Mixing
What is the probability of two people rolling a die and one person rolling higher if one of them adds 1 or 2...
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Two players roll a 6-sided die, it is a 50/50 chance that one will roll higher, but what would the probability be if one player gets to add 1 or 2 to their die's total?
I'm developing a tabletop ancient/medieval battle strategy game. Say a light infantry unit is engaged with a heavy infantry unit, the light unit has no modifications to combat rolls but the heavy infantry gets +2 to combat rolls. What is the probability that the heavy infantry unit will have a higher roll?
Also, since tieing would be a 1/6 chance (which results in no damage being dealt), would that affect the probability? And how would the probability change if rolling a "1" cannot count as more than 1 regardless of modifications?
probability dice
asked Feb 2 at 9:13
Tristan AlexanderTristan Alexander
43
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show 1 more comment
$begingroup$
Two players roll a 6-sided die, it is a 50/50 chance that one will roll higher, but what would the probability be if one player gets to add 1 or 2 to their die's total?
I'm developing a tabletop ancient/medieval battle strategy game. Say a light infantry unit is engaged with a heavy infantry unit, the light unit has no modifications to combat rolls but the heavy infantry gets +2 to combat rolls. What is the probability that the heavy infantry unit will have a higher roll?
Also, since tieing would be a 1/6 chance (which results in no damage being dealt), would that affect the probability? And how would the probability change if rolling a "1" cannot count as more than 1 regardless of modifications?
probability dice
asked Feb 2 at 9:13
Tristan AlexanderTristan Alexander
43
$endgroup$
$begingroup$
If wo players roll a fair 6-sided die, the probability that you score higher than the other is 5/12, which is not 50/50 chance
$endgroup$
– pendermath
Feb 2 at 9:20
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@pendermath Well since both players have a 5/12 chance to roll higher than the other than it is 50/50 isn't it? Edit: so you're right, each player has about a 42% of rolling higher and a 16% chance of tieing. But I really need to know what happens to that probability if one player gets to add a 1 or 2.
$endgroup$
– Tristan Alexander
Feb 2 at 9:30
$begingroup$
I don't understand the question: under which conditions or with which probabilities can one player add 1 or 2? Can either player add 1 or 2 or only one of them? Maybe you can edit your question and add more clarity to it.
$endgroup$
– pendermath
Feb 2 at 9:39
$begingroup$
Only one player gets to add 1 or 2, as I explained in my post, a light infantry unit has no modifications to their roll, but a heavy infantry unit gets +2 to their roll. What is the probability of the heavy infantry unit rolling higher than the light infantry unit? Also, thanks for the response.
$endgroup$
– Tristan Alexander
Feb 2 at 9:43
$begingroup$
The number of cases is small enough that you can easily construct a table of all possible outcomes.
$endgroup$
– amd
Feb 3 at 0:43
|
show 1 more comment
$begingroup$
Two players roll a 6-sided die, it is a 50/50 chance that one will roll higher, but what would the probability be if one player gets to add 1 or 2 to their die's total?
I'm developing a tabletop ancient/medieval battle strategy game. Say a light infantry unit is engaged with a heavy infantry unit, the light unit has no modifications to combat rolls but the heavy infantry gets +2 to combat rolls. What is the probability that the heavy infantry unit will have a higher roll?
Also, since tieing would be a 1/6 chance (which results in no damage being dealt), would that affect the probability? And how would the probability change if rolling a "1" cannot count as more than 1 regardless of modifications?
probability dice
asked Feb 2 at 9:13
Tristan AlexanderTristan Alexander
43
$endgroup$
Two players roll a 6-sided die, it is a 50/50 chance that one will roll higher, but what would the probability be if one player gets to add 1 or 2 to their die's total?
I'm developing a tabletop ancient/medieval battle strategy game. Say a light infantry unit is engaged with a heavy infantry unit, the light unit has no modifications to combat rolls but the heavy infantry gets +2 to combat rolls. What is the probability that the heavy infantry unit will have a higher roll?
