Mixing
Intuition: Why chance of getting atleast one ace when rolling a dice six times is not close to 1?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
So probability of getting 1 ace(the one dot in dice) = 1/6 i.e. in one out of six times we will get an ace.
But when we calculate the probability of getting atleast one ace in six rolls, we get
$$= 1-left(frac56right)^6$$
$$= 0.665$$
I understand how the value is derived.
But what is intuitive explanation for the same?
Since it is so close to 68%, the percentage of population within 1 standard deviation of normal distribution, does it have any relationship with normal distribution?
probability probability-theory statistics probability-distributions dice
asked 1 hour ago
q126y
1986
add a comment |Â
up vote
2
down vote
favorite
So probability of getting 1 ace(the one dot in dice) = 1/6 i.e. in one out of six times we will get an ace.
But when we calculate the probability of getting atleast one ace in six rolls, we get
$$= 1-left(frac56right)^6$$
$$= 0.665$$
I understand how the value is derived.
But what is intuitive explanation for the same?
Since it is so close to 68%, the percentage of population within 1 standard deviation of normal distribution, does it have any relationship with normal distribution?
probability probability-theory statistics probability-distributions dice
asked 1 hour ago
q126y
1986
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
So probability of getting 1 ace(the one dot in dice) = 1/6 i.e. in one out of six times we will get an ace.
But when we calculate the probability of getting atleast one ace in six rolls, we get
$$= 1-left(frac56right)^6$$
$$= 0.665$$
I understand how the value is derived.
But what is intuitive explanation for the same?
Since it is so close to 68%, the percentage of population within 1 standard deviation of normal distribution, does it have any relationship with normal distribution?
probability probability-theory statistics probability-distributions dice
asked 1 hour ago
q126y
1986
So probability of getting 1 ace(the one dot in dice) = 1/6 i.e. in one out of six times we will get an ace.
But when we calculate the probability of getting atleast one ace in six rolls, we get
$$= 1-left(frac56right)^6$$
$$= 0.665$$
I understand how the value is derived.
But what is intuitive explanation for the same?
Since it is so close to 68%, the percentage of population within 1 standard deviation of normal distribution, does it have any relationship with normal distribution?
probability probability-theory statistics probability-distributions dice
probability probability-theory statistics probability-distributions dice
asked 1 hour ago
q126y
1986
asked 1 hour ago
q126y
1986
asked 1 hour ago
q126y
1986
asked 1 hour ago
q126y
1986
asked 1 hour ago
q126y
1986
1986
add a comment |Â
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
3
down vote
The intuitive explanation is that there are $6^6=46656$ different outcomes from rolling a dice 6 times, and $5^6=15625$ of these have no aces. $15625/46656 approx 33.5%$, so $33.5%$ of outcomes have no aces and $67.5%$ of outcomes have at least one ace. If the outcomes are equally likely then the probability of at least one aces is the same as the ratio of the number of favourable outcomes to the total number of outcomes i.e. $67.5%$.
answered 1 hour ago
gandalf61
6,189522
Thanks for the answer. It may be intuitive for you, but I am not able to get it. The answer is analytical to me. I mean if we have 1 in 6 chance of getting an ace, if we roll a dice 6 times, chance of getting an ace should be close to 1.
– q126y
1 hour ago
3
@q126y: Perhaps you could tell us more about this intuition? Do you agree that it doesn't work for a coin? (The chance of getting at least one heads in two tosses is $frac34$.) Do you find that counterintuitive, too? If not, why should it work better for a die?
– joriki
1 hour ago
Perhaps your intuition is really related to the expected number of Aces which we will roll in 6 attempts -- that is 1?. And/or ... perhaps to challenge your own intuition, consider what will happen if we roll 12 times ... should the probability of rolling an Ace now be $frac12 6$ i.e. 2? Obviously not!
– BBO555
48 mins ago
@q126y: There are 15625 cases to have no aces, that's no a small number, at least for me. My answer very similar to Mr. gandalf61 so I deleted mine.
– Isana Yashiro
15 mins ago
add a comment |Â
up vote
2
down vote
When you roll a die six times you shall expect to obtain one ace.
This is an average result, so the probability for obtaining zero aces should counterballance the probability for obtaining more than one aces.
