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What's the probability to start a game of Vintage Dredge with Bazaar of Baghdad in your opening hand?
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I want to pick up Vintage Dredge in the card game Magic - The Gathering and I want to know the probability of starting with Bazaar of Baghdad in your opening hand. We can assume that this is a 60-card deck with 4 Bazaar of Baghdad and 4 Serum Powder, and that we will mulligan aggressively for the Bazaar.
magic-the-gathering
edited 45 mins ago
murgatroid99♦
42.9k7103180
asked 4 hours ago
t.rathjen
41839
add a comment |Â
up vote
1
down vote
favorite
I want to pick up Vintage Dredge in the card game Magic - The Gathering and I want to know the probability of starting with Bazaar of Baghdad in your opening hand. We can assume that this is a 60-card deck with 4 Bazaar of Baghdad and 4 Serum Powder, and that we will mulligan aggressively for the Bazaar.
magic-the-gathering
edited 45 mins ago
murgatroid99♦
42.9k7103180
asked 4 hours ago
t.rathjen
41839
Possible duplicate of How do you calculate the likelihood of drawing certain cards in your opening hand?
– SocioMatt
50 mins ago
This is not a duplicate. It is broader than the linked question because it allows for Serum Powder and mulligans.
– murgatroid99♦
46 mins ago
@murgatroid99 I disagree. This question could have been answered using that question's answer, with different variations of hypergeometric distribution calculations. If anything, the linked question is more broad because it provides the process for getting to the answer.
– SocioMatt
7 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to pick up Vintage Dredge in the card game Magic - The Gathering and I want to know the probability of starting with Bazaar of Baghdad in your opening hand. We can assume that this is a 60-card deck with 4 Bazaar of Baghdad and 4 Serum Powder, and that we will mulligan aggressively for the Bazaar.
magic-the-gathering
edited 45 mins ago
murgatroid99♦
42.9k7103180
asked 4 hours ago
t.rathjen
41839
I want to pick up Vintage Dredge in the card game Magic - The Gathering and I want to know the probability of starting with Bazaar of Baghdad in your opening hand. We can assume that this is a 60-card deck with 4 Bazaar of Baghdad and 4 Serum Powder, and that we will mulligan aggressively for the Bazaar.
magic-the-gathering
magic-the-gathering
edited 45 mins ago
murgatroid99♦
42.9k7103180
asked 4 hours ago
t.rathjen
41839
edited 45 mins ago
murgatroid99♦
42.9k7103180
asked 4 hours ago
t.rathjen
41839
edited 45 mins ago
murgatroid99♦
42.9k7103180
edited 45 mins ago
murgatroid99♦
42.9k7103180
edited 45 mins ago
murgatroid99♦
42.9k7103180
42.9k7103180
asked 4 hours ago
t.rathjen
41839
asked 4 hours ago
t.rathjen
41839
asked 4 hours ago
t.rathjen
41839
41839
Possible duplicate of How do you calculate the likelihood of drawing certain cards in your opening hand?
– SocioMatt
50 mins ago
This is not a duplicate. It is broader than the linked question because it allows for Serum Powder and mulligans.
– murgatroid99♦
46 mins ago
@murgatroid99 I disagree. This question could have been answered using that question's answer, with different variations of hypergeometric distribution calculations. If anything, the linked question is more broad because it provides the process for getting to the answer.
– SocioMatt
7 mins ago
add a comment |Â
Possible duplicate of How do you calculate the likelihood of drawing certain cards in your opening hand?
– SocioMatt
50 mins ago
This is not a duplicate. It is broader than the linked question because it allows for Serum Powder and mulligans.
– murgatroid99♦
46 mins ago
@murgatroid99 I disagree. This question could have been answered using that question's answer, with different variations of hypergeometric distribution calculations. If anything, the linked question is more broad because it provides the process for getting to the answer.
– SocioMatt
7 mins ago
Possible duplicate of How do you calculate the likelihood of drawing certain cards in your opening hand?
– SocioMatt
50 mins ago
Possible duplicate of How do you calculate the likelihood of drawing certain cards in your opening hand?
– SocioMatt
50 mins ago
This is not a duplicate. It is broader than the linked question because it allows for Serum Powder and mulligans.
– murgatroid99♦
46 mins ago
This is not a duplicate. It is broader than the linked question because it allows for Serum Powder and mulligans.