Also, since tieing would be a 1/6 chance (which results in no damage being dealt), would that affect the probability? And how would the probability change if rolling a "1" cannot count as more than 1 regardless of modifications?
probability dice
probability dice
asked Feb 2 at 9:13
Tristan AlexanderTristan Alexander
43
asked Feb 2 at 9:13
Tristan AlexanderTristan Alexander
43
edited Feb 2 at 9:18
Tristan Alexander
asked Feb 2 at 9:13
Tristan AlexanderTristan Alexander
43
asked Feb 2 at 9:13
Tristan AlexanderTristan Alexander
43
asked Feb 2 at 9:13
Tristan AlexanderTristan Alexander
43
43
$begingroup$
If wo players roll a fair 6-sided die, the probability that you score higher than the other is 5/12, which is not 50/50 chance
$endgroup$
– pendermath
Feb 2 at 9:20
$begingroup$
@pendermath Well since both players have a 5/12 chance to roll higher than the other than it is 50/50 isn't it? Edit: so you're right, each player has about a 42% of rolling higher and a 16% chance of tieing. But I really need to know what happens to that probability if one player gets to add a 1 or 2.
$endgroup$
– Tristan Alexander
Feb 2 at 9:30
$begingroup$
I don't understand the question: under which conditions or with which probabilities can one player add 1 or 2? Can either player add 1 or 2 or only one of them? Maybe you can edit your question and add more clarity to it.
$endgroup$
– pendermath
Feb 2 at 9:39
$begingroup$
Only one player gets to add 1 or 2, as I explained in my post, a light infantry unit has no modifications to their roll, but a heavy infantry unit gets +2 to their roll. What is the probability of the heavy infantry unit rolling higher than the light infantry unit? Also, thanks for the response.
$endgroup$
– Tristan Alexander
Feb 2 at 9:43
$begingroup$
The number of cases is small enough that you can easily construct a table of all possible outcomes.
$endgroup$
– amd
Feb 3 at 0:43
|
show 1 more comment
$begingroup$
If wo players roll a fair 6-sided die, the probability that you score higher than the other is 5/12, which is not 50/50 chance
$endgroup$
– pendermath
Feb 2 at 9:20
$begingroup$
@pendermath Well since both players have a 5/12 chance to roll higher than the other than it is 50/50 isn't it? Edit: so you're right, each player has about a 42% of rolling higher and a 16% chance of tieing. But I really need to know what happens to that probability if one player gets to add a 1 or 2.
$endgroup$
– Tristan Alexander
Feb 2 at 9:30
$begingroup$
I don't understand the question: under which conditions or with which probabilities can one player add 1 or 2? Can either player add 1 or 2 or only one of them? Maybe you can edit your question and add more clarity to it.
$endgroup$
– pendermath
Feb 2 at 9:39
$begingroup$
Only one player gets to add 1 or 2, as I explained in my post, a light infantry unit has no modifications to their roll, but a heavy infantry unit gets +2 to their roll. What is the probability of the heavy infantry unit rolling higher than the light infantry unit? Also, thanks for the response.
$endgroup$
– Tristan Alexander
Feb 2 at 9:43
$begingroup$
The number of cases is small enough that you can easily construct a table of all possible outcomes.
$endgroup$
– amd
Feb 3 at 0:43
$begingroup$
If wo players roll a fair 6-sided die, the probability that you score higher than the other is 5/12, which is not 50/50 chance
$endgroup$
– pendermath
Feb 2 at 9:20
$begingroup$
If wo players roll a fair 6-sided die, the probability that you score higher than the other is 5/12, which is not 50/50 chance
$endgroup$
– pendermath
Feb 2 at 9:20
$begingroup$
@pendermath Well since both players have a 5/12 chance to roll higher than the other than it is 50/50 isn't it? Edit: so you're right, each player has about a 42% of rolling higher and a 16% chance of tieing. But I really need to know what happens to that probability if one player gets to add a 1 or 2.
$endgroup$
– Tristan Alexander
Feb 2 at 9:30
$begingroup$
@pendermath Well since both players have a 5/12 chance to roll higher than the other than it is 50/50 isn't it? Edit: so you're right, each player has about a 42% of rolling higher and a 16% chance of tieing. But I really need to know what happens to that probability if one player gets to add a 1 or 2.
