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expected number of rolls to see all 6 dice faces GIVEN we have already seen 4 distinct faces
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Question: given that we have already seen 4 distinct faces of the dice so far, what is the expected number of rolls of the dice required until we see all 6? (The dice is fair.)
EDIT: The question is meant to indicate that we have seen EXACTLY 4 distinct faces so far.
I tried to model my solution in terms of the coupon collector problem/approach. Let $T$ be the number of rolls required until we see all $6$ faces. Let $F$ be the number of faces seen so far. Let $T_i$ be the number of rolls required to get the next new face given that we have already seen $4 + (i - 1) = 3 + i$ faces so far. Then $T_i - Geom((6 - (3 + i))/6) = Geom((3 - i) / 6)$ for $1 leq i leq 2$. Then $E(T | F = 4) = E(T_1) + E(T_2)$ via linearity, so it would be (because the expected value of a variable that takes distribution $Geom(p)$ is $1 / p$), $6/2 + 6/1 = 9$ rolls.
Is this reasoning/answer correct? I don't have too much intution for this kind of problem. If not please let me know where I went wrong!
probability statistics probability-distributions
asked Jan 24 at 21:17
0k330k33
15310
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add a comment |
$begingroup$
Question: given that we have already seen 4 distinct faces of the dice so far, what is the expected number of rolls of the dice required until we see all 6? (The dice is fair.)
EDIT: The question is meant to indicate that we have seen EXACTLY 4 distinct faces so far.
I tried to model my solution in terms of the coupon collector problem/approach. Let $T$ be the number of rolls required until we see all $6$ faces. Let $F$ be the number of faces seen so far. Let $T_i$ be the number of rolls required to get the next new face given that we have already seen $4 + (i - 1) = 3 + i$ faces so far. Then $T_i - Geom((6 - (3 + i))/6) = Geom((3 - i) / 6)$ for $1 leq i leq 2$. Then $E(T | F = 4) = E(T_1) + E(T_2)$ via linearity, so it would be (because the expected value of a variable that takes distribution $Geom(p)$ is $1 / p$), $6/2 + 6/1 = 9$ rolls.
Is this reasoning/answer correct? I don't have too much intution for this kind of problem. If not please let me know where I went wrong!
probability statistics probability-distributions
asked Jan 24 at 21:17
0k330k33
15310
$endgroup$
$begingroup$
@DavidK yes, given that we have seen exactly 4 distinct faces. Thank you for clarifying!
$endgroup$
– 0k33
Jan 24 at 21:20
add a comment |
$begingroup$
Question: given that we have already seen 4 distinct faces of the dice so far, what is the expected number of rolls of the dice required until we see all 6? (The dice is fair.)
EDIT: The question is meant to indicate that we have seen EXACTLY 4 distinct faces so far.
I tried to model my solution in terms of the coupon collector problem/approach. Let $T$ be the number of rolls required until we see all $6$ faces. Let $F$ be the number of faces seen so far. Let $T_i$ be the number of rolls required to get the next new face given that we have already seen $4 + (i - 1) = 3 + i$ faces so far. Then $T_i - Geom((6 - (3 + i))/6) = Geom((3 - i) / 6)$ for $1 leq i leq 2$. Then $E(T | F = 4) = E(T_1) + E(T_2)$ via linearity, so it would be (because the expected value of a variable that takes distribution $Geom(p)$ is $1 / p$), $6/2 + 6/1 = 9$ rolls.
Is this reasoning/answer correct? I don't have too much intution for this kind of problem. If not please let me know where I went wrong!
probability statistics probability-distributions
asked Jan 24 at 21:17
0k330k33
15310
$endgroup$
Question: given that we have already seen 4 distinct faces of the dice so far, what is the expected number of rolls of the dice required until we see all 6? (The dice is fair.)
EDIT: The question is meant to indicate that we have seen EXACTLY 4 distinct faces so far.
I tried to model my solution in terms of the coupon collector problem/approach. Let $T$ be the number of rolls required until we see all $6$ faces. Let $F$ be the number of faces seen so far. Let $T_i$ be the number of rolls required to get the next new face given that we have already seen $4 + (i - 1) = 3 + i$ faces so far. Then $T_i - Geom((6 - (3 + i))/6) = Geom((3 - i) / 6)$ for $1 leq i leq 2$. Then $E(T | F = 4) = E(T_1) + E(T_2)$ via linearity, so it would be (because the expected value of a variable that takes distribution $Geom(p)$ is $1 / p$), $6/2 + 6/1 = 9$ rolls.
Is this reasoning/answer correct? I don't have too much intution for this kind of problem. If not please let me know where I went wrong!
probability statistics probability-distributions
probability statistics probability-distributions
asked Jan 24 at 21:17
0k330k33
15310
asked Jan 24 at 21:17
0k330k33
15310
edited Jan 24 at 21:47
0k33
asked Jan 24 at 21:17
0k330k33
15310
asked Jan 24 at 21:17
0k330k33
15310
asked Jan 24 at 21:17
0k330k33
15310
15310
$begingroup$
@DavidK yes, given that we have seen exactly 4 distinct faces. Thank you for clarifying!
$endgroup$
– 0k33
Jan 24 at 21:20
add a comment |
$begingroup$
@DavidK yes, given that we have seen exactly 4 distinct faces. Thank you for clarifying!
$endgroup$
– 0k33
Jan 24 at 21:20
$begingroup$
@DavidK yes, given that we have seen exactly 4 distinct faces. Thank you for clarifying!
