Number of permutations from 3:2 to 5:4 [on hold] The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)PermutationsCombinations from digitsPermutations from a setOptimal Number of Entries for a Contest of Skill…What is the expected value? Three dice are rolled. For a 1 dollar bet you win 1 dollar for each 6 that appears (plus dollar back). No 6, lose dollar.Two candidates, A & B, are running for president. What is the probability that candidate A beats candidate B?Permutations (Making numbers from digits)Number of possible permutationsCalculating Risk of Ruin for known profit/loss and probabilitiesHow do you solve the Absent Minded Gambler problem?How is the number of permutations of alike objects different from from the normal number of permutations?
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Number of permutations from 3:2 to 5:4 [on hold]
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)PermutationsCombinations from digitsPermutations from a setOptimal Number of Entries for a Contest of Skill…What is the expected value? Three dice are rolled. For a 1 dollar bet you win 1 dollar for each 6 that appears (plus dollar back). No 6, lose dollar.Two candidates, A & B, are running for president. What is the probability that candidate A beats candidate B?Permutations (Making numbers from digits)Number of possible permutationsCalculating Risk of Ruin for known profit/loss and probabilitiesHow do you solve the Absent Minded Gambler problem?How is the number of permutations of alike objects different from from the normal number of permutations?
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I’m trying to make a Rock Paper Scissors game, and also track the percent chance of not losing (tie and win)
I did this previously by seeing the final outcomes of the event and erasing any impossible outcomes. For example, the score is currently 2:1 and its best out of 5. To find the “probability” for the person with the 2, I do this.
(5:0),(4:1),(3:2),(2:3),(1:4),(0:5). Then I see that (1:4),(0:5),(5:0) are impossible. It comes down to (4:1),(3:2),(2:3). In that case, I check for the outcome where the first value is larger than the second. This eliminates (2:3). Then I previously put the amount of remaining possibilities that result in a win (2) over the total amount of logical possibilities (3). 2/3. But I realize now that each of the logical possibilities have different chances of happening. I tried the permutation formula, but I have no clue how to use it in this scenario. Any help would be greatly appreciated! (hinting me toward a formula).
probability permutations
asked Apr 8 at 12:13
nullnull
11
New contributor
null is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as unclear what you're asking by Don Thousand, Lord Shark the Unknown, Shailesh, Javi, José Carlos Santos Apr 9 at 14:33
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I’m trying to make a Rock Paper Scissors game, and also track the percent chance of not losing (tie and win)
I did this previously by seeing the final outcomes of the event and erasing any impossible outcomes. For example, the score is currently 2:1 and its best out of 5. To find the “probability” for the person with the 2, I do this.
(5:0),(4:1),(3:2),(2:3),(1:4),(0:5). Then I see that (1:4),(0:5),(5:0) are impossible. It comes down to (4:1),(3:2),(2:3). In that case, I check for the outcome where the first value is larger than the second. This eliminates (2:3). Then I previously put the amount of remaining possibilities that result in a win (2) over the total amount of logical possibilities (3). 2/3. But I realize now that each of the logical possibilities have different chances of happening. I tried the permutation formula, but I have no clue how to use it in this scenario. Any help would be greatly appreciated! (hinting me toward a formula).
probability permutations
asked Apr 8 at 12:13
nullnull
11
New contributor
null is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as unclear what you're asking by Don Thousand, Lord Shark the Unknown, Shailesh, Javi, José Carlos Santos Apr 9 at 14:33
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
" In that case, I check for the outcome where the first value is tester than the second." What?
$endgroup$
– 5xum
Apr 8 at 12:16
$begingroup$
Sorry am on phone autocorrect
$endgroup$
– null
Apr 8 at 12:21
add a comment |
$begingroup$
I’m trying to make a Rock Paper Scissors game, and also track the percent chance of not losing (tie and win)
I did this previously by seeing the final outcomes of the event and erasing any impossible outcomes. For example, the score is currently 2:1 and its best out of 5. To find the “probability” for the person with the 2, I do this.
(5:0),(4:1),(3:2),(2:3),(1:4),(0:5). Then I see that (1:4),(0:5),(5:0) are impossible. It comes down to (4:1),(3:2),(2:3). In that case, I check for the outcome where the first value is larger than the second. This eliminates (2:3). Then I previously put the amount of remaining possibilities that result in a win (2) over the total amount of logical possibilities (3). 2/3. But I realize now that each of the logical possibilities have different chances of happening. I tried the permutation formula, but I have no clue how to use it in this scenario. Any help would be greatly appreciated! (hinting me toward a formula).
probability permutations
asked Apr 8 at 12:13
nullnull
11
New contributor
null is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I’m trying to make a Rock Paper Scissors game, and also track the percent chance of not losing (tie and win)
I did this previously by seeing the final outcomes of the event and erasing any impossible outcomes. For example, the score is currently 2:1 and its best out of 5. To find the “probability” for the person with the 2, I do this.
(5:0),(4:1),(3:2),(2:3),(1:4),(0:5). Then I see that (1:4),(0:5),(5:0) are impossible. It comes down to (4:1),(3:2),(2:3). In that case, I check for the outcome where the first value is larger than the second. This eliminates (2:3). Then I previously put the amount of remaining possibilities that result in a win (2) over the total amount of logical possibilities (3). 2/3. But I realize now that each of the logical possibilities have different chances of happening. I tried the permutation formula, but I have no clue how to use it in this scenario. Any help would be greatly appreciated! (hinting me toward a formula).
probability permutations
probability permutations
asked Apr 8 at 12:13
nullnull
11
New contributor
null is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 12:13
nullnull
11
New contributor
null is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 8 at 12:21
null
asked Apr 8 at 12:13
nullnull
11
New contributor
null is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 12:13
nullnull
11
asked Apr 8 at 12:13
nullnull
11
11
New contributor
null is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
null is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
null is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as unclear what you're asking by Don Thousand, Lord Shark the Unknown, Shailesh, Javi, José Carlos Santos Apr 9 at 14:33
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Don Thousand, Lord Shark the Unknown, Shailesh, Javi, José Carlos Santos Apr 9 at 14:33
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
" In that case, I check for the outcome where the first value is tester than the second." What?
$endgroup$
– 5xum
Apr 8 at 12:16
$begingroup$
Sorry am on phone autocorrect
$endgroup$
– null
Apr 8 at 12:21
add a comment |
$begingroup$
" In that case, I check for the outcome where the first value is tester than the second." What?
$endgroup$
– 5xum
Apr 8 at 12:16
$begingroup$
Sorry am on phone autocorrect
$endgroup$
– null
Apr 8 at 12:21
$begingroup$
" In that case, I check for the outcome where the first value is tester than the second." What?
$endgroup$
– 5xum
Apr 8 at 12:16
$begingroup$
" In that case, I check for the outcome where the first value is tester than the second." What?
$endgroup$
– 5xum
Apr 8 at 12:16
$begingroup$
Sorry am on phone autocorrect
$endgroup$
– null
Apr 8 at 12:21
$begingroup$
Sorry am on phone autocorrect
$endgroup$
– null
Apr 8 at 12:21
add a comment |
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$begingroup$
" In that case, I check for the outcome where the first value is tester than the second." What?
$endgroup$
– 5xum
Apr 8 at 12:16
$begingroup$
Sorry am on phone autocorrect
$endgroup$
– null
Apr 8 at 12:21