$endgroup$
– Tristan Alexander
Feb 2 at 9:30
$begingroup$
I don't understand the question: under which conditions or with which probabilities can one player add 1 or 2? Can either player add 1 or 2 or only one of them? Maybe you can edit your question and add more clarity to it.
$endgroup$
– pendermath
Feb 2 at 9:39
$begingroup$
I don't understand the question: under which conditions or with which probabilities can one player add 1 or 2? Can either player add 1 or 2 or only one of them? Maybe you can edit your question and add more clarity to it.
$endgroup$
– pendermath
Feb 2 at 9:39
$begingroup$
Only one player gets to add 1 or 2, as I explained in my post, a light infantry unit has no modifications to their roll, but a heavy infantry unit gets +2 to their roll. What is the probability of the heavy infantry unit rolling higher than the light infantry unit? Also, thanks for the response.
$endgroup$
– Tristan Alexander
Feb 2 at 9:43
$begingroup$
Only one player gets to add 1 or 2, as I explained in my post, a light infantry unit has no modifications to their roll, but a heavy infantry unit gets +2 to their roll. What is the probability of the heavy infantry unit rolling higher than the light infantry unit? Also, thanks for the response.
$endgroup$
– Tristan Alexander
Feb 2 at 9:43
$begingroup$
The number of cases is small enough that you can easily construct a table of all possible outcomes.
$endgroup$
– amd
Feb 3 at 0:43
$begingroup$
The number of cases is small enough that you can easily construct a table of all possible outcomes.
$endgroup$
– amd
Feb 3 at 0:43
|
show 1 more comment
1 Answer
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I will handle the case where the die of the high infantry gets an increase of 2. The same approach can then be used to determine the probabilities when adding any other number.
If the light infantry throws a 1 or 2, then the high infantry always wins. If the light infantry throws a 3, the probability of a victory for the high infantry equals $frac{5}{6}$ (throw higher than 1). Throwing a 4, we find a probability of $frac{1}{6}$ for a tie (throw 2) and $frac{4}{6}$ for a victory (throw higher than 2). Applying the same to rolls of 5 and 6, and we find, for the events $V$ (high infantry wins), $T$ (a tie) and $L$ (high infantry loses):
$$P(V) = frac{1}{6} left( 1 + 1 + frac{5}{6} + frac{4}{6} + frac{3}{6} + frac{2}{6} right) = frac{26}{36} = frac{13}{18}$$
$$P(T) = frac{1}{6} left( 0 + 0 + frac{1}{6} + frac{1}{6} + frac{1}{6} + frac{1}{6} right) = frac{4}{36} = frac{1}{9}$$
$$P(L) = 1 - P(V) - P(T) = frac{6}{36} = frac{1}{6}$$
answered Feb 2 at 9:45
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Thank you so much. Looks like the heavy infantry would dominate the light infantry with only a 16% chance of losing. Would it be possible to calculate the heavy infantry's chances against 3 light infantry units (as that is how many I have a heavy infantry being worth)? And can you also calculate the probability if rolling a "1" cannot count as more than 1?
$endgroup$
– Tristan Alexander
Feb 2 at 9:52
$begingroup$
@TristanAlexander You could definitely calculate these probabilities, depending on how you combine the three light infantry dice. If rolling a 1 counts as a 1 only, you can distinguish between this roll and any higher roll to calculate the overall probabilities. The easiest way to go is to draw a 6$times$6 table, where columns have the numbers 1 to 6 for the light infantry, and rows have the numbers 1, 4, 5, 6, 7 and 8 for the heavy infantry. It then simply comes down to counting the number of victories, ties and losses!
$endgroup$
– jvdhooft
Feb 2 at 10:07
$begingroup$
I'm so glad I asked this forum, thank you again! I used the table method and with a "1" not counting as more than 1, the heavy infantry has a 22% chance of losing. If I can ask one last quick question, to calculate the heavy infantry against 3 light infantry, would I just multiply this by 3? The game mechanics make it so anytime a light infantry attacks, the heavy infantry attacks them back using the dice mechanics explained above. Not accounting for flanking and squeezing mechanics, is a heavy infantry actually worth 3 light infantry? Also, heavy infantry has 6 health and light infantry has 4
$endgroup$
– Tristan Alexander
Feb 2 at 10:25
$begingroup$
Nvm I think I got it. A heavy infantry unit can kill three light infantry with an average of 2 health leftover, but accounting for the flanking mechanic, it would only have about 1.6 lives leftover, and with the squeezing mechanic would leave it with -1.5 lives, meaning the light infantry has the advantage if they can squeeze. However, I didn't take into account the at least one turn it would take for the light infantry to manuever into squeezing the heavy infantry though I still think they have a significant advantage. Thanks again!
