How to calculate the expected value when throwing multiple dices?
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I'm starting to learn probability theory and I found an exercise that states:
On a particular move we throw a 20 sided dice one time, and then we throw a 4 sided dice twice. We substract the sum of the 2 throwings of the 4 sided dice to the value we got throwing the 20 sided dice. So if we throw the first one and we get 18, then we throw to times the 4 sided and we get 1 and 2, the final result will be 15.
How can I calculate the expected value? Should I generate all possible throwing combinations and look for the ones who repeat the most?
How can I calculate which range of results will get on the 50% of the times? (I'm a little bit lost on this one)
Thanks for your attention
probability probability-theory probability-distributions
asked Dec 29 '18 at 13:40
Pau MuñozPau Muñoz
154
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add a comment |
$begingroup$
I'm starting to learn probability theory and I found an exercise that states:
On a particular move we throw a 20 sided dice one time, and then we throw a 4 sided dice twice. We substract the sum of the 2 throwings of the 4 sided dice to the value we got throwing the 20 sided dice. So if we throw the first one and we get 18, then we throw to times the 4 sided and we get 1 and 2, the final result will be 15.
How can I calculate the expected value? Should I generate all possible throwing combinations and look for the ones who repeat the most?
How can I calculate which range of results will get on the 50% of the times? (I'm a little bit lost on this one)
Thanks for your attention
probability probability-theory probability-distributions
asked Dec 29 '18 at 13:40
Pau MuñozPau Muñoz
154
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1
$begingroup$
You can use E[X-Y-Z]=E[X]-E[Y]-E[Z]
$endgroup$
– Michael
Dec 29 '18 at 13:47
add a comment |
$begingroup$
I'm starting to learn probability theory and I found an exercise that states:
On a particular move we throw a 20 sided dice one time, and then we throw a 4 sided dice twice. We substract the sum of the 2 throwings of the 4 sided dice to the value we got throwing the 20 sided dice. So if we throw the first one and we get 18, then we throw to times the 4 sided and we get 1 and 2, the final result will be 15.
How can I calculate the expected value? Should I generate all possible throwing combinations and look for the ones who repeat the most?
How can I calculate which range of results will get on the 50% of the times? (I'm a little bit lost on this one)
Thanks for your attention
probability probability-theory probability-distributions
asked Dec 29 '18 at 13:40
Pau MuñozPau Muñoz
154
$endgroup$
I'm starting to learn probability theory and I found an exercise that states:
On a particular move we throw a 20 sided dice one time, and then we throw a 4 sided dice twice. We substract the sum of the 2 throwings of the 4 sided dice to the value we got throwing the 20 sided dice. So if we throw the first one and we get 18, then we throw to times the 4 sided and we get 1 and 2, the final result will be 15.
How can I calculate the expected value? Should I generate all possible throwing combinations and look for the ones who repeat the most?
How can I calculate which range of results will get on the 50% of the times? (I'm a little bit lost on this one)
Thanks for your attention
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Dec 29 '18 at 13:40
Pau MuñozPau Muñoz
154
asked Dec 29 '18 at 13:40
Pau MuñozPau Muñoz
154
asked Dec 29 '18 at 13:40
Pau MuñozPau Muñoz
154
asked Dec 29 '18 at 13:40
Pau MuñozPau Muñoz
154
asked Dec 29 '18 at 13:40
Pau MuñozPau Muñoz
154
154
1
$begingroup$
You can use E[X-Y-Z]=E[X]-E[Y]-E[Z]
$endgroup$
– Michael
Dec 29 '18 at 13:47
add a comment |
1
$begingroup$
You can use E[X-Y-Z]=E[X]-E[Y]-E[Z]
$endgroup$
– Michael
Dec 29 '18 at 13:47
1
1
$begingroup$
You can use E[X-Y-Z]=E[X]-E[Y]-E[Z]
$endgroup$
– Michael
Dec 29 '18 at 13:47
$begingroup$
You can use E[X-Y-Z]=E[X]-E[Y]-E[Z]
$endgroup$
– Michael
Dec 29 '18 at 13:47
add a comment |
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$begingroup$
You can use E[X-Y-Z]=E[X]-E[Y]-E[Z]
$endgroup$
– Michael
Dec 29 '18 at 13:47