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Expected value of two dice
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Consider two dice. One 10-sided, one 6-sided, both fair. Guess a number between 2 and 16, if the sum of the dice shows that number, you get that number of dollars. What is the best number to guess to maximize winnings?
The answer is $11$, but I'm not sure why. I thought that since $mathbbE[D_10] = 5.5$ and $mathbbE[D_6] = 3.5$, the best choice would just be $5.5 + 3.5 = 9$, but I am wrong.
Can someone please explain the solution to this problem for me?
probability expected-value
asked 39 mins ago
Hat
892116
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up vote
1
down vote
favorite
Consider two dice. One 10-sided, one 6-sided, both fair. Guess a number between 2 and 16, if the sum of the dice shows that number, you get that number of dollars. What is the best number to guess to maximize winnings?
The answer is $11$, but I'm not sure why. I thought that since $mathbbE[D_10] = 5.5$ and $mathbbE[D_6] = 3.5$, the best choice would just be $5.5 + 3.5 = 9$, but I am wrong.
Can someone please explain the solution to this problem for me?
probability expected-value
asked 39 mins ago
Hat
892116
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider two dice. One 10-sided, one 6-sided, both fair. Guess a number between 2 and 16, if the sum of the dice shows that number, you get that number of dollars. What is the best number to guess to maximize winnings?
The answer is $11$, but I'm not sure why. I thought that since $mathbbE[D_10] = 5.5$ and $mathbbE[D_6] = 3.5$, the best choice would just be $5.5 + 3.5 = 9$, but I am wrong.
Can someone please explain the solution to this problem for me?
probability expected-value
asked 39 mins ago
Hat
892116
Consider two dice. One 10-sided, one 6-sided, both fair. Guess a number between 2 and 16, if the sum of the dice shows that number, you get that number of dollars. What is the best number to guess to maximize winnings?
The answer is $11$, but I'm not sure why. I thought that since $mathbbE[D_10] = 5.5$ and $mathbbE[D_6] = 3.5$, the best choice would just be $5.5 + 3.5 = 9$, but I am wrong.
Can someone please explain the solution to this problem for me?
probability expected-value
probability expected-value
asked 39 mins ago
Hat
892116
asked 39 mins ago
Hat
892116
asked 39 mins ago
Hat
892116
asked 39 mins ago
Hat
892116
asked 39 mins ago
Hat
892116
892116
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
The number $9$ is the expected value $mathbb E[D_6 + D_10]$, but in general that's not necessarily the likeliest value.
After all, if you were playing the same game with just a single $D_6$, you wouldn't guess $3.5$. If you did, it would take a very long time before that number was rolled.
Here, it turns out that both $9$ and $11$ are equally likely. The distribution of sums is
beginarrayrcccccc
2: & (1,1) \
3: & (1,2) & (2,1) \
4: & (1,3) & (2,2) & (3,1) \
5: & (1,4) & (2,3) & (3,2) & (4,1) \
6: & (1,5) & (2,4) & (3,3) & (4,2) & (5,1) \
7: & (1,6) & (2,5) & (3,4) & (4,3) & (5,2) & (6,1) \
8: & (1,7) & (2,6) & (3,5) & (4,4) & (5,3) & (6,2) \
9: & (1,8) & (2,7) & (3,6) & (4,5) & (5,4) & (6,3) \
10: & (1,9) & (2,8) & (3,7) & (4,6) & (5,5) & (6,4) \
11: & (1,10) & (2,9) & (3,8) & (4,7) & (5,6) & (6,5) \
12: & (2,10) & (3,9) & (4,8) & (5,7) & (6,6) \
13: & (3,10) & (4,9) & (5,8) & (6,7) \
14: & (4,10) & (5,9) & (6,8) \
15: & (5,10) & (6,9) \
16: & (6,10)
endarray
The likeliest value is not necessarily the one you should be picking.
A high value might be unlikely to win, but give you a higher reward if you win, since the number of dollars you get is the number you guessed. We should maximize the expected winnings, not the probability.
