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Probability of failing before a second condition
Calculating the probability that throwing two dice will yield a higher number than throwing one dieProbability with $n$ successes before $m$ failuresThe probability of success if rolling two dice and taking the lower resultGame of Dice and probability of playingProbability of winning this dice gameDifficult ProbabilityProbability that any outcome of a dice roll happens more than X times out of Y trialsRolling two doubles with 4 and 5, 6 sided diceProbability question - clueless about mathI roll two dice, where the first die gets a +1 bonus to it's roll
$begingroup$
This question came up during a D&D game I was playing in and I found that I was unable to come up with an answer. I believe this should be a summation question, but I may be entirely off:
You roll a 20 sided die. A success is defined as rolling 16 or higher and failure is defined as rolling 15 or lower. You roll the die until 60 trials have been completed or you accumulate 6 failures. What is the probability of accumulating 6 failures, and how would you find the probability of lasting an arbitrary number of trials before accumulating 6 failures?
probability dice
asked Mar 29 at 16:35
Rudy BakinRudy Bakin
52
$endgroup$
add a comment |
$begingroup$
This question came up during a D&D game I was playing in and I found that I was unable to come up with an answer. I believe this should be a summation question, but I may be entirely off:
You roll a 20 sided die. A success is defined as rolling 16 or higher and failure is defined as rolling 15 or lower. You roll the die until 60 trials have been completed or you accumulate 6 failures. What is the probability of accumulating 6 failures, and how would you find the probability of lasting an arbitrary number of trials before accumulating 6 failures?
probability dice
asked Mar 29 at 16:35
Rudy BakinRudy Bakin
52
$endgroup$
$begingroup$
Good grief, who are they trying to defeat?! :-o
$endgroup$
– Brian Tung
Mar 29 at 17:25
add a comment |
$begingroup$
This question came up during a D&D game I was playing in and I found that I was unable to come up with an answer. I believe this should be a summation question, but I may be entirely off:
You roll a 20 sided die. A success is defined as rolling 16 or higher and failure is defined as rolling 15 or lower. You roll the die until 60 trials have been completed or you accumulate 6 failures. What is the probability of accumulating 6 failures, and how would you find the probability of lasting an arbitrary number of trials before accumulating 6 failures?
probability dice
asked Mar 29 at 16:35
Rudy BakinRudy Bakin
52
$endgroup$
This question came up during a D&D game I was playing in and I found that I was unable to come up with an answer. I believe this should be a summation question, but I may be entirely off:
You roll a 20 sided die. A success is defined as rolling 16 or higher and failure is defined as rolling 15 or lower. You roll the die until 60 trials have been completed or you accumulate 6 failures. What is the probability of accumulating 6 failures, and how would you find the probability of lasting an arbitrary number of trials before accumulating 6 failures?
probability dice
probability dice
asked Mar 29 at 16:35
Rudy BakinRudy Bakin
52
asked Mar 29 at 16:35
Rudy BakinRudy Bakin
52
asked Mar 29 at 16:35
Rudy BakinRudy Bakin
52
asked Mar 29 at 16:35
Rudy BakinRudy Bakin
52
asked Mar 29 at 16:35
Rudy BakinRudy Bakin
52
52
$begingroup$
Good grief, who are they trying to defeat?! :-o
$endgroup$
– Brian Tung
Mar 29 at 17:25
add a comment |
$begingroup$
Good grief, who are they trying to defeat?! :-o
$endgroup$
– Brian Tung
Mar 29 at 17:25
$begingroup$
Good grief, who are they trying to defeat?! :-o
$endgroup$
– Brian Tung
Mar 29 at 17:25
$begingroup$
Good grief, who are they trying to defeat?! :-o
$endgroup$
– Brian Tung
Mar 29 at 17:25
add a comment |
1 Answer
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oldest
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$begingroup$
I will answer the easiest first. You want the probability that in 60 rolls, you roll 16 or higher all but at most five times. That is a binomial probability. The probability of survival is given by this formula:
$$sum_k=0^5 dbinom60kleft(dfrac14right)^60-kleft(dfrac34right)^k approx 1times 10^-25%$$
Now, the probability that you can survive a given number of trials is:
You have 100% probability of surviving up to 5 trials.