– murgatroid99♦
46 mins ago
@murgatroid99 I disagree. This question could have been answered using that question's answer, with different variations of hypergeometric distribution calculations. If anything, the linked question is more broad because it provides the process for getting to the answer.
– SocioMatt
7 mins ago
@murgatroid99 I disagree. This question could have been answered using that question's answer, with different variations of hypergeometric distribution calculations. If anything, the linked question is more broad because it provides the process for getting to the answer.
– SocioMatt
7 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
I approached this question numerically and drew one billion (1,000,000,000) sample hands.
My assumptions: The deck contains 60 cards, 4 of which are Bazaar of Baghdad and 4 are Serum Powder. No further assumptions regarding the deck building are needed.
Than I draw the opening seven cards. If it contains at least 1 Bazaar, the hand is consider a hit. If I draw a Serum Powder, the hand is removed from the game and I draw the exact same number again. If I miss on both, Bazaar and Powder, than I take a mulligan (e.q. reshuffling the deck and drawing one card less).
This continues until I reach the point where I reshuffle a one-card-hand, since it was considered a miss and start the game with a zero-card-hand. In this situation I consider the somewhat new mulligan scry rule, which let's you scry one after mulligan. So even if I start with zero cards in hand, there is a chance that a Bazaar is in the top 2 cards of my library and (thanks to the scry) can be found within one draw step. Those probability are also included in my end results.
My findings:
Probability of starting with a 7 card hand: 54.2 %
Probability of starting with a 6 card hand: 22.12 %
Probability of starting with a 5 card hand: 9.91 %
Probability of starting with a 4 card hand: 4.79 %
Probability of starting with a 3 card hand: 2.43 %
Probability of starting with a 2 card hand: 1.23 %
Probability of starting with a 1 card hand: 0.52 %
Probability of starting with a 0 card hand: 4.8 % (this means no Bazaar of Baghdad was found)
Probability of starting with a 0 card hand and Bazar on top (including scry 1): 0.9 %
Probability of complete failure: 3.9 %
Probability to hit: 95.2015011 %
Probability to hit with scry: 96.1029072 %
For the curious people, here is a plot how those probabilities converged:
answered 4 hours ago
t.rathjen
41839
You cannot mulligan to zero.
– Neil Meyer
3 hours ago
2
@NeilMeyer You can, it's just normally better to have any 1 card rather than none. Rule 103.4 defines the mulligan process and includes: This process is then repeated until no player takes a mulligan. (Note that if a player’s hand size reaches zero cards, that player must keep that hand.)
– Samthere
3 hours ago
2
Have you tried repeating this experiment with "Mull to 0 before accepting the scry" replaced by "Mull to 1 before accepting the scry"? Intuitively speaking, if your hand is 1 card and it's not Bazaar then the chances of Bazaar being in the top 2 cards of a 59 card deck should be marginally better than the odds of Bazaar being in the top 2 cards of a 60 card deck.
– Kamil Drakari
2 hours ago
1
@KamilDrakari That's a valid argument. I will do it
– t.rathjen
2 hours ago
@Samthere while it is possible, it is inadvisable. As you are relying on the vancouver scry at that point, you are more likely to get what you want on the 59 cards in deck, than you are with all 60 by taking the one card hand.
– Andrew
53 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I approached this question numerically and drew one billion (1,000,000,000) sample hands.
My assumptions: The deck contains 60 cards, 4 of which are Bazaar of Baghdad and 4 are Serum Powder. No further assumptions regarding the deck building are needed.
Than I draw the opening seven cards. If it contains at least 1 Bazaar, the hand is consider a hit. If I draw a Serum Powder, the hand is removed from the game and I draw the exact same number again. If I miss on both, Bazaar and Powder, than I take a mulligan (e.q. reshuffling the deck and drawing one card less).
This continues until I reach the point where I reshuffle a one-card-hand, since it was considered a miss and start the game with a zero-card-hand. In this situation I consider the somewhat new mulligan scry rule, which let's you scry one after mulligan. So even if I start with zero cards in hand, there is a chance that a Bazaar is in the top 2 cards of my library and (thanks to the scry) can be found within one draw step. Those probability are also included in my end results.