From the table above, we see that there's no point in guessing the numbers $2$ through $10$. Rather than a small number like $2$, you should guess a big number with the same probability (like $16$): it has the same chance of winning, but gives you more money if you do. Similarly, since all of $7$ through $11$ have the same chance of winning, the only one of them worth guessing is $11$.
Now we can compute the expected payoffs, multiplying the reward by the probability of getting it.
- For guessing $11$, you get $11 cdot frac660 = frac1110$ on average.
- For guessing $12$, you get $12 cdot frac560 = 1$ on average.
- For guessing $13$, you get $13 cdot frac460 = frac1315$ on average.
- For guessing $14$, you get $14 cdot frac360 = frac710$ on average.
- For guessing $15$, you get $15 cdot frac260 = frac12$ on average.
- For guessing $16$, you get $16 cdot frac160 = frac415$ on average.
You can see that these expected payoffs keep decreasing as the guess becomes higher, and so $11$ is the best value to guess.
(Once you spot that, for values in the range $11dots16$, the expected value is $n cdot frac17-n60$, you can predict this behavior: to maximize the product of two values with a fixed sum, like $n$ and $17-n$, you want to make the values as close to each other as possible.)
answered 25 mins ago
Misha Lavrov
39.5k55197
is there a clever way to do this without listing out every possible outcome? this is an interview question and in a timed setting, I would have only a few seconds to answer.
– Hat
21 mins ago
If you only have a few seconds to answer, then it'd be best to explain $how$ the answer was obtained. That shouldn't take more than 10 seconds.
– Remy
16 mins ago
@Hat The clever way is to imagine what would happen if you did list out every possible outcome, spot the pattern, and use the pattern to figure out the answer. Actually listing the outcomes is just an easy way to demonstrate what happens.
– Misha Lavrov
9 mins ago
I would guess that very few people would have the mental calisthenics to compute this in a few seconds. Von Neumann or Fermi, maybe...
– copper.hat
6 mins ago
add a comment |Â
up vote
2
down vote
beginalign
Pr(D_6 + D_10=x) &= sum_y=1^6Pr(D_6 + D_10=x|D_6=y)Pr(D_6=y) \
&= frac16 sum_y=1^6 Pr(D_10=x-y)\
&= frac16 sum_y=max(1,x-10)^min(x-1,6) Pr(D_10=x-y) \
&= frac160[min(x-1,6)-max(1,x-10)+1]
endalign
Hence, we want to maximize
$$fracx60[min(x-1,6)-max(1,x-10)+1]$$
answered 19 mins ago
Siong Thye Goh
86.3k1459108
Nice closed form expression, I am still struggling...
– copper.hat
15 mins ago
add a comment |Â
up vote
0
down vote
The expected number of winnings when guessing a number $n$ is going to be the dollar value $n$ multiplied by the probability that the sum of the two dice is $n.$ So even though $9$ is the expected value for the sum of the dice, the expected winnings by guessing $9$ is going to be $9 cdot frac110 = $0.90,$ because we can compute that the probability of the sum being $9$ is $frac660 = frac110.$ Using this methodology, we can find that the highest expected winnings occur at $n=11,$ with $11 cdot frac110 = $ 1.10.$
answered 18 mins ago
anonymous helper
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anonymous helper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The number $9$ is the expected value $mathbb E[D_6 + D_10]$, but in general that's not necessarily the likeliest value.
After all, if you were playing the same game with just a single $D_6$, you wouldn't guess $3.5$. If you did, it would take a very long time before that number was rolled.
Here, it turns out that both $9$ and $11$ are equally likely. The distribution of sums is
beginarrayrcccccc
2: & (1,1) \
3: & (1,2) & (2,1) \
4: & (1,3) & (2,2) & (3,1) \
5: & (1,4) & (2,3) & (3,2) & (4,1) \
6: & (1,5) & (2,4) & (3,3) & (4,2) & (5,1) \
7: & (1,6) & (2,5) & (3,4) & (4,3) & (5,2) & (6,1) \
8: & (1,7) & (2,6) & (3,5) & (4,4) & (5,3) & (6,2) \
9: & (1,8) & (2,7) & (3,6) & (4,5) & (5,4) & (6,3) \
10: & (1,9) & (2,8) & (3,7) & (4,6) & (5,5) & (6,4) \
11: & (1,10) & (2,9) & (3,8) & (4,7) & (5,6) & (6,5) \
12: & (2,10) & (3,9) & (4,8) & (5,7) & (6,6) \
13: & (3,10) & (4,9) & (5,8) & (6,7) \
14: & (4,10) & (5,9) & (6,8) \
15: & (5,10) & (6,9) \
16: & (6,10)
endarray
The likeliest value is not necessarily the one you should be picking.