Thus your intuition should anticipate that the probability for obtaining at least one ace is not too much more than one half.
answered 51 mins ago
Graham Kemp
81.6k43275
add a comment |Â
up vote
0
down vote
If you write out all the combinations of rolling one die, you get a row of six numbers. The "border" of this list are the numbers $1$ and $6$. The number of aces is $1$, which is half the number of elements in the border. The length of the list is $s=6$ and the derivative of $s$ is $1$.
If you write out all the combinations of rolling two dice, you get a $6times 6$ table. The rolls with aces are the top row and the left column, which is half the border of the table. In an $stimes s$ square, half the border is $2s$ which is the derivative of $s^2.$ But this counts one roll twice, so we have to subtract $1$ for each intersection of border edges. There's only one $=2choose2$ intersection , so we have $2s-1 = 11$ rolls with aces.
If you make a 3-D array of the all the combinations of rolling 3 dice, you get a $6times 6times 6$ cube of ordered triples. The rolls with aces are the top face, the left face and the front face, so again, half the border, which turns out to be $3s^2$. But again we have to subtract the intersections. This time there are three $=3choose 2$ intersections so we subtract three $stimes 1$ edges. Oh, but we need to add back in all triple intersections. There is one $=3choose 3$, so I end up with $3s^2 - 3s+3$ rolls with aces.
Continue in this fashion till you get to $6$ dimensions. Half the border of your cube will be the derivative of $s^6$, that is $6s^5$. The ratio
$frac6s^5s^6 = 1$ for $s=6$, which is what your intuition wanted. It's the subtraction of the six 5-dimensional faces (and the addition of the sixty 4-dimensional faces times $6choose 2$ and the addition of $6choose 3 times 160 cubic faces etc.) that makes the numerator much smaller.
answered 1 hour ago
B. Goddard
16.8k21338
add a comment |Â
up vote
0
down vote
The expected number of rolls until the first ace is $6$. Do you find this intuitive? If so, consider that there must then be lots of cases where you don't see an ace in the first $6$ rolls, or the expected waiting time would be less than $6.$
answered 49 mins ago
saulspatz
11.6k21324
add a comment |Â
up vote
0
down vote
Consider the following thought experiment:
Take N balls in N containers; pull them out, then assign each ball a new container randomly, without regard to where the previous balls went; pick a container at random. What is the probability that this container is filled? Close to one? That would mean that all of the containers hold a ball, and none contain two.
But there are $n^n$ ways to shuffle the balls, but only $n!$ ways to shuffle them in a way that leaves all containers filled. In fact, the probability that a random shuffle leaves all containers filled is $prod_x=0^n frac x n = frac 1 n * frac 2 n * frac 3 n * dots$. As $n$ grows, this probability quickly drops to zero. In all other cases, there's at least one container empty.
This already tells us that the chance of getting an ace in six rolls should be less than 5/6. The actual value is close to 4/6. Not bad for such a weak argument.
In fact, the probability of getting a probability $frac 1 n$ event in $n$ tries converges to $1-frac 1 e = 0.6321$ (ref).
answered 38 mins ago
John Dvorak
67468
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The intuitive explanation is that there are $6^6=46656$ different outcomes from rolling a dice 6 times, and $5^6=15625$ of these have no aces. $15625/46656 approx 33.5%$, so $33.5%$ of outcomes have no aces and $67.5%$ of outcomes have at least one ace. If the outcomes are equally likely then the probability of at least one aces is the same as the ratio of the number of favourable outcomes to the total number of outcomes i.e. $67.5%$.
answered 1 hour ago
gandalf61
6,189522
Thanks for the answer. It may be intuitive for you, but I am not able to get it. The answer is analytical to me. I mean if we have 1 in 6 chance of getting an ace, if we roll a dice 6 times, chance of getting an ace should be close to 1.
– q126y
1 hour ago
3
@q126y: Perhaps you could tell us more about this intuition? Do you agree that it doesn't work for a coin? (The chance of getting at least one heads in two tosses is $frac34$.) Do you find that counterintuitive, too? If not, why should it work better for a die?
