Probability with loaded and fair dice
I own five different six-sided dice. Four of the dice are fair dice, meaning they have values 1, 2, 3, 4, 5, 6. However, one of the dice is loaded; thus, it never shows 1, 2 or 3, but is equally likely to show the values 4, 5, or 6. For my experiment, I will pick up one random dice and roll it twice.
The first thing I would like to calculate is the probability of getting two sixes. To calculate this, I first calculated the probability of getting one six and multiplied it by two. Suppose $S$ = event that two sixes are rolled.
$$P(S) = 2(frac45(frac16) + frac15(frac13)) = .4 $$
However, I am not sure if this is correct. I need to calculate this because I would also like to calculate $P(L|S)$ where L = event that a loaded die was picked. Additionally, I feel this is incorrect, because if I change the '2' to a '10' to calculate it for 10 rolls instead of 2, I get a value over 1 which makes no sense. To summarize, how can I calculate $P(S)$ properly so I can calculate $P(L|S)$?
probability statistics elementary-set-theory dice
edited Feb 10 '17 at 6:47
Parcly Taxel
41.2k137199
asked Feb 10 '17 at 6:36
Digital VeerDigital Veer
1879
add a comment |
I own five different six-sided dice. Four of the dice are fair dice, meaning they have values 1, 2, 3, 4, 5, 6. However, one of the dice is loaded; thus, it never shows 1, 2 or 3, but is equally likely to show the values 4, 5, or 6. For my experiment, I will pick up one random dice and roll it twice.
The first thing I would like to calculate is the probability of getting two sixes. To calculate this, I first calculated the probability of getting one six and multiplied it by two. Suppose $S$ = event that two sixes are rolled.
$$P(S) = 2(frac45(frac16) + frac15(frac13)) = .4 $$
However, I am not sure if this is correct. I need to calculate this because I would also like to calculate $P(L|S)$ where L = event that a loaded die was picked. Additionally, I feel this is incorrect, because if I change the '2' to a '10' to calculate it for 10 rolls instead of 2, I get a value over 1 which makes no sense. To summarize, how can I calculate $P(S)$ properly so I can calculate $P(L|S)$?
probability statistics elementary-set-theory dice
edited Feb 10 '17 at 6:47
Parcly Taxel
41.2k137199
asked Feb 10 '17 at 6:36
Digital VeerDigital Veer
1879
1
Probability that "two sixes are rolled" is NOT at all equal to 2 * probability of "six is rolled the first time". With your approach you will get that probability of "100 sixes in 100 experiments" is much more that 1, isn't it?
– lesnik
Feb 10 '17 at 6:46
add a comment |
I own five different six-sided dice. Four of the dice are fair dice, meaning they have values 1, 2, 3, 4, 5, 6. However, one of the dice is loaded; thus, it never shows 1, 2 or 3, but is equally likely to show the values 4, 5, or 6. For my experiment, I will pick up one random dice and roll it twice.
The first thing I would like to calculate is the probability of getting two sixes. To calculate this, I first calculated the probability of getting one six and multiplied it by two. Suppose $S$ = event that two sixes are rolled.
$$P(S) = 2(frac45(frac16) + frac15(frac13)) = .4 $$
However, I am not sure if this is correct. I need to calculate this because I would also like to calculate $P(L|S)$ where L = event that a loaded die was picked. Additionally, I feel this is incorrect, because if I change the '2' to a '10' to calculate it for 10 rolls instead of 2, I get a value over 1 which makes no sense. To summarize, how can I calculate $P(S)$ properly so I can calculate $P(L|S)$?
probability statistics elementary-set-theory dice
edited Feb 10 '17 at 6:47
Parcly Taxel
41.2k137199
asked Feb 10 '17 at 6:36
Digital VeerDigital Veer
1879
I own five different six-sided dice. Four of the dice are fair dice, meaning they have values 1, 2, 3, 4, 5, 6. However, one of the dice is loaded; thus, it never shows 1, 2 or 3, but is equally likely to show the values 4, 5, or 6. For my experiment, I will pick up one random dice and roll it twice.
The first thing I would like to calculate is the probability of getting two sixes. To calculate this, I first calculated the probability of getting one six and multiplied it by two. Suppose $S$ = event that two sixes are rolled.
$$P(S) = 2(frac45(frac16) + frac15(frac13)) = .4 $$
However, I am not sure if this is correct. I need to calculate this because I would also like to calculate $P(L|S)$ where L = event that a loaded die was picked. Additionally, I feel this is incorrect, because if I change the '2' to a '10' to calculate it for 10 rolls instead of 2, I get a value over 1 which makes no sense. To summarize, how can I calculate $P(S)$ properly so I can calculate $P(L|S)$?