A high value might be unlikely to win, but give you a higher reward if you win, since the number of dollars you get is the number you guessed. We should maximize the expected winnings, not the probability.
From the table above, we see that there's no point in guessing the numbers $2$ through $10$. Rather than a small number like $2$, you should guess a big number with the same probability (like $16$): it has the same chance of winning, but gives you more money if you do. Similarly, since all of $7$ through $11$ have the same chance of winning, the only one of them worth guessing is $11$.
Now we can compute the expected payoffs, multiplying the reward by the probability of getting it.
- For guessing $11$, you get $11 cdot frac660 = frac1110$ on average.
- For guessing $12$, you get $12 cdot frac560 = 1$ on average.
- For guessing $13$, you get $13 cdot frac460 = frac1315$ on average.
- For guessing $14$, you get $14 cdot frac360 = frac710$ on average.
- For guessing $15$, you get $15 cdot frac260 = frac12$ on average.
- For guessing $16$, you get $16 cdot frac160 = frac415$ on average.
You can see that these expected payoffs keep decreasing as the guess becomes higher, and so $11$ is the best value to guess.
(Once you spot that, for values in the range $11dots16$, the expected value is $n cdot frac17-n60$, you can predict this behavior: to maximize the product of two values with a fixed sum, like $n$ and $17-n$, you want to make the values as close to each other as possible.)
answered 25 mins ago
Misha Lavrov
39.5k55197
is there a clever way to do this without listing out every possible outcome? this is an interview question and in a timed setting, I would have only a few seconds to answer.
– Hat
21 mins ago
If you only have a few seconds to answer, then it'd be best to explain $how$ the answer was obtained. That shouldn't take more than 10 seconds.
– Remy
16 mins ago
@Hat The clever way is to imagine what would happen if you did list out every possible outcome, spot the pattern, and use the pattern to figure out the answer. Actually listing the outcomes is just an easy way to demonstrate what happens.
– Misha Lavrov
9 mins ago
I would guess that very few people would have the mental calisthenics to compute this in a few seconds. Von Neumann or Fermi, maybe...
– copper.hat
6 mins ago
add a comment |Â
up vote
2
down vote
accepted
The number $9$ is the expected value $mathbb E[D_6 + D_10]$, but in general that's not necessarily the likeliest value.
After all, if you were playing the same game with just a single $D_6$, you wouldn't guess $3.5$. If you did, it would take a very long time before that number was rolled.
Here, it turns out that both $9$ and $11$ are equally likely. The distribution of sums is
beginarrayrcccccc
2: & (1,1) \
3: & (1,2) & (2,1) \
4: & (1,3) & (2,2) & (3,1) \
5: & (1,4) & (2,3) & (3,2) & (4,1) \
6: & (1,5) & (2,4) & (3,3) & (4,2) & (5,1) \
7: & (1,6) & (2,5) & (3,4) & (4,3) & (5,2) & (6,1) \
8: & (1,7) & (2,6) & (3,5) & (4,4) & (5,3) & (6,2) \
9: & (1,8) & (2,7) & (3,6) & (4,5) & (5,4) & (6,3) \
10: & (1,9) & (2,8) & (3,7) & (4,6) & (5,5) & (6,4) \
11: & (1,10) & (2,9) & (3,8) & (4,7) & (5,6) & (6,5) \
12: & (2,10) & (3,9) & (4,8) & (5,7) & (6,6) \
13: & (3,10) & (4,9) & (5,8) & (6,7) \
14: & (4,10) & (5,9) & (6,8) \
15: & (5,10) & (6,9) \
16: & (6,10)
endarray
The likeliest value is not necessarily the one you should be picking.