– joriki
1 hour ago
Perhaps your intuition is really related to the expected number of Aces which we will roll in 6 attempts -- that is 1?. And/or ... perhaps to challenge your own intuition, consider what will happen if we roll 12 times ... should the probability of rolling an Ace now be $frac12 6$ i.e. 2? Obviously not!
– BBO555
48 mins ago
@q126y: There are 15625 cases to have no aces, that's no a small number, at least for me. My answer very similar to Mr. gandalf61 so I deleted mine.
– Isana Yashiro
15 mins ago
add a comment |Â
up vote
3
down vote
The intuitive explanation is that there are $6^6=46656$ different outcomes from rolling a dice 6 times, and $5^6=15625$ of these have no aces. $15625/46656 approx 33.5%$, so $33.5%$ of outcomes have no aces and $67.5%$ of outcomes have at least one ace. If the outcomes are equally likely then the probability of at least one aces is the same as the ratio of the number of favourable outcomes to the total number of outcomes i.e. $67.5%$.
answered 1 hour ago
gandalf61
6,189522
Thanks for the answer. It may be intuitive for you, but I am not able to get it. The answer is analytical to me. I mean if we have 1 in 6 chance of getting an ace, if we roll a dice 6 times, chance of getting an ace should be close to 1.
– q126y
1 hour ago
3
@q126y: Perhaps you could tell us more about this intuition? Do you agree that it doesn't work for a coin? (The chance of getting at least one heads in two tosses is $frac34$.) Do you find that counterintuitive, too? If not, why should it work better for a die?
– joriki
1 hour ago
Perhaps your intuition is really related to the expected number of Aces which we will roll in 6 attempts -- that is 1?. And/or ... perhaps to challenge your own intuition, consider what will happen if we roll 12 times ... should the probability of rolling an Ace now be $frac12 6$ i.e. 2? Obviously not!
– BBO555
48 mins ago
@q126y: There are 15625 cases to have no aces, that's no a small number, at least for me. My answer very similar to Mr. gandalf61 so I deleted mine.
– Isana Yashiro
15 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The intuitive explanation is that there are $6^6=46656$ different outcomes from rolling a dice 6 times, and $5^6=15625$ of these have no aces. $15625/46656 approx 33.5%$, so $33.5%$ of outcomes have no aces and $67.5%$ of outcomes have at least one ace. If the outcomes are equally likely then the probability of at least one aces is the same as the ratio of the number of favourable outcomes to the total number of outcomes i.e. $67.5%$.
answered 1 hour ago
gandalf61
6,189522
The intuitive explanation is that there are $6^6=46656$ different outcomes from rolling a dice 6 times, and $5^6=15625$ of these have no aces. $15625/46656 approx 33.5%$, so $33.5%$ of outcomes have no aces and $67.5%$ of outcomes have at least one ace. If the outcomes are equally likely then the probability of at least one aces is the same as the ratio of the number of favourable outcomes to the total number of outcomes i.e. $67.5%$.
answered 1 hour ago
gandalf61
6,189522
edited 1 hour ago
answered 1 hour ago
gandalf61
6,189522
answered 1 hour ago
gandalf61
6,189522
answered 1 hour ago
gandalf61
6,189522
6,189522
Thanks for the answer. It may be intuitive for you, but I am not able to get it. The answer is analytical to me. I mean if we have 1 in 6 chance of getting an ace, if we roll a dice 6 times, chance of getting an ace should be close to 1.
– q126y
1 hour ago
3
@q126y: Perhaps you could tell us more about this intuition? Do you agree that it doesn't work for a coin? (The chance of getting at least one heads in two tosses is $frac34$.) Do you find that counterintuitive, too? If not, why should it work better for a die?
– joriki
1 hour ago
Perhaps your intuition is really related to the expected number of Aces which we will roll in 6 attempts -- that is 1?. And/or ... perhaps to challenge your own intuition, consider what will happen if we roll 12 times ... should the probability of rolling an Ace now be $frac12 6$ i.e. 2? Obviously not!
– BBO555
48 mins ago
@q126y: There are 15625 cases to have no aces, that's no a small number, at least for me. My answer very similar to Mr. gandalf61 so I deleted mine.
– Isana Yashiro
15 mins ago
add a comment |Â
Thanks for the answer. It may be intuitive for you, but I am not able to get it. The answer is analytical to me. I mean if we have 1 in 6 chance of getting an ace, if we roll a dice 6 times, chance of getting an ace should be close to 1.