My findings:
Probability of starting with a 7 card hand: 54.2 %
Probability of starting with a 6 card hand: 22.12 %
Probability of starting with a 5 card hand: 9.91 %
Probability of starting with a 4 card hand: 4.79 %
Probability of starting with a 3 card hand: 2.43 %
Probability of starting with a 2 card hand: 1.23 %
Probability of starting with a 1 card hand: 0.52 %
Probability of starting with a 0 card hand: 4.8 % (this means no Bazaar of Baghdad was found)
Probability of starting with a 0 card hand and Bazar on top (including scry 1): 0.9 %
Probability of complete failure: 3.9 %
Probability to hit: 95.2015011 %
Probability to hit with scry: 96.1029072 %
For the curious people, here is a plot how those probabilities converged:
answered 4 hours ago
t.rathjen
41839
You cannot mulligan to zero.
– Neil Meyer
3 hours ago
2
@NeilMeyer You can, it's just normally better to have any 1 card rather than none. Rule 103.4 defines the mulligan process and includes: This process is then repeated until no player takes a mulligan. (Note that if a player’s hand size reaches zero cards, that player must keep that hand.)
– Samthere
3 hours ago
2
Have you tried repeating this experiment with "Mull to 0 before accepting the scry" replaced by "Mull to 1 before accepting the scry"? Intuitively speaking, if your hand is 1 card and it's not Bazaar then the chances of Bazaar being in the top 2 cards of a 59 card deck should be marginally better than the odds of Bazaar being in the top 2 cards of a 60 card deck.
– Kamil Drakari
2 hours ago
1
@KamilDrakari That's a valid argument. I will do it
– t.rathjen
2 hours ago
@Samthere while it is possible, it is inadvisable. As you are relying on the vancouver scry at that point, you are more likely to get what you want on the 59 cards in deck, than you are with all 60 by taking the one card hand.
– Andrew
53 mins ago
add a comment |Â
up vote
3
down vote
I approached this question numerically and drew one billion (1,000,000,000) sample hands.
My assumptions: The deck contains 60 cards, 4 of which are Bazaar of Baghdad and 4 are Serum Powder. No further assumptions regarding the deck building are needed.
Than I draw the opening seven cards. If it contains at least 1 Bazaar, the hand is consider a hit. If I draw a Serum Powder, the hand is removed from the game and I draw the exact same number again. If I miss on both, Bazaar and Powder, than I take a mulligan (e.q. reshuffling the deck and drawing one card less).
This continues until I reach the point where I reshuffle a one-card-hand, since it was considered a miss and start the game with a zero-card-hand. In this situation I consider the somewhat new mulligan scry rule, which let's you scry one after mulligan. So even if I start with zero cards in hand, there is a chance that a Bazaar is in the top 2 cards of my library and (thanks to the scry) can be found within one draw step. Those probability are also included in my end results.
My findings:
Probability of starting with a 7 card hand: 54.2 %
Probability of starting with a 6 card hand: 22.12 %
Probability of starting with a 5 card hand: 9.91 %
Probability of starting with a 4 card hand: 4.79 %
Probability of starting with a 3 card hand: 2.43 %
Probability of starting with a 2 card hand: 1.23 %
Probability of starting with a 1 card hand: 0.52 %
Probability of starting with a 0 card hand: 4.8 % (this means no Bazaar of Baghdad was found)
Probability of starting with a 0 card hand and Bazar on top (including scry 1): 0.9 %
Probability of complete failure: 3.9 %
Probability to hit: 95.2015011 %
Probability to hit with scry: 96.1029072 %
For the curious people, here is a plot how those probabilities converged:
answered 4 hours ago
t.rathjen
41839
You cannot mulligan to zero.
– Neil Meyer
3 hours ago
2
@NeilMeyer You can, it's just normally better to have any 1 card rather than none. Rule 103.4 defines the mulligan process and includes: This process is then repeated until no player takes a mulligan. (Note that if a player’s hand size reaches zero cards, that player must keep that hand.)
– Samthere
3 hours ago
2
Have you tried repeating this experiment with "Mull to 0 before accepting the scry" replaced by "Mull to 1 before accepting the scry"? Intuitively speaking, if your hand is 1 card and it's not Bazaar then the chances of Bazaar being in the top 2 cards of a 59 card deck should be marginally better than the odds of Bazaar being in the top 2 cards of a 60 card deck.
