Mixing
Probability of matching a hand of 13 random cards with cards dealt from a shuffled deck up to position 35?
$begingroup$
You are given a random assortment of 13 cards from one deck. Another deck is shuffled and cards dealt and called out from the top, the aim is to match all 13 of your cards before reaching 35 cards down into the deck. What is the probability of achieving this?
I realise this is closely related to something like the Birthday Paradox, but I can't seem to use to probability calculator on this site to work out the solution: https://betterexplained.com/articles/understanding-the-birthday-paradox/
The reason I'm asking is because this is a regular thing that happens like a bingo in a pub quiz I frequent and nobody ever wins the prize, which seems unlikely in so many months of playing with a pub full of people all with their own individual random assortment of 13 different cards. After the position of 35 is reached the rest of the deck is read out until someone obviously eventually matches up their cards first, but only win what's in the money pot for the night, rather than big money prize. Me and my friends have worked out various ways to fix the game, but it may be the probability is entirely not in anyone's favour, and the number 35 has been chosen very precisely odds-wise.
Hope someone can provide a solution.
probability combinatorics
asked Jan 4 at 13:59
Steven WilliamsSteven Williams
1
$endgroup$
add a comment |
$begingroup$
You are given a random assortment of 13 cards from one deck. Another deck is shuffled and cards dealt and called out from the top, the aim is to match all 13 of your cards before reaching 35 cards down into the deck. What is the probability of achieving this?
I realise this is closely related to something like the Birthday Paradox, but I can't seem to use to probability calculator on this site to work out the solution: https://betterexplained.com/articles/understanding-the-birthday-paradox/
The reason I'm asking is because this is a regular thing that happens like a bingo in a pub quiz I frequent and nobody ever wins the prize, which seems unlikely in so many months of playing with a pub full of people all with their own individual random assortment of 13 different cards. After the position of 35 is reached the rest of the deck is read out until someone obviously eventually matches up their cards first, but only win what's in the money pot for the night, rather than big money prize. Me and my friends have worked out various ways to fix the game, but it may be the probability is entirely not in anyone's favour, and the number 35 has been chosen very precisely odds-wise.
Hope someone can provide a solution.
probability combinatorics
asked Jan 4 at 13:59
Steven WilliamsSteven Williams
1
$endgroup$
$begingroup$
$C(39,22)/C(52,35)$ = (number of good combinations of 35 cards) / (number of combinations)
$endgroup$
– Ned
Jan 4 at 14:26
add a comment |
$begingroup$
You are given a random assortment of 13 cards from one deck. Another deck is shuffled and cards dealt and called out from the top, the aim is to match all 13 of your cards before reaching 35 cards down into the deck. What is the probability of achieving this?
I realise this is closely related to something like the Birthday Paradox, but I can't seem to use to probability calculator on this site to work out the solution: https://betterexplained.com/articles/understanding-the-birthday-paradox/
The reason I'm asking is because this is a regular thing that happens like a bingo in a pub quiz I frequent and nobody ever wins the prize, which seems unlikely in so many months of playing with a pub full of people all with their own individual random assortment of 13 different cards. After the position of 35 is reached the rest of the deck is read out until someone obviously eventually matches up their cards first, but only win what's in the money pot for the night, rather than big money prize. Me and my friends have worked out various ways to fix the game, but it may be the probability is entirely not in anyone's favour, and the number 35 has been chosen very precisely odds-wise.
Hope someone can provide a solution.
probability combinatorics
asked Jan 4 at 13:59
Steven WilliamsSteven Williams
1
$endgroup$
You are given a random assortment of 13 cards from one deck. Another deck is shuffled and cards dealt and called out from the top, the aim is to match all 13 of your cards before reaching 35 cards down into the deck. What is the probability of achieving this?
I realise this is closely related to something like the Birthday Paradox, but I can't seem to use to probability calculator on this site to work out the solution: https://betterexplained.com/articles/understanding-the-birthday-paradox/
The reason I'm asking is because this is a regular thing that happens like a bingo in a pub quiz I frequent and nobody ever wins the prize, which seems unlikely in so many months of playing with a pub full of people all with their own individual random assortment of 13 different cards. After the position of 35 is reached the rest of the deck is read out until someone obviously eventually matches up their cards first, but only win what's in the money pot for the night, rather than big money prize. Me and my friends have worked out various ways to fix the game, but it may be the probability is entirely not in anyone's favour, and the number 35 has been chosen very precisely odds-wise.
