Slot Machine Math, Probability of hitting X
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My slot machine has 3 slots, and each time the user pulls the lever I want them to be given a different set of symbols in a random order every time including amount of each symbol.
For example, if hitting a row of apples has a chance of winning x100, what would be the math that determines how many apples should be in a row if I wanted the winning amount (in this case x100) to equal to probability of winning (1 in 100). So if there was 10 symbols which would be 10*10*10 to produce 1000 possible combinations (forgetting the fact that an apple can take up more than 1 of those 10 symbols). What would be the math used to determine how much of those 10 symbols needs to be apples to make it exactly a 1 in 100 chance of hitting it?
Would putting only 1 apple in each slot be 1 in 1000? Does that make 2 apples in each slot 1 in 500?
Just did some basic math but don't know if this is right:
A) If I have 10 symbols there is 1000 possibilities.
B) If 2 of those symbols where apples in each slot, there would be 8 out of 1000 possible Apple wins.
C) So does that mean there is an average 125 spins to get all apples? The closest to 1 in 100 chance? I am confused.
probability
asked Sep 7 at 4:27
Ugleh
1034
add a comment |Â
up vote
0
down vote
favorite
My slot machine has 3 slots, and each time the user pulls the lever I want them to be given a different set of symbols in a random order every time including amount of each symbol.
For example, if hitting a row of apples has a chance of winning x100, what would be the math that determines how many apples should be in a row if I wanted the winning amount (in this case x100) to equal to probability of winning (1 in 100). So if there was 10 symbols which would be 10*10*10 to produce 1000 possible combinations (forgetting the fact that an apple can take up more than 1 of those 10 symbols). What would be the math used to determine how much of those 10 symbols needs to be apples to make it exactly a 1 in 100 chance of hitting it?
Would putting only 1 apple in each slot be 1 in 1000? Does that make 2 apples in each slot 1 in 500?
Just did some basic math but don't know if this is right:
A) If I have 10 symbols there is 1000 possibilities.
B) If 2 of those symbols where apples in each slot, there would be 8 out of 1000 possible Apple wins.
C) So does that mean there is an average 125 spins to get all apples? The closest to 1 in 100 chance? I am confused.
probability
asked Sep 7 at 4:27
Ugleh
1034
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My slot machine has 3 slots, and each time the user pulls the lever I want them to be given a different set of symbols in a random order every time including amount of each symbol.
For example, if hitting a row of apples has a chance of winning x100, what would be the math that determines how many apples should be in a row if I wanted the winning amount (in this case x100) to equal to probability of winning (1 in 100). So if there was 10 symbols which would be 10*10*10 to produce 1000 possible combinations (forgetting the fact that an apple can take up more than 1 of those 10 symbols). What would be the math used to determine how much of those 10 symbols needs to be apples to make it exactly a 1 in 100 chance of hitting it?
Would putting only 1 apple in each slot be 1 in 1000? Does that make 2 apples in each slot 1 in 500?
Just did some basic math but don't know if this is right:
A) If I have 10 symbols there is 1000 possibilities.
B) If 2 of those symbols where apples in each slot, there would be 8 out of 1000 possible Apple wins.
C) So does that mean there is an average 125 spins to get all apples? The closest to 1 in 100 chance? I am confused.
probability
asked Sep 7 at 4:27
Ugleh
1034
My slot machine has 3 slots, and each time the user pulls the lever I want them to be given a different set of symbols in a random order every time including amount of each symbol.
For example, if hitting a row of apples has a chance of winning x100, what would be the math that determines how many apples should be in a row if I wanted the winning amount (in this case x100) to equal to probability of winning (1 in 100). So if there was 10 symbols which would be 10*10*10 to produce 1000 possible combinations (forgetting the fact that an apple can take up more than 1 of those 10 symbols). What would be the math used to determine how much of those 10 symbols needs to be apples to make it exactly a 1 in 100 chance of hitting it?
Would putting only 1 apple in each slot be 1 in 1000? Does that make 2 apples in each slot 1 in 500?
Just did some basic math but don't know if this is right:
A) If I have 10 symbols there is 1000 possibilities.
B) If 2 of those symbols where apples in each slot, there would be 8 out of 1000 possible Apple wins.
C) So does that mean there is an average 125 spins to get all apples? The closest to 1 in 100 chance? I am confused.
probability
probability
asked Sep 7 at 4:27
Ugleh
1034
asked Sep 7 at 4:27
Ugleh
1034
edited Sep 7 at 4:38
asked Sep 7 at 4:27
Ugleh
1034
asked Sep 7 at 4:27
Ugleh
1034
asked Sep 7 at 4:27
Ugleh
1034
1034
add a comment |Â
add a comment |Â
1 Answer
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Well, I believe you have the same probability for each slot. $left(fraca^310right)^3=frac1100$ so $a=sqrt[3]10$ so it must occupy $sqrt[3]10$ slots, whatever that means.
If you put $1$ apple in one slot, you would get $frac11000,$ but if you put $2$ apples, then there are $2cdot2cdot2=8$ combinations for a total of $frac1125.$
Note: If you allow different (assuming integral) numbers of slots, it can be shown it must be a permutation if $(1,5,2)$ or $(1,1,10)$ slots.
EDIT: There are a few more questions asked.
A) Yes. There are $10$ combinations for each slot for a total of $10cdot10cdot10=1000$ combinations.
B) Yes. The probability for one slot is $frac210=frac15.$ The probability of three is $left(frac15right)^3=frac1125=frac81000.$
C) Yes, the expected value of spins required is $125.$ It is the closest unless you allow one slot to have more apple slots than another.
answered Sep 7 at 4:34
Jason Kim
548117
Do you mean $frac11000$ if there was just 1 apple?