– q126y
1 hour ago
3
@q126y: Perhaps you could tell us more about this intuition? Do you agree that it doesn't work for a coin? (The chance of getting at least one heads in two tosses is $frac34$.) Do you find that counterintuitive, too? If not, why should it work better for a die?
– joriki
1 hour ago
Perhaps your intuition is really related to the expected number of Aces which we will roll in 6 attempts -- that is 1?. And/or ... perhaps to challenge your own intuition, consider what will happen if we roll 12 times ... should the probability of rolling an Ace now be $frac12 6$ i.e. 2? Obviously not!
– BBO555
48 mins ago
@q126y: There are 15625 cases to have no aces, that's no a small number, at least for me. My answer very similar to Mr. gandalf61 so I deleted mine.
– Isana Yashiro
15 mins ago
Thanks for the answer. It may be intuitive for you, but I am not able to get it. The answer is analytical to me. I mean if we have 1 in 6 chance of getting an ace, if we roll a dice 6 times, chance of getting an ace should be close to 1.
– q126y
1 hour ago
Thanks for the answer. It may be intuitive for you, but I am not able to get it. The answer is analytical to me. I mean if we have 1 in 6 chance of getting an ace, if we roll a dice 6 times, chance of getting an ace should be close to 1.
– q126y
1 hour ago
3
3
@q126y: Perhaps you could tell us more about this intuition? Do you agree that it doesn't work for a coin? (The chance of getting at least one heads in two tosses is $frac34$.) Do you find that counterintuitive, too? If not, why should it work better for a die?
– joriki
1 hour ago
@q126y: Perhaps you could tell us more about this intuition? Do you agree that it doesn't work for a coin? (The chance of getting at least one heads in two tosses is $frac34$.) Do you find that counterintuitive, too? If not, why should it work better for a die?
– joriki
1 hour ago
Perhaps your intuition is really related to the expected number of Aces which we will roll in 6 attempts -- that is 1?. And/or ... perhaps to challenge your own intuition, consider what will happen if we roll 12 times ... should the probability of rolling an Ace now be $frac12 6$ i.e. 2? Obviously not!
– BBO555
48 mins ago
Perhaps your intuition is really related to the expected number of Aces which we will roll in 6 attempts -- that is 1?. And/or ... perhaps to challenge your own intuition, consider what will happen if we roll 12 times ... should the probability of rolling an Ace now be $frac12 6$ i.e. 2? Obviously not!
– BBO555
48 mins ago
@q126y: There are 15625 cases to have no aces, that's no a small number, at least for me. My answer very similar to Mr. gandalf61 so I deleted mine.
– Isana Yashiro
15 mins ago
@q126y: There are 15625 cases to have no aces, that's no a small number, at least for me. My answer very similar to Mr. gandalf61 so I deleted mine.
– Isana Yashiro
15 mins ago
add a comment |Â
up vote
2
down vote
When you roll a die six times you shall expect to obtain one ace.
This is an average result, so the probability for obtaining zero aces should counterballance the probability for obtaining more than one aces.
Thus your intuition should anticipate that the probability for obtaining at least one ace is not too much more than one half.
answered 51 mins ago
Graham Kemp
81.6k43275
add a comment |Â
up vote
2
down vote
When you roll a die six times you shall expect to obtain one ace.
This is an average result, so the probability for obtaining zero aces should counterballance the probability for obtaining more than one aces.
Thus your intuition should anticipate that the probability for obtaining at least one ace is not too much more than one half.
answered 51 mins ago
Graham Kemp
81.6k43275
add a comment |Â
up vote
2
down vote
up vote
2
down vote
When you roll a die six times you shall expect to obtain one ace.
This is an average result, so the probability for obtaining zero aces should counterballance the probability for obtaining more than one aces.
Thus your intuition should anticipate that the probability for obtaining at least one ace is not too much more than one half.
answered 51 mins ago
Graham Kemp
81.6k43275
When you roll a die six times you shall expect to obtain one ace.
This is an average result, so the probability for obtaining zero aces should counterballance the probability for obtaining more than one aces.