Hope someone can provide a solution.
probability combinatorics
probability combinatorics
asked Jan 4 at 13:59
Steven WilliamsSteven Williams
1
asked Jan 4 at 13:59
Steven WilliamsSteven Williams
1
asked Jan 4 at 13:59
Steven WilliamsSteven Williams
1
asked Jan 4 at 13:59
Steven WilliamsSteven Williams
1
asked Jan 4 at 13:59
Steven WilliamsSteven Williams
1
1
$begingroup$
$C(39,22)/C(52,35)$ = (number of good combinations of 35 cards) / (number of combinations)
$endgroup$
– Ned
Jan 4 at 14:26
add a comment |
$begingroup$
$C(39,22)/C(52,35)$ = (number of good combinations of 35 cards) / (number of combinations)
$endgroup$
– Ned
Jan 4 at 14:26
$begingroup$
$C(39,22)/C(52,35)$ = (number of good combinations of 35 cards) / (number of combinations)
$endgroup$
– Ned
Jan 4 at 14:26
$begingroup$
$C(39,22)/C(52,35)$ = (number of good combinations of 35 cards) / (number of combinations)
$endgroup$
– Ned
Jan 4 at 14:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If I understand well then your question can be translated into: "if $13$ cards are labeled and then after shuffling $35$ cards are drawn what is the probability that all labeled cards are among them?"
The answer is: $$frac{binom{35}{13}}{binom{52}{13}}=frac{35!39!}{22!52!}=frac{23}{40}timescdotstimesfrac{39}{52}approx0.002325$$So a very small chance.
answered Jan 4 at 14:29
drhabdrhab
99.1k544130
$endgroup$
$begingroup$
Thanks. So about 0.2%? It seems like it would be a lot more likely than that, but my brain had been coloured by reading about The Birthday Paradox which is only having one match between 23 from 365. Not 13 matches at once, which obviously makes it much less likely. At which number card through the deck would the probability become 50%?
$endgroup$
– Steven Williams
Jan 4 at 16:50
1
$begingroup$
You need to go to the 50th card to exceed 50%. A way to understand why it is so high is to ask the complementary question (imagine the 13 special cards are the spades to make it easy talk about): What is the probability that a selection of 3 cards from the deck contains at least one spade? Well that's close to 59% (not surprising that it's more than 50% since each card has 1/4 chance of being a spade), so the probability that all 13 spades are in the other 49 cards is about 41%.
$endgroup$
– Ned
Jan 4 at 20:27
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If I understand well then your question can be translated into: "if $13$ cards are labeled and then after shuffling $35$ cards are drawn what is the probability that all labeled cards are among them?"
The answer is: $$frac{binom{35}{13}}{binom{52}{13}}=frac{35!39!}{22!52!}=frac{23}{40}timescdotstimesfrac{39}{52}approx0.002325$$So a very small chance.
answered Jan 4 at 14:29
drhabdrhab
99.1k544130
$endgroup$
$begingroup$
Thanks. So about 0.2%? It seems like it would be a lot more likely than that, but my brain had been coloured by reading about The Birthday Paradox which is only having one match between 23 from 365. Not 13 matches at once, which obviously makes it much less likely. At which number card through the deck would the probability become 50%?
$endgroup$
– Steven Williams
Jan 4 at 16:50
1
$begingroup$
You need to go to the 50th card to exceed 50%. A way to understand why it is so high is to ask the complementary question (imagine the 13 special cards are the spades to make it easy talk about): What is the probability that a selection of 3 cards from the deck contains at least one spade? Well that's close to 59% (not surprising that it's more than 50% since each card has 1/4 chance of being a spade), so the probability that all 13 spades are in the other 49 cards is about 41%.
$endgroup$
– Ned
Jan 4 at 20:27
add a comment |
$begingroup$
If I understand well then your question can be translated into: "if $13$ cards are labeled and then after shuffling $35$ cards are drawn what is the probability that all labeled cards are among them?"
The answer is: $$frac{binom{35}{13}}{binom{52}{13}}=frac{35!39!}{22!52!}=frac{23}{40}timescdotstimesfrac{39}{52}approx0.002325$$So a very small chance.
answered Jan 4 at 14:29
drhabdrhab
99.1k544130
$endgroup$
$begingroup$
Thanks. So about 0.2%? It seems like it would be a lot more likely than that, but my brain had been coloured by reading about The Birthday Paradox which is only having one match between 23 from 365. Not 13 matches at once, which obviously makes it much less likely. At which number card through the deck would the probability become 50%?
$endgroup$
– Steven Williams
Jan 4 at 16:50
1
$begingroup$
You need to go to the 50th card to exceed 50%. A way to understand why it is so high is to ask the complementary question (imagine the 13 special cards are the spades to make it easy talk about): What is the probability that a selection of 3 cards from the deck contains at least one spade? Well that's close to 59% (not surprising that it's more than 50% since each card has 1/4 chance of being a spade), so the probability that all 13 spades are in the other 49 cards is about 41%.