$endgroup$
– 0k33
Jan 24 at 21:20
$begingroup$
@DavidK yes, given that we have seen exactly 4 distinct faces. Thank you for clarifying!
$endgroup$
– 0k33
Jan 24 at 21:20
add a comment |
1 Answer
1
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If you already observed 4 faces, since the probability to observe one of the two faces remaining is $p = 2/6 = 1/3$, the expected number dice throw before we observe one of the two faces remaining is $3$ and with the same reasoning. When we will have observed the fifth face, the probability to observe the last face at the next dice throw is $p = 1/6$ so the expected number of dice throw before we observe the sixth given we observed the five first faces is $6$. In conclusion the number of dice throw before we observe the two last faces is $6 + 3 = 9$. Was writing it on my way home, so if it doesn'make sense please tell me and I will rewrite it in a clearer manner.
answered Jan 24 at 21:33
Robin NicoleRobin Nicole
20112
$endgroup$
$begingroup$
yes, this makes sense! Above I somehow managed to write 6/2 + 6/1 = 4, when it is in fact 9. Thanks!
$endgroup$
– 0k33
Jan 24 at 21:47
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If you already observed 4 faces, since the probability to observe one of the two faces remaining is $p = 2/6 = 1/3$, the expected number dice throw before we observe one of the two faces remaining is $3$ and with the same reasoning. When we will have observed the fifth face, the probability to observe the last face at the next dice throw is $p = 1/6$ so the expected number of dice throw before we observe the sixth given we observed the five first faces is $6$. In conclusion the number of dice throw before we observe the two last faces is $6 + 3 = 9$. Was writing it on my way home, so if it doesn'make sense please tell me and I will rewrite it in a clearer manner.
answered Jan 24 at 21:33
Robin NicoleRobin Nicole
20112
$endgroup$
$begingroup$
yes, this makes sense! Above I somehow managed to write 6/2 + 6/1 = 4, when it is in fact 9. Thanks!
$endgroup$
– 0k33
Jan 24 at 21:47
add a comment |
$begingroup$
If you already observed 4 faces, since the probability to observe one of the two faces remaining is $p = 2/6 = 1/3$, the expected number dice throw before we observe one of the two faces remaining is $3$ and with the same reasoning. When we will have observed the fifth face, the probability to observe the last face at the next dice throw is $p = 1/6$ so the expected number of dice throw before we observe the sixth given we observed the five first faces is $6$. In conclusion the number of dice throw before we observe the two last faces is $6 + 3 = 9$. Was writing it on my way home, so if it doesn'make sense please tell me and I will rewrite it in a clearer manner.
answered Jan 24 at 21:33
Robin NicoleRobin Nicole
20112
$endgroup$
$begingroup$
yes, this makes sense! Above I somehow managed to write 6/2 + 6/1 = 4, when it is in fact 9. Thanks!
$endgroup$
– 0k33
Jan 24 at 21:47
add a comment |
$begingroup$
If you already observed 4 faces, since the probability to observe one of the two faces remaining is $p = 2/6 = 1/3$, the expected number dice throw before we observe one of the two faces remaining is $3$ and with the same reasoning. When we will have observed the fifth face, the probability to observe the last face at the next dice throw is $p = 1/6$ so the expected number of dice throw before we observe the sixth given we observed the five first faces is $6$. In conclusion the number of dice throw before we observe the two last faces is $6 + 3 = 9$. Was writing it on my way home, so if it doesn'make sense please tell me and I will rewrite it in a clearer manner.
answered Jan 24 at 21:33
Robin NicoleRobin Nicole
20112
$endgroup$
If you already observed 4 faces, since the probability to observe one of the two faces remaining is $p = 2/6 = 1/3$, the expected number dice throw before we observe one of the two faces remaining is $3$ and with the same reasoning. When we will have observed the fifth face, the probability to observe the last face at the next dice throw is $p = 1/6$ so the expected number of dice throw before we observe the sixth given we observed the five first faces is $6$. In conclusion the number of dice throw before we observe the two last faces is $6 + 3 = 9$. Was writing it on my way home, so if it doesn'make sense please tell me and I will rewrite it in a clearer manner.
answered Jan 24 at 21:33
Robin NicoleRobin Nicole
20112
answered Jan 24 at 21:33
Robin NicoleRobin Nicole
20112
answered Jan 24 at 21:33
Robin NicoleRobin Nicole
20112
answered Jan 24 at 21:33
Robin NicoleRobin Nicole
20112
20112
$begingroup$
yes, this makes sense! Above I somehow managed to write 6/2 + 6/1 = 4, when it is in fact 9. Thanks!
$endgroup$
– 0k33
Jan 24 at 21:47
add a comment |
$begingroup$
yes, this makes sense! Above I somehow managed to write 6/2 + 6/1 = 4, when it is in fact 9. Thanks!
$endgroup$
– 0k33
Jan 24 at 21:47
$begingroup$
yes, this makes sense! Above I somehow managed to write 6/2 + 6/1 = 4, when it is in fact 9. Thanks!
$endgroup$
– 0k33
Jan 24 at 21:47
$begingroup$
yes, this makes sense! Above I somehow managed to write 6/2 + 6/1 = 4, when it is in fact 9. Thanks!
$endgroup$
– 0k33
Jan 24 at 21:47
add a comment |
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$begingroup$
@DavidK yes, given that we have seen exactly 4 distinct faces. Thank you for clarifying!
$endgroup$
– 0k33
Jan 24 at 21:20