A high value might be unlikely to win, but give you a higher reward if you win, since the number of dollars you get is the number you guessed. We should maximize the expected winnings, not the probability.
From the table above, we see that there's no point in guessing the numbers $2$ through $10$. Rather than a small number like $2$, you should guess a big number with the same probability (like $16$): it has the same chance of winning, but gives you more money if you do. Similarly, since all of $7$ through $11$ have the same chance of winning, the only one of them worth guessing is $11$.
Now we can compute the expected payoffs, multiplying the reward by the probability of getting it.
- For guessing $11$, you get $11 cdot frac660 = frac1110$ on average.
- For guessing $12$, you get $12 cdot frac560 = 1$ on average.
- For guessing $13$, you get $13 cdot frac460 = frac1315$ on average.
- For guessing $14$, you get $14 cdot frac360 = frac710$ on average.
- For guessing $15$, you get $15 cdot frac260 = frac12$ on average.
- For guessing $16$, you get $16 cdot frac160 = frac415$ on average.
You can see that these expected payoffs keep decreasing as the guess becomes higher, and so $11$ is the best value to guess.
(Once you spot that, for values in the range $11dots16$, the expected value is $n cdot frac17-n60$, you can predict this behavior: to maximize the product of two values with a fixed sum, like $n$ and $17-n$, you want to make the values as close to each other as possible.)
answered 25 mins ago
Misha Lavrov
39.5k55197
is there a clever way to do this without listing out every possible outcome? this is an interview question and in a timed setting, I would have only a few seconds to answer.
– Hat
21 mins ago
If you only have a few seconds to answer, then it'd be best to explain $how$ the answer was obtained. That shouldn't take more than 10 seconds.
– Remy
16 mins ago
@Hat The clever way is to imagine what would happen if you did list out every possible outcome, spot the pattern, and use the pattern to figure out the answer. Actually listing the outcomes is just an easy way to demonstrate what happens.
– Misha Lavrov
9 mins ago
I would guess that very few people would have the mental calisthenics to compute this in a few seconds. Von Neumann or Fermi, maybe...
– copper.hat
6 mins ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The number $9$ is the expected value $mathbb E[D_6 + D_10]$, but in general that's not necessarily the likeliest value.
After all, if you were playing the same game with just a single $D_6$, you wouldn't guess $3.5$. If you did, it would take a very long time before that number was rolled.
Here, it turns out that both $9$ and $11$ are equally likely. The distribution of sums is
beginarrayrcccccc
2: & (1,1) \
3: & (1,2) & (2,1) \
4: & (1,3) & (2,2) & (3,1) \
5: & (1,4) & (2,3) & (3,2) & (4,1) \
6: & (1,5) & (2,4) & (3,3) & (4,2) & (5,1) \
7: & (1,6) & (2,5) & (3,4) & (4,3) & (5,2) & (6,1) \
8: & (1,7) & (2,6) & (3,5) & (4,4) & (5,3) & (6,2) \
9: & (1,8) & (2,7) & (3,6) & (4,5) & (5,4) & (6,3) \
10: & (1,9) & (2,8) & (3,7) & (4,6) & (5,5) & (6,4) \
11: & (1,10) & (2,9) & (3,8) & (4,7) & (5,6) & (6,5) \
12: & (2,10) & (3,9) & (4,8) & (5,7) & (6,6) \
13: & (3,10) & (4,9) & (5,8) & (6,7) \
14: & (4,10) & (5,9) & (6,8) \
15: & (5,10) & (6,9) \
16: & (6,10)
endarray
The likeliest value is not necessarily the one you should be picking.
A high value might be unlikely to win, but give you a higher reward if you win, since the number of dollars you get is the number you guessed. We should maximize the expected winnings, not the probability.
From the table above, we see that there's no point in guessing the numbers $2$ through $10$. Rather than a small number like $2$, you should guess a big number with the same probability (like $16$): it has the same chance of winning, but gives you more money if you do. Similarly, since all of $7$ through $11$ have the same chance of winning, the only one of them worth guessing is $11$.
Now we can compute the expected payoffs, multiplying the reward by the probability of getting it.
- For guessing $11$, you get $11 cdot frac660 = frac1110$ on average.