Thus your intuition should anticipate that the probability for obtaining at least one ace is not too much more than one half.
answered 51 mins ago
Graham Kemp
81.6k43275
answered 51 mins ago
Graham Kemp
81.6k43275
answered 51 mins ago
Graham Kemp
81.6k43275
answered 51 mins ago
Graham Kemp
81.6k43275
81.6k43275
add a comment |Â
add a comment |Â
up vote
0
down vote
If you write out all the combinations of rolling one die, you get a row of six numbers. The "border" of this list are the numbers $1$ and $6$. The number of aces is $1$, which is half the number of elements in the border. The length of the list is $s=6$ and the derivative of $s$ is $1$.
If you write out all the combinations of rolling two dice, you get a $6times 6$ table. The rolls with aces are the top row and the left column, which is half the border of the table. In an $stimes s$ square, half the border is $2s$ which is the derivative of $s^2.$ But this counts one roll twice, so we have to subtract $1$ for each intersection of border edges. There's only one $=2choose2$ intersection , so we have $2s-1 = 11$ rolls with aces.
If you make a 3-D array of the all the combinations of rolling 3 dice, you get a $6times 6times 6$ cube of ordered triples. The rolls with aces are the top face, the left face and the front face, so again, half the border, which turns out to be $3s^2$. But again we have to subtract the intersections. This time there are three $=3choose 2$ intersections so we subtract three $stimes 1$ edges. Oh, but we need to add back in all triple intersections. There is one $=3choose 3$, so I end up with $3s^2 - 3s+3$ rolls with aces.
Continue in this fashion till you get to $6$ dimensions. Half the border of your cube will be the derivative of $s^6$, that is $6s^5$. The ratio
$frac6s^5s^6 = 1$ for $s=6$, which is what your intuition wanted. It's the subtraction of the six 5-dimensional faces (and the addition of the sixty 4-dimensional faces times $6choose 2$ and the addition of $6choose 3 times 160 cubic faces etc.) that makes the numerator much smaller.
answered 1 hour ago
B. Goddard
16.8k21338
add a comment |Â
up vote
0
down vote
If you write out all the combinations of rolling one die, you get a row of six numbers. The "border" of this list are the numbers $1$ and $6$. The number of aces is $1$, which is half the number of elements in the border. The length of the list is $s=6$ and the derivative of $s$ is $1$.
If you write out all the combinations of rolling two dice, you get a $6times 6$ table. The rolls with aces are the top row and the left column, which is half the border of the table. In an $stimes s$ square, half the border is $2s$ which is the derivative of $s^2.$ But this counts one roll twice, so we have to subtract $1$ for each intersection of border edges. There's only one $=2choose2$ intersection , so we have $2s-1 = 11$ rolls with aces.
If you make a 3-D array of the all the combinations of rolling 3 dice, you get a $6times 6times 6$ cube of ordered triples. The rolls with aces are the top face, the left face and the front face, so again, half the border, which turns out to be $3s^2$. But again we have to subtract the intersections. This time there are three $=3choose 2$ intersections so we subtract three $stimes 1$ edges. Oh, but we need to add back in all triple intersections. There is one $=3choose 3$, so I end up with $3s^2 - 3s+3$ rolls with aces.
Continue in this fashion till you get to $6$ dimensions. Half the border of your cube will be the derivative of $s^6$, that is $6s^5$. The ratio
$frac6s^5s^6 = 1$ for $s=6$, which is what your intuition wanted. It's the subtraction of the six 5-dimensional faces (and the addition of the sixty 4-dimensional faces times $6choose 2$ and the addition of $6choose 3 times 160 cubic faces etc.) that makes the numerator much smaller.
answered 1 hour ago
B. Goddard
16.8k21338
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If you write out all the combinations of rolling one die, you get a row of six numbers. The "border" of this list are the numbers $1$ and $6$. The number of aces is $1$, which is half the number of elements in the border. The length of the list is $s=6$ and the derivative of $s$ is $1$.
If you write out all the combinations of rolling two dice, you get a $6times 6$ table. The rolls with aces are the top row and the left column, which is half the border of the table. In an $stimes s$ square, half the border is $2s$ which is the derivative of $s^2.$ But this counts one roll twice, so we have to subtract $1$ for each intersection of border edges. There's only one $=2choose2$ intersection , so we have $2s-1 = 11$ rolls with aces.