– Kamil Drakari
2 hours ago
1
@KamilDrakari That's a valid argument. I will do it
– t.rathjen
2 hours ago
@Samthere while it is possible, it is inadvisable. As you are relying on the vancouver scry at that point, you are more likely to get what you want on the 59 cards in deck, than you are with all 60 by taking the one card hand.
– Andrew
53 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I approached this question numerically and drew one billion (1,000,000,000) sample hands.
My assumptions: The deck contains 60 cards, 4 of which are Bazaar of Baghdad and 4 are Serum Powder. No further assumptions regarding the deck building are needed.
Than I draw the opening seven cards. If it contains at least 1 Bazaar, the hand is consider a hit. If I draw a Serum Powder, the hand is removed from the game and I draw the exact same number again. If I miss on both, Bazaar and Powder, than I take a mulligan (e.q. reshuffling the deck and drawing one card less).
This continues until I reach the point where I reshuffle a one-card-hand, since it was considered a miss and start the game with a zero-card-hand. In this situation I consider the somewhat new mulligan scry rule, which let's you scry one after mulligan. So even if I start with zero cards in hand, there is a chance that a Bazaar is in the top 2 cards of my library and (thanks to the scry) can be found within one draw step. Those probability are also included in my end results.
My findings:
Probability of starting with a 7 card hand: 54.2 %
Probability of starting with a 6 card hand: 22.12 %
Probability of starting with a 5 card hand: 9.91 %
Probability of starting with a 4 card hand: 4.79 %
Probability of starting with a 3 card hand: 2.43 %
Probability of starting with a 2 card hand: 1.23 %
Probability of starting with a 1 card hand: 0.52 %
Probability of starting with a 0 card hand: 4.8 % (this means no Bazaar of Baghdad was found)
Probability of starting with a 0 card hand and Bazar on top (including scry 1): 0.9 %
Probability of complete failure: 3.9 %
Probability to hit: 95.2015011 %
Probability to hit with scry: 96.1029072 %
For the curious people, here is a plot how those probabilities converged:
answered 4 hours ago
t.rathjen
41839
I approached this question numerically and drew one billion (1,000,000,000) sample hands.
My assumptions: The deck contains 60 cards, 4 of which are Bazaar of Baghdad and 4 are Serum Powder. No further assumptions regarding the deck building are needed.
Than I draw the opening seven cards. If it contains at least 1 Bazaar, the hand is consider a hit. If I draw a Serum Powder, the hand is removed from the game and I draw the exact same number again. If I miss on both, Bazaar and Powder, than I take a mulligan (e.q. reshuffling the deck and drawing one card less).
This continues until I reach the point where I reshuffle a one-card-hand, since it was considered a miss and start the game with a zero-card-hand. In this situation I consider the somewhat new mulligan scry rule, which let's you scry one after mulligan. So even if I start with zero cards in hand, there is a chance that a Bazaar is in the top 2 cards of my library and (thanks to the scry) can be found within one draw step. Those probability are also included in my end results.
My findings:
Probability of starting with a 7 card hand: 54.2 %
Probability of starting with a 6 card hand: 22.12 %
Probability of starting with a 5 card hand: 9.91 %
Probability of starting with a 4 card hand: 4.79 %
Probability of starting with a 3 card hand: 2.43 %
Probability of starting with a 2 card hand: 1.23 %
Probability of starting with a 1 card hand: 0.52 %
Probability of starting with a 0 card hand: 4.8 % (this means no Bazaar of Baghdad was found)
Probability of starting with a 0 card hand and Bazar on top (including scry 1): 0.9 %
Probability of complete failure: 3.9 %
Probability to hit: 95.2015011 %
Probability to hit with scry: 96.1029072 %
For the curious people, here is a plot how those probabilities converged:
answered 4 hours ago
t.rathjen
41839
answered 4 hours ago
t.rathjen
41839
answered 4 hours ago
t.rathjen
41839
answered 4 hours ago
t.rathjen
41839
41839
You cannot mulligan to zero.
– Neil Meyer
3 hours ago
2
@NeilMeyer You can, it's just normally better to have any 1 card rather than none. Rule 103.4 defines the mulligan process and includes: This process is then repeated until no player takes a mulligan. (Note that if a player’s hand size reaches zero cards, that player must keep that hand.)