probability statistics elementary-set-theory dice
probability statistics elementary-set-theory dice
edited Feb 10 '17 at 6:47
Parcly Taxel
41.2k137199
asked Feb 10 '17 at 6:36
Digital VeerDigital Veer
1879
edited Feb 10 '17 at 6:47
Parcly Taxel
41.2k137199
asked Feb 10 '17 at 6:36
Digital VeerDigital Veer
1879
edited Feb 10 '17 at 6:47
Parcly Taxel
41.2k137199
edited Feb 10 '17 at 6:47
Parcly Taxel
41.2k137199
edited Feb 10 '17 at 6:47
Parcly Taxel
41.2k137199
41.2k137199
asked Feb 10 '17 at 6:36
Digital VeerDigital Veer
1879
asked Feb 10 '17 at 6:36
Digital VeerDigital Veer
1879
asked Feb 10 '17 at 6:36
Digital VeerDigital Veer
1879
1879
1
Probability that "two sixes are rolled" is NOT at all equal to 2 * probability of "six is rolled the first time". With your approach you will get that probability of "100 sixes in 100 experiments" is much more that 1, isn't it?
– lesnik
Feb 10 '17 at 6:46
add a comment |
1
Probability that "two sixes are rolled" is NOT at all equal to 2 * probability of "six is rolled the first time". With your approach you will get that probability of "100 sixes in 100 experiments" is much more that 1, isn't it?
– lesnik
Feb 10 '17 at 6:46
1
1
Probability that "two sixes are rolled" is NOT at all equal to 2 * probability of "six is rolled the first time". With your approach you will get that probability of "100 sixes in 100 experiments" is much more that 1, isn't it?
– lesnik
Feb 10 '17 at 6:46
Probability that "two sixes are rolled" is NOT at all equal to 2 * probability of "six is rolled the first time". With your approach you will get that probability of "100 sixes in 100 experiments" is much more that 1, isn't it?
– lesnik
Feb 10 '17 at 6:46
add a comment |
1 Answer
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The $frac16$ and $frac13$ should be squared at first, not doubled at the end. The correct calculation for $P(S)$ is
$$P(S)=frac45cdotfrac1{6^2}+frac15cdotfrac1{3^2}=frac1{45}+frac1{45}=frac2{45}$$
The second term above is $P(Lcap S)$, so
$$P(Lmid S)=frac{P(Lcap S)}{P(S)}=frac{1/45}{2/45}=frac12$$
answered Feb 10 '17 at 6:46
Parcly TaxelParcly Taxel
41.2k137199
add a comment |
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The $frac16$ and $frac13$ should be squared at first, not doubled at the end. The correct calculation for $P(S)$ is
$$P(S)=frac45cdotfrac1{6^2}+frac15cdotfrac1{3^2}=frac1{45}+frac1{45}=frac2{45}$$
The second term above is $P(Lcap S)$, so
$$P(Lmid S)=frac{P(Lcap S)}{P(S)}=frac{1/45}{2/45}=frac12$$
answered Feb 10 '17 at 6:46
Parcly TaxelParcly Taxel
41.2k137199
add a comment |
The $frac16$ and $frac13$ should be squared at first, not doubled at the end. The correct calculation for $P(S)$ is
$$P(S)=frac45cdotfrac1{6^2}+frac15cdotfrac1{3^2}=frac1{45}+frac1{45}=frac2{45}$$
The second term above is $P(Lcap S)$, so
$$P(Lmid S)=frac{P(Lcap S)}{P(S)}=frac{1/45}{2/45}=frac12$$
answered Feb 10 '17 at 6:46
Parcly TaxelParcly Taxel
41.2k137199
add a comment |
The $frac16$ and $frac13$ should be squared at first, not doubled at the end. The correct calculation for $P(S)$ is
$$P(S)=frac45cdotfrac1{6^2}+frac15cdotfrac1{3^2}=frac1{45}+frac1{45}=frac2{45}$$
The second term above is $P(Lcap S)$, so
$$P(Lmid S)=frac{P(Lcap S)}{P(S)}=frac{1/45}{2/45}=frac12$$
answered Feb 10 '17 at 6:46
Parcly TaxelParcly Taxel
41.2k137199
The $frac16$ and $frac13$ should be squared at first, not doubled at the end. The correct calculation for $P(S)$ is
$$P(S)=frac45cdotfrac1{6^2}+frac15cdotfrac1{3^2}=frac1{45}+frac1{45}=frac2{45}$$
The second term above is $P(Lcap S)$, so
$$P(Lmid S)=frac{P(Lcap S)}{P(S)}=frac{1/45}{2/45}=frac12$$
answered Feb 10 '17 at 6:46
Parcly TaxelParcly Taxel
41.2k137199
answered Feb 10 '17 at 6:46
Parcly TaxelParcly Taxel
41.2k137199
answered Feb 10 '17 at 6:46
Parcly TaxelParcly Taxel
41.2k137199
answered Feb 10 '17 at 6:46
Parcly TaxelParcly Taxel
41.2k137199
41.2k137199
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Probability that "two sixes are rolled" is NOT at all equal to 2 * probability of "six is rolled the first time". With your approach you will get that probability of "100 sixes in 100 experiments" is much more that 1, isn't it?
– lesnik
Feb 10 '17 at 6:46