- For guessing $12$, you get $12 cdot frac560 = 1$ on average.
- For guessing $13$, you get $13 cdot frac460 = frac1315$ on average.
- For guessing $14$, you get $14 cdot frac360 = frac710$ on average.
- For guessing $15$, you get $15 cdot frac260 = frac12$ on average.
- For guessing $16$, you get $16 cdot frac160 = frac415$ on average.
You can see that these expected payoffs keep decreasing as the guess becomes higher, and so $11$ is the best value to guess.
(Once you spot that, for values in the range $11dots16$, the expected value is $n cdot frac17-n60$, you can predict this behavior: to maximize the product of two values with a fixed sum, like $n$ and $17-n$, you want to make the values as close to each other as possible.)
answered 25 mins ago
Misha Lavrov
39.5k55197
The number $9$ is the expected value $mathbb E[D_6 + D_10]$, but in general that's not necessarily the likeliest value.
After all, if you were playing the same game with just a single $D_6$, you wouldn't guess $3.5$. If you did, it would take a very long time before that number was rolled.
Here, it turns out that both $9$ and $11$ are equally likely. The distribution of sums is
beginarrayrcccccc
2: & (1,1) \
3: & (1,2) & (2,1) \
4: & (1,3) & (2,2) & (3,1) \
5: & (1,4) & (2,3) & (3,2) & (4,1) \
6: & (1,5) & (2,4) & (3,3) & (4,2) & (5,1) \
7: & (1,6) & (2,5) & (3,4) & (4,3) & (5,2) & (6,1) \
8: & (1,7) & (2,6) & (3,5) & (4,4) & (5,3) & (6,2) \
9: & (1,8) & (2,7) & (3,6) & (4,5) & (5,4) & (6,3) \
10: & (1,9) & (2,8) & (3,7) & (4,6) & (5,5) & (6,4) \
11: & (1,10) & (2,9) & (3,8) & (4,7) & (5,6) & (6,5) \
12: & (2,10) & (3,9) & (4,8) & (5,7) & (6,6) \
13: & (3,10) & (4,9) & (5,8) & (6,7) \
14: & (4,10) & (5,9) & (6,8) \
15: & (5,10) & (6,9) \
16: & (6,10)
endarray
The likeliest value is not necessarily the one you should be picking.
A high value might be unlikely to win, but give you a higher reward if you win, since the number of dollars you get is the number you guessed. We should maximize the expected winnings, not the probability.
From the table above, we see that there's no point in guessing the numbers $2$ through $10$. Rather than a small number like $2$, you should guess a big number with the same probability (like $16$): it has the same chance of winning, but gives you more money if you do. Similarly, since all of $7$ through $11$ have the same chance of winning, the only one of them worth guessing is $11$.
Now we can compute the expected payoffs, multiplying the reward by the probability of getting it.
- For guessing $11$, you get $11 cdot frac660 = frac1110$ on average.
- For guessing $12$, you get $12 cdot frac560 = 1$ on average.
- For guessing $13$, you get $13 cdot frac460 = frac1315$ on average.
- For guessing $14$, you get $14 cdot frac360 = frac710$ on average.
- For guessing $15$, you get $15 cdot frac260 = frac12$ on average.
- For guessing $16$, you get $16 cdot frac160 = frac415$ on average.
You can see that these expected payoffs keep decreasing as the guess becomes higher, and so $11$ is the best value to guess.
(Once you spot that, for values in the range $11dots16$, the expected value is $n cdot frac17-n60$, you can predict this behavior: to maximize the product of two values with a fixed sum, like $n$ and $17-n$, you want to make the values as close to each other as possible.)
answered 25 mins ago
Misha Lavrov
39.5k55197
edited 3 mins ago
answered 25 mins ago
Misha Lavrov
39.5k55197
answered 25 mins ago
Misha Lavrov
39.5k55197
answered 25 mins ago
Misha Lavrov
39.5k55197
39.5k55197
is there a clever way to do this without listing out every possible outcome? this is an interview question and in a timed setting, I would have only a few seconds to answer.
– Hat
21 mins ago
If you only have a few seconds to answer, then it'd be best to explain $how$ the answer was obtained. That shouldn't take more than 10 seconds.