$endgroup$
– Tristan Alexander
Feb 2 at 10:48
$begingroup$
@TristanAlexander In the end, it all depends on how you combine the three dice. If you attack one by one, and deal damage based on winning or losing, you can just repeat this procedure three times. Best of luck with your game!
$endgroup$
– jvdhooft
Feb 2 at 10:49
|
show 1 more comment
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1 Answer
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I will handle the case where the die of the high infantry gets an increase of 2. The same approach can then be used to determine the probabilities when adding any other number.
If the light infantry throws a 1 or 2, then the high infantry always wins. If the light infantry throws a 3, the probability of a victory for the high infantry equals $frac{5}{6}$ (throw higher than 1). Throwing a 4, we find a probability of $frac{1}{6}$ for a tie (throw 2) and $frac{4}{6}$ for a victory (throw higher than 2). Applying the same to rolls of 5 and 6, and we find, for the events $V$ (high infantry wins), $T$ (a tie) and $L$ (high infantry loses):
$$P(V) = frac{1}{6} left( 1 + 1 + frac{5}{6} + frac{4}{6} + frac{3}{6} + frac{2}{6} right) = frac{26}{36} = frac{13}{18}$$
$$P(T) = frac{1}{6} left( 0 + 0 + frac{1}{6} + frac{1}{6} + frac{1}{6} + frac{1}{6} right) = frac{4}{36} = frac{1}{9}$$
$$P(L) = 1 - P(V) - P(T) = frac{6}{36} = frac{1}{6}$$
answered Feb 2 at 9:45
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Thank you so much. Looks like the heavy infantry would dominate the light infantry with only a 16% chance of losing. Would it be possible to calculate the heavy infantry's chances against 3 light infantry units (as that is how many I have a heavy infantry being worth)? And can you also calculate the probability if rolling a "1" cannot count as more than 1?
$endgroup$
– Tristan Alexander
Feb 2 at 9:52
$begingroup$
@TristanAlexander You could definitely calculate these probabilities, depending on how you combine the three light infantry dice. If rolling a 1 counts as a 1 only, you can distinguish between this roll and any higher roll to calculate the overall probabilities. The easiest way to go is to draw a 6$times$6 table, where columns have the numbers 1 to 6 for the light infantry, and rows have the numbers 1, 4, 5, 6, 7 and 8 for the heavy infantry. It then simply comes down to counting the number of victories, ties and losses!
$endgroup$
– jvdhooft
Feb 2 at 10:07
$begingroup$
I'm so glad I asked this forum, thank you again! I used the table method and with a "1" not counting as more than 1, the heavy infantry has a 22% chance of losing. If I can ask one last quick question, to calculate the heavy infantry against 3 light infantry, would I just multiply this by 3? The game mechanics make it so anytime a light infantry attacks, the heavy infantry attacks them back using the dice mechanics explained above. Not accounting for flanking and squeezing mechanics, is a heavy infantry actually worth 3 light infantry? Also, heavy infantry has 6 health and light infantry has 4
$endgroup$
– Tristan Alexander
Feb 2 at 10:25
$begingroup$
Nvm I think I got it. A heavy infantry unit can kill three light infantry with an average of 2 health leftover, but accounting for the flanking mechanic, it would only have about 1.6 lives leftover, and with the squeezing mechanic would leave it with -1.5 lives, meaning the light infantry has the advantage if they can squeeze. However, I didn't take into account the at least one turn it would take for the light infantry to manuever into squeezing the heavy infantry though I still think they have a significant advantage. Thanks again!
$endgroup$
– Tristan Alexander
Feb 2 at 10:48
$begingroup$
@TristanAlexander In the end, it all depends on how you combine the three dice. If you attack one by one, and deal damage based on winning or losing, you can just repeat this procedure three times. Best of luck with your game!
