my intuition is incorrect. I ran simulations (verified correct, 20000 trials) that say the probability of getting 3 of a kind on the first roll is roughly 20.06%, and the probability of getting a straight, given you get 3 of a kind on the first roll, is roughly 4.34%.
However, plugging in these values to my conditional probability results in:
$$P(A mid C) = frac{0.0434}{0.2006} = 0.2163$$
I am confused - isn't $P(A mid C)$ the probability of getting a straight in 3 rolls, given you roll a 3 of a kind on the first try?
Any guidance would be very helpful - I wrote code simulations first, and now need to write a general verification of my result (0.0434) by hand.
Thanks!
probability dice conditional-probability
probability dice conditional-probability
edited Dec 4 '18 at 9:15
Tki Deneb
32710
asked Dec 4 '18 at 8:12
Megan ByersMegan Byers
2616
edited Dec 4 '18 at 9:15
Tki Deneb
32710
asked Dec 4 '18 at 8:12
Megan ByersMegan Byers
2616
edited Dec 4 '18 at 9:15
Tki Deneb
32710
edited Dec 4 '18 at 9:15
Tki Deneb
32710
edited Dec 4 '18 at 9:15
Tki Deneb
32710
32710
asked Dec 4 '18 at 8:12
Megan ByersMegan Byers
2616
asked Dec 4 '18 at 8:12
Megan ByersMegan Byers
2616
asked Dec 4 '18 at 8:12
Megan ByersMegan Byers
2616
2616
1
$begingroup$
I think you are confusing $P(A cap C)$ and $P(A | C)$ in the calculation after your simulation. The numerator ($P(A cap C)$) shouldn't be $0.0434$, but the number of times you got $A$ and $C$, divided by the total amount of simulations. The result should then be $0.0434$. For the mathematical calculation of this probability, you should be more precise on how we choose which dice to reroll after each round.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:19
$begingroup$
@TkiDeneb Thank you - this helped clear some stuff up. Is it right to say that this is question is asking a conditional probability? Do you think that testing for the probability of A (get a straight) given C (get 3 of a kind on first roll) is the right verbal terminology?
$endgroup$
– Megan Byers
Dec 4 '18 at 19:11
1
$begingroup$
That depends on what you want. If you are looking for the probability to get $3$ of a kind in the first roll and then a straight in any attempt, you want $P(A cap C)$. (Imagine placing yourself before the first roll.) If you are looking for the probability to get a straight after you have already gotten $3$ of a kind, then you want $P(A mid C)$. (Imagine placing yourself after a first roll that had $3$ of a kind.)
$endgroup$
– Tki Deneb
Dec 4 '18 at 19:56
add a comment |
1
$begingroup$
I think you are confusing $P(A cap C)$ and $P(A | C)$ in the calculation after your simulation. The numerator ($P(A cap C)$) shouldn't be $0.0434$, but the number of times you got $A$ and $C$, divided by the total amount of simulations. The result should then be $0.0434$. For the mathematical calculation of this probability, you should be more precise on how we choose which dice to reroll after each round.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:19
$begingroup$
@TkiDeneb Thank you - this helped clear some stuff up. Is it right to say that this is question is asking a conditional probability? Do you think that testing for the probability of A (get a straight) given C (get 3 of a kind on first roll) is the right verbal terminology?
$endgroup$
– Megan Byers
Dec 4 '18 at 19:11
1
$begingroup$
That depends on what you want. If you are looking for the probability to get $3$ of a kind in the first roll and then a straight in any attempt, you want $P(A cap C)$. (Imagine placing yourself before the first roll.) If you are looking for the probability to get a straight after you have already gotten $3$ of a kind, then you want $P(A mid C)$. (Imagine placing yourself after a first roll that had $3$ of a kind.)
$endgroup$
– Tki Deneb
Dec 4 '18 at 19:56
1
1
$begingroup$
I think you are confusing $P(A cap C)$ and $P(A | C)$ in the calculation after your simulation. The numerator ($P(A cap C)$) shouldn't be $0.0434$, but the number of times you got $A$ and $C$, divided by the total amount of simulations. The result should then be $0.0434$. For the mathematical calculation of this probability, you should be more precise on how we choose which dice to reroll after each round.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:19
$begingroup$
I think you are confusing $P(A cap C)$ and $P(A | C)$ in the calculation after your simulation. The numerator ($P(A cap C)$) shouldn't be $0.0434$, but the number of times you got $A$ and $C$, divided by the total amount of simulations. The result should then be $0.0434$. For the mathematical calculation of this probability, you should be more precise on how we choose which dice to reroll after each round.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:19
$begingroup$
@TkiDeneb Thank you - this helped clear some stuff up. Is it right to say that this is question is asking a conditional probability? Do you think that testing for the probability of A (get a straight) given C (get 3 of a kind on first roll) is the right verbal terminology?
$endgroup$
– Megan Byers
Dec 4 '18 at 19:11
$begingroup$
@TkiDeneb Thank you - this helped clear some stuff up. Is it right to say that this is question is asking a conditional probability? Do you think that testing for the probability of A (get a straight) given C (get 3 of a kind on first roll) is the right verbal terminology?
$endgroup$
– Megan Byers
Dec 4 '18 at 19:11
1
1
$begingroup$
That depends on what you want. If you are looking for the probability to get $3$ of a kind in the first roll and then a straight in any attempt, you want $P(A cap C)$. (Imagine placing yourself before the first roll.) If you are looking for the probability to get a straight after you have already gotten $3$ of a kind, then you want $P(A mid C)$. (Imagine placing yourself after a first roll that had $3$ of a kind.)
$endgroup$
– Tki Deneb
Dec 4 '18 at 19:56
$begingroup$
That depends on what you want. If you are looking for the probability to get $3$ of a kind in the first roll and then a straight in any attempt, you want $P(A cap C)$. (Imagine placing yourself before the first roll.) If you are looking for the probability to get a straight after you have already gotten $3$ of a kind, then you want $P(A mid C)$. (Imagine placing yourself after a first roll that had $3$ of a kind.)
$endgroup$
– Tki Deneb
Dec 4 '18 at 19:56
add a comment |
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$begingroup$
I think you are confusing $P(A cap C)$ and $P(A | C)$ in the calculation after your simulation. The numerator ($P(A cap C)$) shouldn't be $0.0434$, but the number of times you got $A$ and $C$, divided by the total amount of simulations. The result should then be $0.0434$. For the mathematical calculation of this probability, you should be more precise on how we choose which dice to reroll after each round.
$endgroup$
– Tki Deneb
Dec 4 '18 at 9:19
$begingroup$
@TkiDeneb Thank you - this helped clear some stuff up. Is it right to say that this is question is asking a conditional probability? Do you think that testing for the probability of A (get a straight) given C (get 3 of a kind on first roll) is the right verbal terminology?
$endgroup$
– Megan Byers
Dec 4 '18 at 19:11
1
$begingroup$
That depends on what you want. If you are looking for the probability to get $3$ of a kind in the first roll and then a straight in any attempt, you want $P(A cap C)$. (Imagine placing yourself before the first roll.) If you are looking for the probability to get a straight after you have already gotten $3$ of a kind, then you want $P(A mid C)$. (Imagine placing yourself after a first roll that had $3$ of a kind.)
$endgroup$
– Tki Deneb
Dec 4 '18 at 19:56