– Samthere
3 hours ago
2
Have you tried repeating this experiment with "Mull to 0 before accepting the scry" replaced by "Mull to 1 before accepting the scry"? Intuitively speaking, if your hand is 1 card and it's not Bazaar then the chances of Bazaar being in the top 2 cards of a 59 card deck should be marginally better than the odds of Bazaar being in the top 2 cards of a 60 card deck.
– Kamil Drakari
2 hours ago
1
@KamilDrakari That's a valid argument. I will do it
– t.rathjen
2 hours ago
@Samthere while it is possible, it is inadvisable. As you are relying on the vancouver scry at that point, you are more likely to get what you want on the 59 cards in deck, than you are with all 60 by taking the one card hand.
– Andrew
53 mins ago
add a comment |Â
You cannot mulligan to zero.
– Neil Meyer
3 hours ago
2
@NeilMeyer You can, it's just normally better to have any 1 card rather than none. Rule 103.4 defines the mulligan process and includes: This process is then repeated until no player takes a mulligan. (Note that if a player’s hand size reaches zero cards, that player must keep that hand.)
– Samthere
3 hours ago
2
Have you tried repeating this experiment with "Mull to 0 before accepting the scry" replaced by "Mull to 1 before accepting the scry"? Intuitively speaking, if your hand is 1 card and it's not Bazaar then the chances of Bazaar being in the top 2 cards of a 59 card deck should be marginally better than the odds of Bazaar being in the top 2 cards of a 60 card deck.
– Kamil Drakari
2 hours ago
1
@KamilDrakari That's a valid argument. I will do it
– t.rathjen
2 hours ago
@Samthere while it is possible, it is inadvisable. As you are relying on the vancouver scry at that point, you are more likely to get what you want on the 59 cards in deck, than you are with all 60 by taking the one card hand.
– Andrew
53 mins ago
You cannot mulligan to zero.
– Neil Meyer
3 hours ago
You cannot mulligan to zero.
– Neil Meyer
3 hours ago
2
2
@NeilMeyer You can, it's just normally better to have any 1 card rather than none. Rule 103.4 defines the mulligan process and includes: This process is then repeated until no player takes a mulligan. (Note that if a player’s hand size reaches zero cards, that player must keep that hand.)
– Samthere
3 hours ago
@NeilMeyer You can, it's just normally better to have any 1 card rather than none. Rule 103.4 defines the mulligan process and includes: This process is then repeated until no player takes a mulligan. (Note that if a player’s hand size reaches zero cards, that player must keep that hand.)
– Samthere
3 hours ago
2
2
Have you tried repeating this experiment with "Mull to 0 before accepting the scry" replaced by "Mull to 1 before accepting the scry"? Intuitively speaking, if your hand is 1 card and it's not Bazaar then the chances of Bazaar being in the top 2 cards of a 59 card deck should be marginally better than the odds of Bazaar being in the top 2 cards of a 60 card deck.
– Kamil Drakari
2 hours ago
Have you tried repeating this experiment with "Mull to 0 before accepting the scry" replaced by "Mull to 1 before accepting the scry"? Intuitively speaking, if your hand is 1 card and it's not Bazaar then the chances of Bazaar being in the top 2 cards of a 59 card deck should be marginally better than the odds of Bazaar being in the top 2 cards of a 60 card deck.
– Kamil Drakari
2 hours ago
1
1
@KamilDrakari That's a valid argument. I will do it
– t.rathjen
2 hours ago
@KamilDrakari That's a valid argument. I will do it
– t.rathjen
2 hours ago
@Samthere while it is possible, it is inadvisable. As you are relying on the vancouver scry at that point, you are more likely to get what you want on the 59 cards in deck, than you are with all 60 by taking the one card hand.
– Andrew
53 mins ago
@Samthere while it is possible, it is inadvisable. As you are relying on the vancouver scry at that point, you are more likely to get what you want on the 59 cards in deck, than you are with all 60 by taking the one card hand.
– Andrew
53 mins ago
add a comment |Â
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Possible duplicate of How do you calculate the likelihood of drawing certain cards in your opening hand?
– SocioMatt
50 mins ago
This is not a duplicate. It is broader than the linked question because it allows for Serum Powder and mulligans.
– murgatroid99♦
46 mins ago
@murgatroid99 I disagree. This question could have been answered using that question's answer, with different variations of hypergeometric distribution calculations. If anything, the linked question is more broad because it provides the process for getting to the answer.
– SocioMatt
7 mins ago