Mixing
Expected winnings of a slot machine
$begingroup$
Of the three wheels on a slot machine where each wheel has ten items:
- First wheel has 2 flowers, 4 dogs, 4 houses.
- Second wheel has 6 flowers, 3 dogs, 1 house.
- Third wheel has 2 flowers, 3 dogs, 5 houses.
You win $20$ dollars per dollar if $3$ flowers show, $10$ dollars if $3$ dogs show, and $5$ dollars if $3$ houses show (and lose otherwise). We're asked to find the expected winning per dollar spent.
I tried solving it this way:
$p(text{flowers})=(2/10)(6/10)(2/10) = 3/125$
$p(text{dogs})=(4/10)(3/10)(3/10) = 9/250$
$p(text{houses})=(4/10)(1/10)(5/10) = 1/50$
$p(text{rest})=1-[p(text{flower})+p(text{dogs})+p(text{houses})]=23/25$
So expected winning is;
$20(3/125)+10(9/250)+5(1/50)-1(23/25)=0.92$
I thought I did it correctly, but the answer key says it is $0.94. Did I do something wrong?
probability expectation
edited Nov 13 '14 at 10:48
user642796
44.9k565119
asked Nov 13 '14 at 10:21
user192198user192198
61
$endgroup$
add a comment |
$begingroup$
Of the three wheels on a slot machine where each wheel has ten items:
- First wheel has 2 flowers, 4 dogs, 4 houses.
- Second wheel has 6 flowers, 3 dogs, 1 house.
- Third wheel has 2 flowers, 3 dogs, 5 houses.
You win $20$ dollars per dollar if $3$ flowers show, $10$ dollars if $3$ dogs show, and $5$ dollars if $3$ houses show (and lose otherwise). We're asked to find the expected winning per dollar spent.
I tried solving it this way:
$p(text{flowers})=(2/10)(6/10)(2/10) = 3/125$
$p(text{dogs})=(4/10)(3/10)(3/10) = 9/250$
$p(text{houses})=(4/10)(1/10)(5/10) = 1/50$
$p(text{rest})=1-[p(text{flower})+p(text{dogs})+p(text{houses})]=23/25$
So expected winning is;
$20(3/125)+10(9/250)+5(1/50)-1(23/25)=0.92$
I thought I did it correctly, but the answer key says it is $0.94. Did I do something wrong?
probability expectation
edited Nov 13 '14 at 10:48
user642796
44.9k565119
asked Nov 13 '14 at 10:21
user192198user192198
61
$endgroup$
1
$begingroup$
Especially subtracting $1(23/25)$ is incorrect. You are asked to calculate the expected winning on a dollar spent. Maybe it is not the only thing. Just compare with my answer.
$endgroup$
– drhab
Nov 13 '14 at 11:03
add a comment |
$begingroup$
Of the three wheels on a slot machine where each wheel has ten items:
- First wheel has 2 flowers, 4 dogs, 4 houses.
- Second wheel has 6 flowers, 3 dogs, 1 house.
- Third wheel has 2 flowers, 3 dogs, 5 houses.
You win $20$ dollars per dollar if $3$ flowers show, $10$ dollars if $3$ dogs show, and $5$ dollars if $3$ houses show (and lose otherwise). We're asked to find the expected winning per dollar spent.
I tried solving it this way:
$p(text{flowers})=(2/10)(6/10)(2/10) = 3/125$
$p(text{dogs})=(4/10)(3/10)(3/10) = 9/250$
$p(text{houses})=(4/10)(1/10)(5/10) = 1/50$
$p(text{rest})=1-[p(text{flower})+p(text{dogs})+p(text{houses})]=23/25$
So expected winning is;
$20(3/125)+10(9/250)+5(1/50)-1(23/25)=0.92$
I thought I did it correctly, but the answer key says it is $0.94. Did I do something wrong?
probability expectation
edited Nov 13 '14 at 10:48
user642796
44.9k565119
asked Nov 13 '14 at 10:21
user192198user192198
61
$endgroup$
Of the three wheels on a slot machine where each wheel has ten items:
- First wheel has 2 flowers, 4 dogs, 4 houses.
- Second wheel has 6 flowers, 3 dogs, 1 house.
- Third wheel has 2 flowers, 3 dogs, 5 houses.
You win $20$ dollars per dollar if $3$ flowers show, $10$ dollars if $3$ dogs show, and $5$ dollars if $3$ houses show (and lose otherwise). We're asked to find the expected winning per dollar spent.
I tried solving it this way:
$p(text{flowers})=(2/10)(6/10)(2/10) = 3/125$
$p(text{dogs})=(4/10)(3/10)(3/10) = 9/250$
$p(text{houses})=(4/10)(1/10)(5/10) = 1/50$
$p(text{rest})=1-[p(text{flower})+p(text{dogs})+p(text{houses})]=23/25$
So expected winning is;
$20(3/125)+10(9/250)+5(1/50)-1(23/25)=0.92$
I thought I did it correctly, but the answer key says it is $0.94. Did I do something wrong?
probability expectation
probability expectation
edited Nov 13 '14 at 10:48
user642796
44.9k565119
asked Nov 13 '14 at 10:21
user192198user192198
61
edited Nov 13 '14 at 10:48
user642796
44.9k565119
asked Nov 13 '14 at 10:21
user192198user192198
61
edited Nov 13 '14 at 10:48
user642796
44.9k565119
edited Nov 13 '14 at 10:48
user642796
44.9k565119
edited Nov 13 '14 at 10:48
user642796
44.9k565119
44.9k565119
asked Nov 13 '14 at 10:21
user192198user192198
61
asked Nov 13 '14 at 10:21
user192198user192198
61
asked Nov 13 '14 at 10:21
user192198user192198
61
61
1
$begingroup$
Especially subtracting $1(23/25)$ is incorrect. You are asked to calculate the expected winning on a dollar spent. Maybe it is not the only thing. Just compare with my answer.