– Remy
16 mins ago
@Hat The clever way is to imagine what would happen if you did list out every possible outcome, spot the pattern, and use the pattern to figure out the answer. Actually listing the outcomes is just an easy way to demonstrate what happens.
– Misha Lavrov
9 mins ago
I would guess that very few people would have the mental calisthenics to compute this in a few seconds. Von Neumann or Fermi, maybe...
– copper.hat
6 mins ago
add a comment |Â
is there a clever way to do this without listing out every possible outcome? this is an interview question and in a timed setting, I would have only a few seconds to answer.
– Hat
21 mins ago
If you only have a few seconds to answer, then it'd be best to explain $how$ the answer was obtained. That shouldn't take more than 10 seconds.
– Remy
16 mins ago
@Hat The clever way is to imagine what would happen if you did list out every possible outcome, spot the pattern, and use the pattern to figure out the answer. Actually listing the outcomes is just an easy way to demonstrate what happens.
– Misha Lavrov
9 mins ago
I would guess that very few people would have the mental calisthenics to compute this in a few seconds. Von Neumann or Fermi, maybe...
– copper.hat
6 mins ago
is there a clever way to do this without listing out every possible outcome? this is an interview question and in a timed setting, I would have only a few seconds to answer.
– Hat
21 mins ago
is there a clever way to do this without listing out every possible outcome? this is an interview question and in a timed setting, I would have only a few seconds to answer.
– Hat
21 mins ago
If you only have a few seconds to answer, then it'd be best to explain $how$ the answer was obtained. That shouldn't take more than 10 seconds.
– Remy
16 mins ago
If you only have a few seconds to answer, then it'd be best to explain $how$ the answer was obtained. That shouldn't take more than 10 seconds.
– Remy
16 mins ago
@Hat The clever way is to imagine what would happen if you did list out every possible outcome, spot the pattern, and use the pattern to figure out the answer. Actually listing the outcomes is just an easy way to demonstrate what happens.
– Misha Lavrov
9 mins ago
@Hat The clever way is to imagine what would happen if you did list out every possible outcome, spot the pattern, and use the pattern to figure out the answer. Actually listing the outcomes is just an easy way to demonstrate what happens.
– Misha Lavrov
9 mins ago
I would guess that very few people would have the mental calisthenics to compute this in a few seconds. Von Neumann or Fermi, maybe...
– copper.hat
6 mins ago
I would guess that very few people would have the mental calisthenics to compute this in a few seconds. Von Neumann or Fermi, maybe...
– copper.hat
6 mins ago
add a comment |Â
up vote
2
down vote
beginalign
Pr(D_6 + D_10=x) &= sum_y=1^6Pr(D_6 + D_10=x|D_6=y)Pr(D_6=y) \
&= frac16 sum_y=1^6 Pr(D_10=x-y)\
&= frac16 sum_y=max(1,x-10)^min(x-1,6) Pr(D_10=x-y) \
&= frac160[min(x-1,6)-max(1,x-10)+1]
endalign
Hence, we want to maximize
$$fracx60[min(x-1,6)-max(1,x-10)+1]$$
answered 19 mins ago
Siong Thye Goh
86.3k1459108
Nice closed form expression, I am still struggling...
– copper.hat
15 mins ago
add a comment |Â
up vote
2
down vote
beginalign
Pr(D_6 + D_10=x) &= sum_y=1^6Pr(D_6 + D_10=x|D_6=y)Pr(D_6=y) \
&= frac16 sum_y=1^6 Pr(D_10=x-y)\
&= frac16 sum_y=max(1,x-10)^min(x-1,6) Pr(D_10=x-y) \
&= frac160[min(x-1,6)-max(1,x-10)+1]
endalign
Hence, we want to maximize
$$fracx60[min(x-1,6)-max(1,x-10)+1]$$
answered 19 mins ago
Siong Thye Goh
86.3k1459108
Nice closed form expression, I am still struggling...
