Mixing
Combinatorics deck/dice
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A player has 52 cards and dice. Every time he picks a card (without seeing it) throws the dice. How many cards at least does he have to draw to make sure that discovering them there are at least 4 cards of the same suit and that the same number appears at least 3 times in the sequence of the dice throw?
I'm struggling with this problem. Can anyone help me?
combinatorics dice card-games
edited Feb 10 at 7:59
jvdhooft
5,67561641
asked Jan 24 at 9:28
JackJack
345
$endgroup$
add a comment |
$begingroup$
A player has 52 cards and dice. Every time he picks a card (without seeing it) throws the dice. How many cards at least does he have to draw to make sure that discovering them there are at least 4 cards of the same suit and that the same number appears at least 3 times in the sequence of the dice throw?
I'm struggling with this problem. Can anyone help me?
combinatorics dice card-games
edited Feb 10 at 7:59
jvdhooft
5,67561641
asked Jan 24 at 9:28
JackJack
345
$endgroup$
1
$begingroup$
If the player only had the die, or only the deck of cards, could you solve the problem?
$endgroup$
– Arthur
Jan 24 at 9:33
$begingroup$
@Arthur Yes, I suppose
$endgroup$
– Jack
Jan 24 at 9:34
$begingroup$
And once you have the solution to both of those, can you see how to use that to solve this combined problem?
$endgroup$
– Arthur
Jan 24 at 9:35
add a comment |
$begingroup$
A player has 52 cards and dice. Every time he picks a card (without seeing it) throws the dice. How many cards at least does he have to draw to make sure that discovering them there are at least 4 cards of the same suit and that the same number appears at least 3 times in the sequence of the dice throw?
I'm struggling with this problem. Can anyone help me?
combinatorics dice card-games
edited Feb 10 at 7:59
jvdhooft
5,67561641
asked Jan 24 at 9:28
JackJack
345
$endgroup$
A player has 52 cards and dice. Every time he picks a card (without seeing it) throws the dice. How many cards at least does he have to draw to make sure that discovering them there are at least 4 cards of the same suit and that the same number appears at least 3 times in the sequence of the dice throw?
I'm struggling with this problem. Can anyone help me?
combinatorics dice card-games
combinatorics dice card-games
edited Feb 10 at 7:59
jvdhooft
5,67561641
asked Jan 24 at 9:28
JackJack
345
edited Feb 10 at 7:59
jvdhooft
5,67561641
asked Jan 24 at 9:28
JackJack
345
edited Feb 10 at 7:59
jvdhooft
5,67561641
edited Feb 10 at 7:59
jvdhooft
5,67561641
edited Feb 10 at 7:59
jvdhooft
5,67561641
5,67561641
asked Jan 24 at 9:28
JackJack
345
asked Jan 24 at 9:28
JackJack
345
asked Jan 24 at 9:28
JackJack
345
345
1
$begingroup$
If the player only had the die, or only the deck of cards, could you solve the problem?
$endgroup$
– Arthur
Jan 24 at 9:33
$begingroup$
@Arthur Yes, I suppose
$endgroup$
– Jack
Jan 24 at 9:34
$begingroup$
And once you have the solution to both of those, can you see how to use that to solve this combined problem?
$endgroup$
– Arthur
Jan 24 at 9:35
add a comment |
1
$begingroup$
If the player only had the die, or only the deck of cards, could you solve the problem?
$endgroup$
– Arthur
Jan 24 at 9:33
$begingroup$
@Arthur Yes, I suppose
$endgroup$
– Jack
Jan 24 at 9:34
$begingroup$
And once you have the solution to both of those, can you see how to use that to solve this combined problem?
$endgroup$
– Arthur
Jan 24 at 9:35
1
1
$begingroup$
If the player only had the die, or only the deck of cards, could you solve the problem?
$endgroup$
– Arthur
Jan 24 at 9:33
$begingroup$
If the player only had the die, or only the deck of cards, could you solve the problem?
$endgroup$
– Arthur
Jan 24 at 9:33
$begingroup$
@Arthur Yes, I suppose
$endgroup$
– Jack
Jan 24 at 9:34
$begingroup$
@Arthur Yes, I suppose
$endgroup$
– Jack
Jan 24 at 9:34
$begingroup$
And once you have the solution to both of those, can you see how to use that to solve this combined problem?
