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Picking a specific domino last
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Deck Building Probability Stats for Fun/Beginners (HEX tcg)Probability of drawing 6 specific letters from a set of 144 with 21 drawsProbability of two cot deaths in one familyConditional probability of picking particular second letters in Scrabble, given the first letter pickedPicking balls blindfolded without replacementStatistics: Probability of two cases under specific conditions.In a shuffled deck, are the probabilities of finding one pair of cards and a second pair of cards independent events?Probability of picking the right answer in a yes no question.probability of having specific numbers in an array selected from another arrayFind a value for a specific conditional probability
$begingroup$
My family recently had a domino game (double 12 set = 91 unique) in which each of 5 players started with 10 in their hand. We needed to start the game with the double "1", but no one had it. Everyone took turns and continued to draw for the double "1" until it turned out to be the last one drawn from the remaining 41. We all joked that the odds of that happening are probably one in a million. What were the true odds?
conditional-probability
asked Mar 26 at 18:24
KCRJKCRJ
61
$endgroup$
add a comment |
$begingroup$
My family recently had a domino game (double 12 set = 91 unique) in which each of 5 players started with 10 in their hand. We needed to start the game with the double "1", but no one had it. Everyone took turns and continued to draw for the double "1" until it turned out to be the last one drawn from the remaining 41. We all joked that the odds of that happening are probably one in a million. What were the true odds?
conditional-probability
asked Mar 26 at 18:24
KCRJKCRJ
61
$endgroup$
add a comment |
$begingroup$
My family recently had a domino game (double 12 set = 91 unique) in which each of 5 players started with 10 in their hand. We needed to start the game with the double "1", but no one had it. Everyone took turns and continued to draw for the double "1" until it turned out to be the last one drawn from the remaining 41. We all joked that the odds of that happening are probably one in a million. What were the true odds?
conditional-probability
asked Mar 26 at 18:24
KCRJKCRJ
61
$endgroup$
My family recently had a domino game (double 12 set = 91 unique) in which each of 5 players started with 10 in their hand. We needed to start the game with the double "1", but no one had it. Everyone took turns and continued to draw for the double "1" until it turned out to be the last one drawn from the remaining 41. We all joked that the odds of that happening are probably one in a million. What were the true odds?
conditional-probability
conditional-probability
asked Mar 26 at 18:24
KCRJKCRJ
61
asked Mar 26 at 18:24
KCRJKCRJ
61
asked Mar 26 at 18:24
KCRJKCRJ
61
asked Mar 26 at 18:24
KCRJKCRJ
61
asked Mar 26 at 18:24
KCRJKCRJ
61
61
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
From the start any domino has the same chance as any other to be picked last, so $1$ in $91$. Once you know the double-$1$ isn't among the first $50$ it has a $1$ in $41$ chance to be picked last from the remaining 41. So not one in a million, once in every $41$ such games.
The chance that no one has the double-$1$ after taking the first $50$ is just $41/91$, so that will happen in just under half the games.
Edit in response to a comment:
Each time you pick a domino and don't get the double-$1$ the probability that you get it the next time increases. When it's the only one left the probability is, of course, $1$. But that does not change the fact that at the moment you have $50$ drawn and not that one, the probability that it will be last is just $1/41$. There is nothing to multiply.
You can do an experiment. Pick six of your favorite dominoes - say 12, 11, 21, 22, 31 and 32 Now draw two hands of one each. Each time you don't see the 11 drawn, turn the other four over one at a time. The 11 will be last in about $1/4$ of the games - do this lots of times so the randomness has a chance to work. (I know those $6$ don't make for a good domino game.)
answered Mar 26 at 18:28
Ethan BolkerEthan Bolker
46.2k555121
$endgroup$
$begingroup$
Doesn't probability change with every picked domino? I would think we need to multiply probabilities that a particular piece is not picked during first, second, third try and so forth.
$endgroup$
– Vasya
Mar 26 at 18:39
$begingroup$
The double "1" was the last domino after the 40 others were picked. The "odds" of it being picked at each draw was 1 in 41, but the probability of it being picked last would be much smaller.
$endgroup$
– KCRJ
Mar 26 at 21:19
$begingroup$
@Vasya No. See my edit.