$endgroup$
– jvdhooft
Feb 2 at 10:49
|
show 1 more comment
$begingroup$
I will handle the case where the die of the high infantry gets an increase of 2. The same approach can then be used to determine the probabilities when adding any other number.
If the light infantry throws a 1 or 2, then the high infantry always wins. If the light infantry throws a 3, the probability of a victory for the high infantry equals $frac{5}{6}$ (throw higher than 1). Throwing a 4, we find a probability of $frac{1}{6}$ for a tie (throw 2) and $frac{4}{6}$ for a victory (throw higher than 2). Applying the same to rolls of 5 and 6, and we find, for the events $V$ (high infantry wins), $T$ (a tie) and $L$ (high infantry loses):
$$P(V) = frac{1}{6} left( 1 + 1 + frac{5}{6} + frac{4}{6} + frac{3}{6} + frac{2}{6} right) = frac{26}{36} = frac{13}{18}$$
$$P(T) = frac{1}{6} left( 0 + 0 + frac{1}{6} + frac{1}{6} + frac{1}{6} + frac{1}{6} right) = frac{4}{36} = frac{1}{9}$$
$$P(L) = 1 - P(V) - P(T) = frac{6}{36} = frac{1}{6}$$
answered Feb 2 at 9:45
jvdhooftjvdhooft
5,65961641
$endgroup$
$begingroup$
Thank you so much. Looks like the heavy infantry would dominate the light infantry with only a 16% chance of losing. Would it be possible to calculate the heavy infantry's chances against 3 light infantry units (as that is how many I have a heavy infantry being worth)? And can you also calculate the probability if rolling a "1" cannot count as more than 1?
$endgroup$
– Tristan Alexander
Feb 2 at 9:52
$begingroup$
@TristanAlexander You could definitely calculate these probabilities, depending on how you combine the three light infantry dice. If rolling a 1 counts as a 1 only, you can distinguish between this roll and any higher roll to calculate the overall probabilities. The easiest way to go is to draw a 6$times$6 table, where columns have the numbers 1 to 6 for the light infantry, and rows have the numbers 1, 4, 5, 6, 7 and 8 for the heavy infantry. It then simply comes down to counting the number of victories, ties and losses!
$endgroup$
– jvdhooft
Feb 2 at 10:07
$begingroup$
I'm so glad I asked this forum, thank you again! I used the table method and with a "1" not counting as more than 1, the heavy infantry has a 22% chance of losing. If I can ask one last quick question, to calculate the heavy infantry against 3 light infantry, would I just multiply this by 3? The game mechanics make it so anytime a light infantry attacks, the heavy infantry attacks them back using the dice mechanics explained above. Not accounting for flanking and squeezing mechanics, is a heavy infantry actually worth 3 light infantry? Also, heavy infantry has 6 health and light infantry has 4
$endgroup$
– Tristan Alexander
Feb 2 at 10:25
$begingroup$
Nvm I think I got it. A heavy infantry unit can kill three light infantry with an average of 2 health leftover, but accounting for the flanking mechanic, it would only have about 1.6 lives leftover, and with the squeezing mechanic would leave it with -1.5 lives, meaning the light infantry has the advantage if they can squeeze. However, I didn't take into account the at least one turn it would take for the light infantry to manuever into squeezing the heavy infantry though I still think they have a significant advantage. Thanks again!
$endgroup$
– Tristan Alexander
Feb 2 at 10:48
$begingroup$
@TristanAlexander In the end, it all depends on how you combine the three dice. If you attack one by one, and deal damage based on winning or losing, you can just repeat this procedure three times. Best of luck with your game!
$endgroup$
– jvdhooft
Feb 2 at 10:49
|
show 1 more comment
$begingroup$
I will handle the case where the die of the high infantry gets an increase of 2. The same approach can then be used to determine the probabilities when adding any other number.