If you make a 3-D array of the all the combinations of rolling 3 dice, you get a $6times 6times 6$ cube of ordered triples. The rolls with aces are the top face, the left face and the front face, so again, half the border, which turns out to be $3s^2$. But again we have to subtract the intersections. This time there are three $=3choose 2$ intersections so we subtract three $stimes 1$ edges. Oh, but we need to add back in all triple intersections. There is one $=3choose 3$, so I end up with $3s^2 - 3s+3$ rolls with aces.
Continue in this fashion till you get to $6$ dimensions. Half the border of your cube will be the derivative of $s^6$, that is $6s^5$. The ratio
$frac6s^5s^6 = 1$ for $s=6$, which is what your intuition wanted. It's the subtraction of the six 5-dimensional faces (and the addition of the sixty 4-dimensional faces times $6choose 2$ and the addition of $6choose 3 times 160 cubic faces etc.) that makes the numerator much smaller.
answered 1 hour ago
B. Goddard
16.8k21338
If you write out all the combinations of rolling one die, you get a row of six numbers. The "border" of this list are the numbers $1$ and $6$. The number of aces is $1$, which is half the number of elements in the border. The length of the list is $s=6$ and the derivative of $s$ is $1$.
If you write out all the combinations of rolling two dice, you get a $6times 6$ table. The rolls with aces are the top row and the left column, which is half the border of the table. In an $stimes s$ square, half the border is $2s$ which is the derivative of $s^2.$ But this counts one roll twice, so we have to subtract $1$ for each intersection of border edges. There's only one $=2choose2$ intersection , so we have $2s-1 = 11$ rolls with aces.
If you make a 3-D array of the all the combinations of rolling 3 dice, you get a $6times 6times 6$ cube of ordered triples. The rolls with aces are the top face, the left face and the front face, so again, half the border, which turns out to be $3s^2$. But again we have to subtract the intersections. This time there are three $=3choose 2$ intersections so we subtract three $stimes 1$ edges. Oh, but we need to add back in all triple intersections. There is one $=3choose 3$, so I end up with $3s^2 - 3s+3$ rolls with aces.
Continue in this fashion till you get to $6$ dimensions. Half the border of your cube will be the derivative of $s^6$, that is $6s^5$. The ratio
$frac6s^5s^6 = 1$ for $s=6$, which is what your intuition wanted. It's the subtraction of the six 5-dimensional faces (and the addition of the sixty 4-dimensional faces times $6choose 2$ and the addition of $6choose 3 times 160 cubic faces etc.) that makes the numerator much smaller.
answered 1 hour ago
B. Goddard
16.8k21338
answered 1 hour ago
B. Goddard
16.8k21338
answered 1 hour ago
B. Goddard
16.8k21338
answered 1 hour ago
B. Goddard
16.8k21338
16.8k21338
add a comment |Â
add a comment |Â
up vote
0
down vote
The expected number of rolls until the first ace is $6$. Do you find this intuitive? If so, consider that there must then be lots of cases where you don't see an ace in the first $6$ rolls, or the expected waiting time would be less than $6.$
answered 49 mins ago
saulspatz
11.6k21324
add a comment |Â
up vote
0
down vote
The expected number of rolls until the first ace is $6$. Do you find this intuitive? If so, consider that there must then be lots of cases where you don't see an ace in the first $6$ rolls, or the expected waiting time would be less than $6.$
answered 49 mins ago
saulspatz
11.6k21324
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The expected number of rolls until the first ace is $6$. Do you find this intuitive? If so, consider that there must then be lots of cases where you don't see an ace in the first $6$ rolls, or the expected waiting time would be less than $6.$
answered 49 mins ago
saulspatz
11.6k21324
The expected number of rolls until the first ace is $6$. Do you find this intuitive? If so, consider that there must then be lots of cases where you don't see an ace in the first $6$ rolls, or the expected waiting time would be less than $6.$
answered 49 mins ago
saulspatz
11.6k21324
answered 49 mins ago
saulspatz
11.6k21324
answered 49 mins ago
saulspatz
11.6k21324
answered 49 mins ago
saulspatz
11.6k21324
11.6k21324
add a comment |Â
add a comment |Â
up vote
0
down vote
Consider the following thought experiment:
Take N balls in N containers; pull them out, then assign each ball a new container randomly, without regard to where the previous balls went; pick a container at random. What is the probability that this container is filled? Close to one? That would mean that all of the containers hold a ball, and none contain two.