– Ugleh
Sep 7 at 4:40
Yes... edited :)
– Jason Kim
Sep 7 at 4:50
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Well, I believe you have the same probability for each slot. $left(fraca^310right)^3=frac1100$ so $a=sqrt[3]10$ so it must occupy $sqrt[3]10$ slots, whatever that means.
If you put $1$ apple in one slot, you would get $frac11000,$ but if you put $2$ apples, then there are $2cdot2cdot2=8$ combinations for a total of $frac1125.$
Note: If you allow different (assuming integral) numbers of slots, it can be shown it must be a permutation if $(1,5,2)$ or $(1,1,10)$ slots.
EDIT: There are a few more questions asked.
A) Yes. There are $10$ combinations for each slot for a total of $10cdot10cdot10=1000$ combinations.
B) Yes. The probability for one slot is $frac210=frac15.$ The probability of three is $left(frac15right)^3=frac1125=frac81000.$
C) Yes, the expected value of spins required is $125.$ It is the closest unless you allow one slot to have more apple slots than another.
answered Sep 7 at 4:34
Jason Kim
548117
Do you mean $frac11000$ if there was just 1 apple?
– Ugleh
Sep 7 at 4:40
Yes... edited :)
– Jason Kim
Sep 7 at 4:50
add a comment |Â
up vote
2
down vote
accepted
Well, I believe you have the same probability for each slot. $left(fraca^310right)^3=frac1100$ so $a=sqrt[3]10$ so it must occupy $sqrt[3]10$ slots, whatever that means.
If you put $1$ apple in one slot, you would get $frac11000,$ but if you put $2$ apples, then there are $2cdot2cdot2=8$ combinations for a total of $frac1125.$
Note: If you allow different (assuming integral) numbers of slots, it can be shown it must be a permutation if $(1,5,2)$ or $(1,1,10)$ slots.
EDIT: There are a few more questions asked.
A) Yes. There are $10$ combinations for each slot for a total of $10cdot10cdot10=1000$ combinations.
B) Yes. The probability for one slot is $frac210=frac15.$ The probability of three is $left(frac15right)^3=frac1125=frac81000.$
C) Yes, the expected value of spins required is $125.$ It is the closest unless you allow one slot to have more apple slots than another.
answered Sep 7 at 4:34
Jason Kim
548117
Do you mean $frac11000$ if there was just 1 apple?
– Ugleh
Sep 7 at 4:40
Yes... edited :)
– Jason Kim
Sep 7 at 4:50
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Well, I believe you have the same probability for each slot. $left(fraca^310right)^3=frac1100$ so $a=sqrt[3]10$ so it must occupy $sqrt[3]10$ slots, whatever that means.
If you put $1$ apple in one slot, you would get $frac11000,$ but if you put $2$ apples, then there are $2cdot2cdot2=8$ combinations for a total of $frac1125.$
Note: If you allow different (assuming integral) numbers of slots, it can be shown it must be a permutation if $(1,5,2)$ or $(1,1,10)$ slots.
EDIT: There are a few more questions asked.
A) Yes. There are $10$ combinations for each slot for a total of $10cdot10cdot10=1000$ combinations.
B) Yes. The probability for one slot is $frac210=frac15.$ The probability of three is $left(frac15right)^3=frac1125=frac81000.$
C) Yes, the expected value of spins required is $125.$ It is the closest unless you allow one slot to have more apple slots than another.
answered Sep 7 at 4:34
Jason Kim
548117
Well, I believe you have the same probability for each slot. $left(fraca^310right)^3=frac1100$ so $a=sqrt[3]10$ so it must occupy $sqrt[3]10$ slots, whatever that means.
If you put $1$ apple in one slot, you would get $frac11000,$ but if you put $2$ apples, then there are $2cdot2cdot2=8$ combinations for a total of $frac1125.$
Note: If you allow different (assuming integral) numbers of slots, it can be shown it must be a permutation if $(1,5,2)$ or $(1,1,10)$ slots.
EDIT: There are a few more questions asked.
A) Yes. There are $10$ combinations for each slot for a total of $10cdot10cdot10=1000$ combinations.
B) Yes. The probability for one slot is $frac210=frac15.$ The probability of three is $left(frac15right)^3=frac1125=frac81000.$
C) Yes, the expected value of spins required is $125.$ It is the closest unless you allow one slot to have more apple slots than another.
answered Sep 7 at 4:34
Jason Kim
548117
edited Sep 8 at 1:17
answered Sep 7 at 4:34
Jason Kim
548117
answered Sep 7 at 4:34
Jason Kim
548117
answered Sep 7 at 4:34
Jason Kim
548117
548117
Do you mean $frac11000$ if there was just 1 apple?
– Ugleh
Sep 7 at 4:40
Yes... edited :)
– Jason Kim
Sep 7 at 4:50
add a comment |Â
Do you mean $frac11000$ if there was just 1 apple?
– Ugleh
Sep 7 at 4:40
Yes... edited :)
– Jason Kim
Sep 7 at 4:50
Do you mean $frac11000$ if there was just 1 apple?
– Ugleh
Sep 7 at 4:40
Do you mean $frac11000$ if there was just 1 apple?
– Ugleh
Sep 7 at 4:40
Yes... edited :)
– Jason Kim
Sep 7 at 4:50
Yes... edited :)
– Jason Kim
Sep 7 at 4:50
add a comment |Â
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