– copper.hat
15 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
beginalign
Pr(D_6 + D_10=x) &= sum_y=1^6Pr(D_6 + D_10=x|D_6=y)Pr(D_6=y) \
&= frac16 sum_y=1^6 Pr(D_10=x-y)\
&= frac16 sum_y=max(1,x-10)^min(x-1,6) Pr(D_10=x-y) \
&= frac160[min(x-1,6)-max(1,x-10)+1]
endalign
Hence, we want to maximize
$$fracx60[min(x-1,6)-max(1,x-10)+1]$$
answered 19 mins ago
Siong Thye Goh
86.3k1459108
beginalign
Pr(D_6 + D_10=x) &= sum_y=1^6Pr(D_6 + D_10=x|D_6=y)Pr(D_6=y) \
&= frac16 sum_y=1^6 Pr(D_10=x-y)\
&= frac16 sum_y=max(1,x-10)^min(x-1,6) Pr(D_10=x-y) \
&= frac160[min(x-1,6)-max(1,x-10)+1]
endalign
Hence, we want to maximize
$$fracx60[min(x-1,6)-max(1,x-10)+1]$$
answered 19 mins ago
Siong Thye Goh
86.3k1459108
answered 19 mins ago
Siong Thye Goh
86.3k1459108
answered 19 mins ago
Siong Thye Goh
86.3k1459108
answered 19 mins ago
Siong Thye Goh
86.3k1459108
86.3k1459108
Nice closed form expression, I am still struggling...
– copper.hat
15 mins ago
add a comment |Â
Nice closed form expression, I am still struggling...
– copper.hat
15 mins ago
Nice closed form expression, I am still struggling...
– copper.hat
15 mins ago
Nice closed form expression, I am still struggling...
– copper.hat
15 mins ago
add a comment |Â
up vote
0
down vote
The expected number of winnings when guessing a number $n$ is going to be the dollar value $n$ multiplied by the probability that the sum of the two dice is $n.$ So even though $9$ is the expected value for the sum of the dice, the expected winnings by guessing $9$ is going to be $9 cdot frac110 = $0.90,$ because we can compute that the probability of the sum being $9$ is $frac660 = frac110.$ Using this methodology, we can find that the highest expected winnings occur at $n=11,$ with $11 cdot frac110 = $ 1.10.$
answered 18 mins ago
anonymous helper
1
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anonymous helper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
The expected number of winnings when guessing a number $n$ is going to be the dollar value $n$ multiplied by the probability that the sum of the two dice is $n.$ So even though $9$ is the expected value for the sum of the dice, the expected winnings by guessing $9$ is going to be $9 cdot frac110 = $0.90,$ because we can compute that the probability of the sum being $9$ is $frac660 = frac110.$ Using this methodology, we can find that the highest expected winnings occur at $n=11,$ with $11 cdot frac110 = $ 1.10.$
answered 18 mins ago
anonymous helper
1
New contributor
anonymous helper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The expected number of winnings when guessing a number $n$ is going to be the dollar value $n$ multiplied by the probability that the sum of the two dice is $n.$ So even though $9$ is the expected value for the sum of the dice, the expected winnings by guessing $9$ is going to be $9 cdot frac110 = $0.90,$ because we can compute that the probability of the sum being $9$ is $frac660 = frac110.$ Using this methodology, we can find that the highest expected winnings occur at $n=11,$ with $11 cdot frac110 = $ 1.10.$
answered 18 mins ago
anonymous helper
1
New contributor
anonymous helper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The expected number of winnings when guessing a number $n$ is going to be the dollar value $n$ multiplied by the probability that the sum of the two dice is $n.$ So even though $9$ is the expected value for the sum of the dice, the expected winnings by guessing $9$ is going to be $9 cdot frac110 = $0.90,$ because we can compute that the probability of the sum being $9$ is $frac660 = frac110.$ Using this methodology, we can find that the highest expected winnings occur at $n=11,$ with $11 cdot frac110 = $ 1.10.$
answered 18 mins ago
anonymous helper
1
New contributor
anonymous helper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 18 mins ago
anonymous helper
1
New contributor
anonymous helper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 18 mins ago
anonymous helper
1
answered 18 mins ago
anonymous helper
1
1
New contributor
anonymous helper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
anonymous helper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
anonymous helper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
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