$endgroup$
– Arthur
Jan 24 at 9:35
$begingroup$
And once you have the solution to both of those, can you see how to use that to solve this combined problem?
$endgroup$
– Arthur
Jan 24 at 9:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The player wants to be sure to get 3 times the same dice number and 4 times the same suit. Then, we can look for the longest series without 3 times any number nor 4 times the same suit. Next die and next card will add will add a repeated 3rd time number or a repeated 4t time suit.
With a 12 dice series we could get the numbers 1 to 6 tow times each - worst case. We can be sure that 13th die will show one of those 6 numbers that have already appeared.
Similarly, with a 12 cards series we could get at worst three card of each suit. The 13th card will be a repetition of one of those suits.
Therefore, a 13 cards and 13 dice series is enough.
answered Jan 24 at 9:36
PerePere
46637
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The player wants to be sure to get 3 times the same dice number and 4 times the same suit. Then, we can look for the longest series without 3 times any number nor 4 times the same suit. Next die and next card will add will add a repeated 3rd time number or a repeated 4t time suit.
With a 12 dice series we could get the numbers 1 to 6 tow times each - worst case. We can be sure that 13th die will show one of those 6 numbers that have already appeared.
Similarly, with a 12 cards series we could get at worst three card of each suit. The 13th card will be a repetition of one of those suits.
Therefore, a 13 cards and 13 dice series is enough.
answered Jan 24 at 9:36
PerePere
46637
$endgroup$
add a comment |
$begingroup$
The player wants to be sure to get 3 times the same dice number and 4 times the same suit. Then, we can look for the longest series without 3 times any number nor 4 times the same suit. Next die and next card will add will add a repeated 3rd time number or a repeated 4t time suit.
With a 12 dice series we could get the numbers 1 to 6 tow times each - worst case. We can be sure that 13th die will show one of those 6 numbers that have already appeared.
Similarly, with a 12 cards series we could get at worst three card of each suit. The 13th card will be a repetition of one of those suits.
Therefore, a 13 cards and 13 dice series is enough.
answered Jan 24 at 9:36
PerePere
46637
$endgroup$
add a comment |
$begingroup$
The player wants to be sure to get 3 times the same dice number and 4 times the same suit. Then, we can look for the longest series without 3 times any number nor 4 times the same suit. Next die and next card will add will add a repeated 3rd time number or a repeated 4t time suit.
With a 12 dice series we could get the numbers 1 to 6 tow times each - worst case. We can be sure that 13th die will show one of those 6 numbers that have already appeared.
Similarly, with a 12 cards series we could get at worst three card of each suit. The 13th card will be a repetition of one of those suits.
Therefore, a 13 cards and 13 dice series is enough.
answered Jan 24 at 9:36
PerePere
46637
$endgroup$
The player wants to be sure to get 3 times the same dice number and 4 times the same suit. Then, we can look for the longest series without 3 times any number nor 4 times the same suit. Next die and next card will add will add a repeated 3rd time number or a repeated 4t time suit.
With a 12 dice series we could get the numbers 1 to 6 tow times each - worst case. We can be sure that 13th die will show one of those 6 numbers that have already appeared.
Similarly, with a 12 cards series we could get at worst three card of each suit. The 13th card will be a repetition of one of those suits.
Therefore, a 13 cards and 13 dice series is enough.
answered Jan 24 at 9:36
PerePere
46637
answered Jan 24 at 9:36
PerePere
46637
answered Jan 24 at 9:36
PerePere
46637
answered Jan 24 at 9:36
PerePere
46637
46637
add a comment |
add a comment |
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$begingroup$
If the player only had the die, or only the deck of cards, could you solve the problem?
$endgroup$
– Arthur
Jan 24 at 9:33
$begingroup$
@Arthur Yes, I suppose
$endgroup$
– Jack
Jan 24 at 9:34
$begingroup$
And once you have the solution to both of those, can you see how to use that to solve this combined problem?
$endgroup$
– Arthur
Jan 24 at 9:35