If the light infantry throws a 1 or 2, then the high infantry always wins. If the light infantry throws a 3, the probability of a victory for the high infantry equals $frac{5}{6}$ (throw higher than 1). Throwing a 4, we find a probability of $frac{1}{6}$ for a tie (throw 2) and $frac{4}{6}$ for a victory (throw higher than 2). Applying the same to rolls of 5 and 6, and we find, for the events $V$ (high infantry wins), $T$ (a tie) and $L$ (high infantry loses):
$$P(V) = frac{1}{6} left( 1 + 1 + frac{5}{6} + frac{4}{6} + frac{3}{6} + frac{2}{6} right) = frac{26}{36} = frac{13}{18}$$
$$P(T) = frac{1}{6} left( 0 + 0 + frac{1}{6} + frac{1}{6} + frac{1}{6} + frac{1}{6} right) = frac{4}{36} = frac{1}{9}$$
$$P(L) = 1 - P(V) - P(T) = frac{6}{36} = frac{1}{6}$$
answered Feb 2 at 9:45
jvdhooftjvdhooft
5,65961641
$endgroup$
I will handle the case where the die of the high infantry gets an increase of 2. The same approach can then be used to determine the probabilities when adding any other number.
If the light infantry throws a 1 or 2, then the high infantry always wins. If the light infantry throws a 3, the probability of a victory for the high infantry equals $frac{5}{6}$ (throw higher than 1). Throwing a 4, we find a probability of $frac{1}{6}$ for a tie (throw 2) and $frac{4}{6}$ for a victory (throw higher than 2). Applying the same to rolls of 5 and 6, and we find, for the events $V$ (high infantry wins), $T$ (a tie) and $L$ (high infantry loses):
$$P(V) = frac{1}{6} left( 1 + 1 + frac{5}{6} + frac{4}{6} + frac{3}{6} + frac{2}{6} right) = frac{26}{36} = frac{13}{18}$$
$$P(T) = frac{1}{6} left( 0 + 0 + frac{1}{6} + frac{1}{6} + frac{1}{6} + frac{1}{6} right) = frac{4}{36} = frac{1}{9}$$
$$P(L) = 1 - P(V) - P(T) = frac{6}{36} = frac{1}{6}$$
answered Feb 2 at 9:45
jvdhooftjvdhooft
5,65961641
answered Feb 2 at 9:45
jvdhooftjvdhooft
5,65961641
answered Feb 2 at 9:45
jvdhooftjvdhooft
5,65961641
answered Feb 2 at 9:45
jvdhooftjvdhooft
5,65961641
5,65961641
$begingroup$
Thank you so much. Looks like the heavy infantry would dominate the light infantry with only a 16% chance of losing. Would it be possible to calculate the heavy infantry's chances against 3 light infantry units (as that is how many I have a heavy infantry being worth)? And can you also calculate the probability if rolling a "1" cannot count as more than 1?
$endgroup$
– Tristan Alexander
Feb 2 at 9:52
$begingroup$
@TristanAlexander You could definitely calculate these probabilities, depending on how you combine the three light infantry dice. If rolling a 1 counts as a 1 only, you can distinguish between this roll and any higher roll to calculate the overall probabilities. The easiest way to go is to draw a 6$times$6 table, where columns have the numbers 1 to 6 for the light infantry, and rows have the numbers 1, 4, 5, 6, 7 and 8 for the heavy infantry. It then simply comes down to counting the number of victories, ties and losses!
$endgroup$
– jvdhooft
Feb 2 at 10:07
$begingroup$
I'm so glad I asked this forum, thank you again! I used the table method and with a "1" not counting as more than 1, the heavy infantry has a 22% chance of losing. If I can ask one last quick question, to calculate the heavy infantry against 3 light infantry, would I just multiply this by 3? The game mechanics make it so anytime a light infantry attacks, the heavy infantry attacks them back using the dice mechanics explained above. Not accounting for flanking and squeezing mechanics, is a heavy infantry actually worth 3 light infantry? Also, heavy infantry has 6 health and light infantry has 4
$endgroup$
– Tristan Alexander
Feb 2 at 10:25
$begingroup$
Nvm I think I got it. A heavy infantry unit can kill three light infantry with an average of 2 health leftover, but accounting for the flanking mechanic, it would only have about 1.6 lives leftover, and with the squeezing mechanic would leave it with -1.5 lives, meaning the light infantry has the advantage if they can squeeze. However, I didn't take into account the at least one turn it would take for the light infantry to manuever into squeezing the heavy infantry though I still think they have a significant advantage. Thanks again!