$endgroup$
– Ethan Bolker
Mar 26 at 21:29
$begingroup$
@EthanBolker: I think I've got it. The number of ways to pick remaining $41$ dominoes is $41!$ and if we put one-one aside, it's $40!$. So the probability that one-one will remain is $frac40!41!=frac141$
$endgroup$
– Vasya
Mar 27 at 0:57
$begingroup$
@Vasya Yes, that is one way to see it. You can see it without the formulas since there's no way any particular one of the $41$ is more likely to be last, so the probability for each one must by $1/41$. (If the answer satisfies you, you can accept it (the check mark) and upvote it.)
$endgroup$
– Ethan Bolker
Mar 27 at 1:23
add a comment |
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1 Answer
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$begingroup$
From the start any domino has the same chance as any other to be picked last, so $1$ in $91$. Once you know the double-$1$ isn't among the first $50$ it has a $1$ in $41$ chance to be picked last from the remaining 41. So not one in a million, once in every $41$ such games.
The chance that no one has the double-$1$ after taking the first $50$ is just $41/91$, so that will happen in just under half the games.
Edit in response to a comment:
Each time you pick a domino and don't get the double-$1$ the probability that you get it the next time increases. When it's the only one left the probability is, of course, $1$. But that does not change the fact that at the moment you have $50$ drawn and not that one, the probability that it will be last is just $1/41$. There is nothing to multiply.
You can do an experiment. Pick six of your favorite dominoes - say 12, 11, 21, 22, 31 and 32 Now draw two hands of one each. Each time you don't see the 11 drawn, turn the other four over one at a time. The 11 will be last in about $1/4$ of the games - do this lots of times so the randomness has a chance to work. (I know those $6$ don't make for a good domino game.)
answered Mar 26 at 18:28
Ethan BolkerEthan Bolker
46.2k555121
$endgroup$
$begingroup$
Doesn't probability change with every picked domino? I would think we need to multiply probabilities that a particular piece is not picked during first, second, third try and so forth.
$endgroup$
– Vasya
Mar 26 at 18:39
$begingroup$
The double "1" was the last domino after the 40 others were picked. The "odds" of it being picked at each draw was 1 in 41, but the probability of it being picked last would be much smaller.
$endgroup$
– KCRJ
Mar 26 at 21:19
$begingroup$
@Vasya No. See my edit.
$endgroup$
– Ethan Bolker
Mar 26 at 21:29
$begingroup$
@EthanBolker: I think I've got it. The number of ways to pick remaining $41$ dominoes is $41!$ and if we put one-one aside, it's $40!$. So the probability that one-one will remain is $frac40!41!=frac141$
$endgroup$
– Vasya
Mar 27 at 0:57
$begingroup$
@Vasya Yes, that is one way to see it. You can see it without the formulas since there's no way any particular one of the $41$ is more likely to be last, so the probability for each one must by $1/41$. (If the answer satisfies you, you can accept it (the check mark) and upvote it.)
$endgroup$
– Ethan Bolker
Mar 27 at 1:23
add a comment |
$begingroup$
From the start any domino has the same chance as any other to be picked last, so $1$ in $91$. Once you know the double-$1$ isn't among the first $50$ it has a $1$ in $41$ chance to be picked last from the remaining 41. So not one in a million, once in every $41$ such games.
The chance that no one has the double-$1$ after taking the first $50$ is just $41/91$, so that will happen in just under half the games.
Edit in response to a comment:
Each time you pick a domino and don't get the double-$1$ the probability that you get it the next time increases. When it's the only one left the probability is, of course, $1$. But that does not change the fact that at the moment you have $50$ drawn and not that one, the probability that it will be last is just $1/41$. There is nothing to multiply.
You can do an experiment. Pick six of your favorite dominoes - say 12, 11, 21, 22, 31 and 32 Now draw two hands of one each. Each time you don't see the 11 drawn, turn the other four over one at a time. The 11 will be last in about $1/4$ of the games - do this lots of times so the randomness has a chance to work. (I know those $6$ don't make for a good domino game.)
answered Mar 26 at 18:28
Ethan BolkerEthan Bolker
46.2k555121
$endgroup$
$begingroup$
Doesn't probability change with every picked domino? I would think we need to multiply probabilities that a particular piece is not picked during first, second, third try and so forth.