$endgroup$
– drhab
Nov 13 '14 at 11:03
add a comment |
1
$begingroup$
Especially subtracting $1(23/25)$ is incorrect. You are asked to calculate the expected winning on a dollar spent. Maybe it is not the only thing. Just compare with my answer.
$endgroup$
– drhab
Nov 13 '14 at 11:03
1
1
$begingroup$
Especially subtracting $1(23/25)$ is incorrect. You are asked to calculate the expected winning on a dollar spent. Maybe it is not the only thing. Just compare with my answer.
$endgroup$
– drhab
Nov 13 '14 at 11:03
$begingroup$
Especially subtracting $1(23/25)$ is incorrect. You are asked to calculate the expected winning on a dollar spent. Maybe it is not the only thing. Just compare with my answer.
$endgroup$
– drhab
Nov 13 '14 at 11:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It concerns purely the winning on a dollar spent.
$p_{f}=frac{2}{10}frac{6}{10}frac{2}{10}=frac{24}{1000}$
$p_{d}=frac{4}{10}frac{3}{10}frac{3}{10}=frac{36}{1000}$
$p_{h}=frac{4}{10}frac{1}{10}frac{5}{10}=frac{20}{1000}$
$text{expected winning}=20timesfrac{24}{1000}+10timesfrac{36}{1000}+5timesfrac{20}{1000}=frac{940}{1000}=0.94$
If it had concerned 'rendement' in the sense of winning minus loosing then the correct answer would have been $0.94-1.00$. In all cases you must pay a dollar. Not only when you do not win.
answered Nov 13 '14 at 10:30
drhabdrhab
104k545136
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It concerns purely the winning on a dollar spent.
$p_{f}=frac{2}{10}frac{6}{10}frac{2}{10}=frac{24}{1000}$
$p_{d}=frac{4}{10}frac{3}{10}frac{3}{10}=frac{36}{1000}$
$p_{h}=frac{4}{10}frac{1}{10}frac{5}{10}=frac{20}{1000}$
$text{expected winning}=20timesfrac{24}{1000}+10timesfrac{36}{1000}+5timesfrac{20}{1000}=frac{940}{1000}=0.94$
If it had concerned 'rendement' in the sense of winning minus loosing then the correct answer would have been $0.94-1.00$. In all cases you must pay a dollar. Not only when you do not win.
answered Nov 13 '14 at 10:30
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
It concerns purely the winning on a dollar spent.
$p_{f}=frac{2}{10}frac{6}{10}frac{2}{10}=frac{24}{1000}$
$p_{d}=frac{4}{10}frac{3}{10}frac{3}{10}=frac{36}{1000}$
$p_{h}=frac{4}{10}frac{1}{10}frac{5}{10}=frac{20}{1000}$
$text{expected winning}=20timesfrac{24}{1000}+10timesfrac{36}{1000}+5timesfrac{20}{1000}=frac{940}{1000}=0.94$
If it had concerned 'rendement' in the sense of winning minus loosing then the correct answer would have been $0.94-1.00$. In all cases you must pay a dollar. Not only when you do not win.
answered Nov 13 '14 at 10:30
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
It concerns purely the winning on a dollar spent.
$p_{f}=frac{2}{10}frac{6}{10}frac{2}{10}=frac{24}{1000}$
$p_{d}=frac{4}{10}frac{3}{10}frac{3}{10}=frac{36}{1000}$
$p_{h}=frac{4}{10}frac{1}{10}frac{5}{10}=frac{20}{1000}$
$text{expected winning}=20timesfrac{24}{1000}+10timesfrac{36}{1000}+5timesfrac{20}{1000}=frac{940}{1000}=0.94$
If it had concerned 'rendement' in the sense of winning minus loosing then the correct answer would have been $0.94-1.00$. In all cases you must pay a dollar. Not only when you do not win.
answered Nov 13 '14 at 10:30
drhabdrhab
104k545136
$endgroup$
It concerns purely the winning on a dollar spent.
$p_{f}=frac{2}{10}frac{6}{10}frac{2}{10}=frac{24}{1000}$
$p_{d}=frac{4}{10}frac{3}{10}frac{3}{10}=frac{36}{1000}$
$p_{h}=frac{4}{10}frac{1}{10}frac{5}{10}=frac{20}{1000}$
$text{expected winning}=20timesfrac{24}{1000}+10timesfrac{36}{1000}+5timesfrac{20}{1000}=frac{940}{1000}=0.94$
If it had concerned 'rendement' in the sense of winning minus loosing then the correct answer would have been $0.94-1.00$. In all cases you must pay a dollar. Not only when you do not win.
answered Nov 13 '14 at 10:30
drhabdrhab
104k545136
edited Nov 13 '14 at 11:10
answered Nov 13 '14 at 10:30
drhabdrhab
104k545136
answered Nov 13 '14 at 10:30
drhabdrhab
104k545136
answered Nov 13 '14 at 10:30
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
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$begingroup$
Especially subtracting $1(23/25)$ is incorrect. You are asked to calculate the expected winning on a dollar spent. Maybe it is not the only thing. Just compare with my answer.
$endgroup$
– drhab
Nov 13 '14 at 11:03