$endgroup$
– Tristan Alexander
Feb 2 at 10:48
$begingroup$
@TristanAlexander In the end, it all depends on how you combine the three dice. If you attack one by one, and deal damage based on winning or losing, you can just repeat this procedure three times. Best of luck with your game!
$endgroup$
– jvdhooft
Feb 2 at 10:49
|
show 1 more comment
$begingroup$
Thank you so much. Looks like the heavy infantry would dominate the light infantry with only a 16% chance of losing. Would it be possible to calculate the heavy infantry's chances against 3 light infantry units (as that is how many I have a heavy infantry being worth)? And can you also calculate the probability if rolling a "1" cannot count as more than 1?
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– Tristan Alexander
Feb 2 at 9:52
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@TristanAlexander You could definitely calculate these probabilities, depending on how you combine the three light infantry dice. If rolling a 1 counts as a 1 only, you can distinguish between this roll and any higher roll to calculate the overall probabilities. The easiest way to go is to draw a 6$times$6 table, where columns have the numbers 1 to 6 for the light infantry, and rows have the numbers 1, 4, 5, 6, 7 and 8 for the heavy infantry. It then simply comes down to counting the number of victories, ties and losses!
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– jvdhooft
Feb 2 at 10:07
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I'm so glad I asked this forum, thank you again! I used the table method and with a "1" not counting as more than 1, the heavy infantry has a 22% chance of losing. If I can ask one last quick question, to calculate the heavy infantry against 3 light infantry, would I just multiply this by 3? The game mechanics make it so anytime a light infantry attacks, the heavy infantry attacks them back using the dice mechanics explained above. Not accounting for flanking and squeezing mechanics, is a heavy infantry actually worth 3 light infantry? Also, heavy infantry has 6 health and light infantry has 4
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– Tristan Alexander
Feb 2 at 10:25
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Nvm I think I got it. A heavy infantry unit can kill three light infantry with an average of 2 health leftover, but accounting for the flanking mechanic, it would only have about 1.6 lives leftover, and with the squeezing mechanic would leave it with -1.5 lives, meaning the light infantry has the advantage if they can squeeze. However, I didn't take into account the at least one turn it would take for the light infantry to manuever into squeezing the heavy infantry though I still think they have a significant advantage. Thanks again!
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– Tristan Alexander
Feb 2 at 10:48
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@TristanAlexander In the end, it all depends on how you combine the three dice. If you attack one by one, and deal damage based on winning or losing, you can just repeat this procedure three times. Best of luck with your game!
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– jvdhooft
Feb 2 at 10:49
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Thank you so much. Looks like the heavy infantry would dominate the light infantry with only a 16% chance of losing. Would it be possible to calculate the heavy infantry's chances against 3 light infantry units (as that is how many I have a heavy infantry being worth)? And can you also calculate the probability if rolling a "1" cannot count as more than 1?
$endgroup$
– Tristan Alexander
Feb 2 at 9:52
$begingroup$
Thank you so much. Looks like the heavy infantry would dominate the light infantry with only a 16% chance of losing. Would it be possible to calculate the heavy infantry's chances against 3 light infantry units (as that is how many I have a heavy infantry being worth)? And can you also calculate the probability if rolling a "1" cannot count as more than 1?
$endgroup$
– Tristan Alexander
Feb 2 at 9:52
$begingroup$
@TristanAlexander You could definitely calculate these probabilities, depending on how you combine the three light infantry dice. If rolling a 1 counts as a 1 only, you can distinguish between this roll and any higher roll to calculate the overall probabilities. The easiest way to go is to draw a 6$times$6 table, where columns have the numbers 1 to 6 for the light infantry, and rows have the numbers 1, 4, 5, 6, 7 and 8 for the heavy infantry. It then simply comes down to counting the number of victories, ties and losses!
$endgroup$
– jvdhooft
Feb 2 at 10:07
$begingroup$
@TristanAlexander You could definitely calculate these probabilities, depending on how you combine the three light infantry dice. If rolling a 1 counts as a 1 only, you can distinguish between this roll and any higher roll to calculate the overall probabilities. The easiest way to go is to draw a 6$times$6 table, where columns have the numbers 1 to 6 for the light infantry, and rows have the numbers 1, 4, 5, 6, 7 and 8 for the heavy infantry. It then simply comes down to counting the number of victories, ties and losses!