$endgroup$
– Vasya
Mar 26 at 18:39
$begingroup$
The double "1" was the last domino after the 40 others were picked. The "odds" of it being picked at each draw was 1 in 41, but the probability of it being picked last would be much smaller.
$endgroup$
– KCRJ
Mar 26 at 21:19
$begingroup$
@Vasya No. See my edit.
$endgroup$
– Ethan Bolker
Mar 26 at 21:29
$begingroup$
@EthanBolker: I think I've got it. The number of ways to pick remaining $41$ dominoes is $41!$ and if we put one-one aside, it's $40!$. So the probability that one-one will remain is $frac40!41!=frac141$
$endgroup$
– Vasya
Mar 27 at 0:57
$begingroup$
@Vasya Yes, that is one way to see it. You can see it without the formulas since there's no way any particular one of the $41$ is more likely to be last, so the probability for each one must by $1/41$. (If the answer satisfies you, you can accept it (the check mark) and upvote it.)
$endgroup$
– Ethan Bolker
Mar 27 at 1:23
add a comment |
$begingroup$
From the start any domino has the same chance as any other to be picked last, so $1$ in $91$. Once you know the double-$1$ isn't among the first $50$ it has a $1$ in $41$ chance to be picked last from the remaining 41. So not one in a million, once in every $41$ such games.
The chance that no one has the double-$1$ after taking the first $50$ is just $41/91$, so that will happen in just under half the games.
Edit in response to a comment:
Each time you pick a domino and don't get the double-$1$ the probability that you get it the next time increases. When it's the only one left the probability is, of course, $1$. But that does not change the fact that at the moment you have $50$ drawn and not that one, the probability that it will be last is just $1/41$. There is nothing to multiply.
You can do an experiment. Pick six of your favorite dominoes - say 12, 11, 21, 22, 31 and 32 Now draw two hands of one each. Each time you don't see the 11 drawn, turn the other four over one at a time. The 11 will be last in about $1/4$ of the games - do this lots of times so the randomness has a chance to work. (I know those $6$ don't make for a good domino game.)
answered Mar 26 at 18:28
Ethan BolkerEthan Bolker
46.2k555121
$endgroup$
From the start any domino has the same chance as any other to be picked last, so $1$ in $91$. Once you know the double-$1$ isn't among the first $50$ it has a $1$ in $41$ chance to be picked last from the remaining 41. So not one in a million, once in every $41$ such games.
The chance that no one has the double-$1$ after taking the first $50$ is just $41/91$, so that will happen in just under half the games.
Edit in response to a comment:
Each time you pick a domino and don't get the double-$1$ the probability that you get it the next time increases. When it's the only one left the probability is, of course, $1$. But that does not change the fact that at the moment you have $50$ drawn and not that one, the probability that it will be last is just $1/41$. There is nothing to multiply.
You can do an experiment. Pick six of your favorite dominoes - say 12, 11, 21, 22, 31 and 32 Now draw two hands of one each. Each time you don't see the 11 drawn, turn the other four over one at a time. The 11 will be last in about $1/4$ of the games - do this lots of times so the randomness has a chance to work. (I know those $6$ don't make for a good domino game.)
answered Mar 26 at 18:28
Ethan BolkerEthan Bolker
46.2k555121
edited Mar 26 at 21:28
answered Mar 26 at 18:28
Ethan BolkerEthan Bolker
46.2k555121
answered Mar 26 at 18:28
Ethan BolkerEthan Bolker
46.2k555121
answered Mar 26 at 18:28
Ethan BolkerEthan Bolker
46.2k555121
46.2k555121
$begingroup$
Doesn't probability change with every picked domino? I would think we need to multiply probabilities that a particular piece is not picked during first, second, third try and so forth.
$endgroup$
– Vasya
Mar 26 at 18:39
$begingroup$
The double "1" was the last domino after the 40 others were picked. The "odds" of it being picked at each draw was 1 in 41, but the probability of it being picked last would be much smaller.
$endgroup$
– KCRJ
Mar 26 at 21:19
$begingroup$
@Vasya No. See my edit.