But there are $n^n$ ways to shuffle the balls, but only $n!$ ways to shuffle them in a way that leaves all containers filled. In fact, the probability that a random shuffle leaves all containers filled is $prod_x=0^n frac x n = frac 1 n * frac 2 n * frac 3 n * dots$. As $n$ grows, this probability quickly drops to zero. In all other cases, there's at least one container empty.
This already tells us that the chance of getting an ace in six rolls should be less than 5/6. The actual value is close to 4/6. Not bad for such a weak argument.
In fact, the probability of getting a probability $frac 1 n$ event in $n$ tries converges to $1-frac 1 e = 0.6321$ (ref).
answered 38 mins ago
John Dvorak
67468
add a comment |Â
up vote
0
down vote
Consider the following thought experiment:
Take N balls in N containers; pull them out, then assign each ball a new container randomly, without regard to where the previous balls went; pick a container at random. What is the probability that this container is filled? Close to one? That would mean that all of the containers hold a ball, and none contain two.
But there are $n^n$ ways to shuffle the balls, but only $n!$ ways to shuffle them in a way that leaves all containers filled. In fact, the probability that a random shuffle leaves all containers filled is $prod_x=0^n frac x n = frac 1 n * frac 2 n * frac 3 n * dots$. As $n$ grows, this probability quickly drops to zero. In all other cases, there's at least one container empty.
This already tells us that the chance of getting an ace in six rolls should be less than 5/6. The actual value is close to 4/6. Not bad for such a weak argument.
In fact, the probability of getting a probability $frac 1 n$ event in $n$ tries converges to $1-frac 1 e = 0.6321$ (ref).
answered 38 mins ago
John Dvorak
67468
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider the following thought experiment:
Take N balls in N containers; pull them out, then assign each ball a new container randomly, without regard to where the previous balls went; pick a container at random. What is the probability that this container is filled? Close to one? That would mean that all of the containers hold a ball, and none contain two.
But there are $n^n$ ways to shuffle the balls, but only $n!$ ways to shuffle them in a way that leaves all containers filled. In fact, the probability that a random shuffle leaves all containers filled is $prod_x=0^n frac x n = frac 1 n * frac 2 n * frac 3 n * dots$. As $n$ grows, this probability quickly drops to zero. In all other cases, there's at least one container empty.
This already tells us that the chance of getting an ace in six rolls should be less than 5/6. The actual value is close to 4/6. Not bad for such a weak argument.
In fact, the probability of getting a probability $frac 1 n$ event in $n$ tries converges to $1-frac 1 e = 0.6321$ (ref).
answered 38 mins ago
John Dvorak
67468
Consider the following thought experiment:
Take N balls in N containers; pull them out, then assign each ball a new container randomly, without regard to where the previous balls went; pick a container at random. What is the probability that this container is filled? Close to one? That would mean that all of the containers hold a ball, and none contain two.
But there are $n^n$ ways to shuffle the balls, but only $n!$ ways to shuffle them in a way that leaves all containers filled. In fact, the probability that a random shuffle leaves all containers filled is $prod_x=0^n frac x n = frac 1 n * frac 2 n * frac 3 n * dots$. As $n$ grows, this probability quickly drops to zero. In all other cases, there's at least one container empty.
This already tells us that the chance of getting an ace in six rolls should be less than 5/6. The actual value is close to 4/6. Not bad for such a weak argument.
In fact, the probability of getting a probability $frac 1 n$ event in $n$ tries converges to $1-frac 1 e = 0.6321$ (ref).
answered 38 mins ago
John Dvorak
67468
answered 38 mins ago
John Dvorak
67468
answered 38 mins ago
John Dvorak
67468
answered 38 mins ago
John Dvorak
67468
67468
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2925338%2fintuition-why-chance-of-getting-atleast-one-ace-when-rolling-a-dice-six-times-i%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