$endgroup$
– jvdhooft
Feb 2 at 10:07
$begingroup$
I'm so glad I asked this forum, thank you again! I used the table method and with a "1" not counting as more than 1, the heavy infantry has a 22% chance of losing. If I can ask one last quick question, to calculate the heavy infantry against 3 light infantry, would I just multiply this by 3? The game mechanics make it so anytime a light infantry attacks, the heavy infantry attacks them back using the dice mechanics explained above. Not accounting for flanking and squeezing mechanics, is a heavy infantry actually worth 3 light infantry? Also, heavy infantry has 6 health and light infantry has 4
$endgroup$
– Tristan Alexander
Feb 2 at 10:25
$begingroup$
I'm so glad I asked this forum, thank you again! I used the table method and with a "1" not counting as more than 1, the heavy infantry has a 22% chance of losing. If I can ask one last quick question, to calculate the heavy infantry against 3 light infantry, would I just multiply this by 3? The game mechanics make it so anytime a light infantry attacks, the heavy infantry attacks them back using the dice mechanics explained above. Not accounting for flanking and squeezing mechanics, is a heavy infantry actually worth 3 light infantry? Also, heavy infantry has 6 health and light infantry has 4
$endgroup$
– Tristan Alexander
Feb 2 at 10:25
$begingroup$
Nvm I think I got it. A heavy infantry unit can kill three light infantry with an average of 2 health leftover, but accounting for the flanking mechanic, it would only have about 1.6 lives leftover, and with the squeezing mechanic would leave it with -1.5 lives, meaning the light infantry has the advantage if they can squeeze. However, I didn't take into account the at least one turn it would take for the light infantry to manuever into squeezing the heavy infantry though I still think they have a significant advantage. Thanks again!
$endgroup$
– Tristan Alexander
Feb 2 at 10:48
$begingroup$
Nvm I think I got it. A heavy infantry unit can kill three light infantry with an average of 2 health leftover, but accounting for the flanking mechanic, it would only have about 1.6 lives leftover, and with the squeezing mechanic would leave it with -1.5 lives, meaning the light infantry has the advantage if they can squeeze. However, I didn't take into account the at least one turn it would take for the light infantry to manuever into squeezing the heavy infantry though I still think they have a significant advantage. Thanks again!
$endgroup$
– Tristan Alexander
Feb 2 at 10:48
$begingroup$
@TristanAlexander In the end, it all depends on how you combine the three dice. If you attack one by one, and deal damage based on winning or losing, you can just repeat this procedure three times. Best of luck with your game!
$endgroup$
– jvdhooft
Feb 2 at 10:49
$begingroup$
@TristanAlexander In the end, it all depends on how you combine the three dice. If you attack one by one, and deal damage based on winning or losing, you can just repeat this procedure three times. Best of luck with your game!
$endgroup$
– jvdhooft
Feb 2 at 10:49
|
show 1 more comment
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$begingroup$
If wo players roll a fair 6-sided die, the probability that you score higher than the other is 5/12, which is not 50/50 chance
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– pendermath
Feb 2 at 9:20
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@pendermath Well since both players have a 5/12 chance to roll higher than the other than it is 50/50 isn't it? Edit: so you're right, each player has about a 42% of rolling higher and a 16% chance of tieing. But I really need to know what happens to that probability if one player gets to add a 1 or 2.
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– Tristan Alexander
Feb 2 at 9:30
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I don't understand the question: under which conditions or with which probabilities can one player add 1 or 2? Can either player add 1 or 2 or only one of them? Maybe you can edit your question and add more clarity to it.
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– pendermath
Feb 2 at 9:39
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Only one player gets to add 1 or 2, as I explained in my post, a light infantry unit has no modifications to their roll, but a heavy infantry unit gets +2 to their roll. What is the probability of the heavy infantry unit rolling higher than the light infantry unit? Also, thanks for the response.
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– Tristan Alexander
Feb 2 at 9:43
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The number of cases is small enough that you can easily construct a table of all possible outcomes.
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– amd
Feb 3 at 0:43