$endgroup$
– Ethan Bolker
Mar 26 at 21:29
$begingroup$
@EthanBolker: I think I've got it. The number of ways to pick remaining $41$ dominoes is $41!$ and if we put one-one aside, it's $40!$. So the probability that one-one will remain is $frac40!41!=frac141$
$endgroup$
– Vasya
Mar 27 at 0:57
$begingroup$
@Vasya Yes, that is one way to see it. You can see it without the formulas since there's no way any particular one of the $41$ is more likely to be last, so the probability for each one must by $1/41$. (If the answer satisfies you, you can accept it (the check mark) and upvote it.)
$endgroup$
– Ethan Bolker
Mar 27 at 1:23
add a comment |
$begingroup$
Doesn't probability change with every picked domino? I would think we need to multiply probabilities that a particular piece is not picked during first, second, third try and so forth.
$endgroup$
– Vasya
Mar 26 at 18:39
$begingroup$
The double "1" was the last domino after the 40 others were picked. The "odds" of it being picked at each draw was 1 in 41, but the probability of it being picked last would be much smaller.
$endgroup$
– KCRJ
Mar 26 at 21:19
$begingroup$
@Vasya No. See my edit.
$endgroup$
– Ethan Bolker
Mar 26 at 21:29
$begingroup$
@EthanBolker: I think I've got it. The number of ways to pick remaining $41$ dominoes is $41!$ and if we put one-one aside, it's $40!$. So the probability that one-one will remain is $frac40!41!=frac141$
$endgroup$
– Vasya
Mar 27 at 0:57
$begingroup$
@Vasya Yes, that is one way to see it. You can see it without the formulas since there's no way any particular one of the $41$ is more likely to be last, so the probability for each one must by $1/41$. (If the answer satisfies you, you can accept it (the check mark) and upvote it.)
$endgroup$
– Ethan Bolker
Mar 27 at 1:23
$begingroup$
Doesn't probability change with every picked domino? I would think we need to multiply probabilities that a particular piece is not picked during first, second, third try and so forth.
$endgroup$
– Vasya
Mar 26 at 18:39
$begingroup$
Doesn't probability change with every picked domino? I would think we need to multiply probabilities that a particular piece is not picked during first, second, third try and so forth.
$endgroup$
– Vasya
Mar 26 at 18:39
$begingroup$
The double "1" was the last domino after the 40 others were picked. The "odds" of it being picked at each draw was 1 in 41, but the probability of it being picked last would be much smaller.
$endgroup$
– KCRJ
Mar 26 at 21:19
$begingroup$
The double "1" was the last domino after the 40 others were picked. The "odds" of it being picked at each draw was 1 in 41, but the probability of it being picked last would be much smaller.
$endgroup$
– KCRJ
Mar 26 at 21:19
$begingroup$
@Vasya No. See my edit.
$endgroup$
– Ethan Bolker
Mar 26 at 21:29
$begingroup$
@Vasya No. See my edit.
$endgroup$
– Ethan Bolker
Mar 26 at 21:29
$begingroup$
@EthanBolker: I think I've got it. The number of ways to pick remaining $41$ dominoes is $41!$ and if we put one-one aside, it's $40!$. So the probability that one-one will remain is $frac40!41!=frac141$
$endgroup$
– Vasya
Mar 27 at 0:57
$begingroup$
@EthanBolker: I think I've got it. The number of ways to pick remaining $41$ dominoes is $41!$ and if we put one-one aside, it's $40!$. So the probability that one-one will remain is $frac40!41!=frac141$
$endgroup$
– Vasya
Mar 27 at 0:57
$begingroup$
@Vasya Yes, that is one way to see it. You can see it without the formulas since there's no way any particular one of the $41$ is more likely to be last, so the probability for each one must by $1/41$. (If the answer satisfies you, you can accept it (the check mark) and upvote it.)
$endgroup$
– Ethan Bolker
Mar 27 at 1:23
$begingroup$
@Vasya Yes, that is one way to see it. You can see it without the formulas since there's no way any particular one of the $41$ is more likely to be last, so the probability for each one must by $1/41$. (If the answer satisfies you, you can accept it (the check mark) and upvote it.)
$endgroup$
– Ethan Bolker
Mar 27 at 1:23
add a comment |
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