$endgroup$
– Ned
Jan 4 at 20:27
add a comment |
$begingroup$
If I understand well then your question can be translated into: "if $13$ cards are labeled and then after shuffling $35$ cards are drawn what is the probability that all labeled cards are among them?"
The answer is: $$frac{binom{35}{13}}{binom{52}{13}}=frac{35!39!}{22!52!}=frac{23}{40}timescdotstimesfrac{39}{52}approx0.002325$$So a very small chance.
answered Jan 4 at 14:29
drhabdrhab
99.1k544130
$endgroup$
If I understand well then your question can be translated into: "if $13$ cards are labeled and then after shuffling $35$ cards are drawn what is the probability that all labeled cards are among them?"
The answer is: $$frac{binom{35}{13}}{binom{52}{13}}=frac{35!39!}{22!52!}=frac{23}{40}timescdotstimesfrac{39}{52}approx0.002325$$So a very small chance.
answered Jan 4 at 14:29
drhabdrhab
99.1k544130
answered Jan 4 at 14:29
drhabdrhab
99.1k544130
answered Jan 4 at 14:29
drhabdrhab
99.1k544130
answered Jan 4 at 14:29
drhabdrhab
99.1k544130
99.1k544130
$begingroup$
Thanks. So about 0.2%? It seems like it would be a lot more likely than that, but my brain had been coloured by reading about The Birthday Paradox which is only having one match between 23 from 365. Not 13 matches at once, which obviously makes it much less likely. At which number card through the deck would the probability become 50%?
$endgroup$
– Steven Williams
Jan 4 at 16:50
1
$begingroup$
You need to go to the 50th card to exceed 50%. A way to understand why it is so high is to ask the complementary question (imagine the 13 special cards are the spades to make it easy talk about): What is the probability that a selection of 3 cards from the deck contains at least one spade? Well that's close to 59% (not surprising that it's more than 50% since each card has 1/4 chance of being a spade), so the probability that all 13 spades are in the other 49 cards is about 41%.
$endgroup$
– Ned
Jan 4 at 20:27
add a comment |
$begingroup$
Thanks. So about 0.2%? It seems like it would be a lot more likely than that, but my brain had been coloured by reading about The Birthday Paradox which is only having one match between 23 from 365. Not 13 matches at once, which obviously makes it much less likely. At which number card through the deck would the probability become 50%?
$endgroup$
– Steven Williams
Jan 4 at 16:50
1
$begingroup$
You need to go to the 50th card to exceed 50%. A way to understand why it is so high is to ask the complementary question (imagine the 13 special cards are the spades to make it easy talk about): What is the probability that a selection of 3 cards from the deck contains at least one spade? Well that's close to 59% (not surprising that it's more than 50% since each card has 1/4 chance of being a spade), so the probability that all 13 spades are in the other 49 cards is about 41%.
$endgroup$
– Ned
Jan 4 at 20:27
$begingroup$
Thanks. So about 0.2%? It seems like it would be a lot more likely than that, but my brain had been coloured by reading about The Birthday Paradox which is only having one match between 23 from 365. Not 13 matches at once, which obviously makes it much less likely. At which number card through the deck would the probability become 50%?
$endgroup$
– Steven Williams
Jan 4 at 16:50
$begingroup$
Thanks. So about 0.2%? It seems like it would be a lot more likely than that, but my brain had been coloured by reading about The Birthday Paradox which is only having one match between 23 from 365. Not 13 matches at once, which obviously makes it much less likely. At which number card through the deck would the probability become 50%?
$endgroup$
– Steven Williams
Jan 4 at 16:50
1
1
$begingroup$
You need to go to the 50th card to exceed 50%. A way to understand why it is so high is to ask the complementary question (imagine the 13 special cards are the spades to make it easy talk about): What is the probability that a selection of 3 cards from the deck contains at least one spade? Well that's close to 59% (not surprising that it's more than 50% since each card has 1/4 chance of being a spade), so the probability that all 13 spades are in the other 49 cards is about 41%.
$endgroup$
– Ned
Jan 4 at 20:27
$begingroup$
You need to go to the 50th card to exceed 50%. A way to understand why it is so high is to ask the complementary question (imagine the 13 special cards are the spades to make it easy talk about): What is the probability that a selection of 3 cards from the deck contains at least one spade? Well that's close to 59% (not surprising that it's more than 50% since each card has 1/4 chance of being a spade), so the probability that all 13 spades are in the other 49 cards is about 41%.
$endgroup$
– Ned
Jan 4 at 20:27
add a comment |
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$begingroup$
$C(39,22)/C(52,35)$ = (number of good combinations of 35 cards) / (number of combinations)
$endgroup$
– Ned
Jan 4 at 14:26