Probability that you survive up to $n$ trials (for $nge 6$):
$$sum_k=0^5 dbinomnkleft(dfrac14right)^n-kleft(dfrac34right)^k$$
Probability you lose on the $n$-th trial:
$$dbinomn-15left(dfrac14right)^n-6left(dfrac34right)^6$$
Expected number of trials before finally succumbing to luck:
$$sum_nge 6ndbinomn-15left(dfrac14right)^n-6left(dfrac34right)^6 = 8$$
You can expect to fail by the eighth trial (on average).
answered Mar 29 at 16:49
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
I will answer the easiest first. You want the probability that in 60 rolls, you roll 16 or higher all but at most five times. That is a binomial probability. The probability of survival is given by this formula:
$$sum_k=0^5 dbinom60kleft(dfrac14right)^60-kleft(dfrac34right)^k approx 1times 10^-25%$$
Now, the probability that you can survive a given number of trials is:
You have 100% probability of surviving up to 5 trials.
Probability that you survive up to $n$ trials (for $nge 6$):
$$sum_k=0^5 dbinomnkleft(dfrac14right)^n-kleft(dfrac34right)^k$$
Probability you lose on the $n$-th trial:
$$dbinomn-15left(dfrac14right)^n-6left(dfrac34right)^6$$
Expected number of trials before finally succumbing to luck:
$$sum_nge 6ndbinomn-15left(dfrac14right)^n-6left(dfrac34right)^6 = 8$$
You can expect to fail by the eighth trial (on average).
answered Mar 29 at 16:49
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
add a comment |
$begingroup$
I will answer the easiest first. You want the probability that in 60 rolls, you roll 16 or higher all but at most five times. That is a binomial probability. The probability of survival is given by this formula:
$$sum_k=0^5 dbinom60kleft(dfrac14right)^60-kleft(dfrac34right)^k approx 1times 10^-25%$$
Now, the probability that you can survive a given number of trials is:
You have 100% probability of surviving up to 5 trials.
Probability that you survive up to $n$ trials (for $nge 6$):
$$sum_k=0^5 dbinomnkleft(dfrac14right)^n-kleft(dfrac34right)^k$$
Probability you lose on the $n$-th trial:
$$dbinomn-15left(dfrac14right)^n-6left(dfrac34right)^6$$
Expected number of trials before finally succumbing to luck:
$$sum_nge 6ndbinomn-15left(dfrac14right)^n-6left(dfrac34right)^6 = 8$$
You can expect to fail by the eighth trial (on average).
answered Mar 29 at 16:49
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
add a comment |
$begingroup$
I will answer the easiest first. You want the probability that in 60 rolls, you roll 16 or higher all but at most five times. That is a binomial probability. The probability of survival is given by this formula:
$$sum_k=0^5 dbinom60kleft(dfrac14right)^60-kleft(dfrac34right)^k approx 1times 10^-25%$$
Now, the probability that you can survive a given number of trials is:
You have 100% probability of surviving up to 5 trials.
Probability that you survive up to $n$ trials (for $nge 6$):
$$sum_k=0^5 dbinomnkleft(dfrac14right)^n-kleft(dfrac34right)^k$$
Probability you lose on the $n$-th trial:
$$dbinomn-15left(dfrac14right)^n-6left(dfrac34right)^6$$
Expected number of trials before finally succumbing to luck:
$$sum_nge 6ndbinomn-15left(dfrac14right)^n-6left(dfrac34right)^6 = 8$$
You can expect to fail by the eighth trial (on average).
answered Mar 29 at 16:49
InterstellarProbeInterstellarProbe
3,154728
$endgroup$
I will answer the easiest first. You want the probability that in 60 rolls, you roll 16 or higher all but at most five times. That is a binomial probability. The probability of survival is given by this formula:
$$sum_k=0^5 dbinom60kleft(dfrac14right)^60-kleft(dfrac34right)^k approx 1times 10^-25%$$
Now, the probability that you can survive a given number of trials is:
You have 100% probability of surviving up to 5 trials.
Probability that you survive up to $n$ trials (for $nge 6$):
$$sum_k=0^5 dbinomnkleft(dfrac14right)^n-kleft(dfrac34right)^k$$
Probability you lose on the $n$-th trial:
$$dbinomn-15left(dfrac14right)^n-6left(dfrac34right)^6$$
Expected number of trials before finally succumbing to luck:
$$sum_nge 6ndbinomn-15left(dfrac14right)^n-6left(dfrac34right)^6 = 8$$
You can expect to fail by the eighth trial (on average).
answered Mar 29 at 16:49
InterstellarProbeInterstellarProbe
3,154728
answered Mar 29 at 16:49
InterstellarProbeInterstellarProbe
3,154728
answered Mar 29 at 16:49
InterstellarProbeInterstellarProbe
3,154728
answered Mar 29 at 16:49
InterstellarProbeInterstellarProbe
3,154728
3,154728
add a comment |
add a comment |
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$begingroup$
Good grief, who are they trying to defeat?! :-o
$endgroup$
– Brian Tung
Mar 29 at 17:25