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Two players pick cards from standard 52 card deck without replacement…
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I am struggling with this interview prep question... SOS
Two players pick cards from standard 52 card deck without replacement: 1st player
picks a card, then 2nd, then again 1st, then 2nd etc. They stop once somebody picks a king (of any suit),
the player who picks the king wins. What is the probability that the first player wins? the second player?
Which player has more chances of winning: the first or the second? It is sufficient to set up the correct
formula for the probabilities, no need to evaluate it numerically. The last question can be answered without
numerical evaluation, by analysis of the formulas
What is the probability that the first player wins? the second player? couldn't be far of from 50% for both... right?
Which player has more chances of winning: the first or the second? Intuition tells me the first person but I'm not sure if that follows or how to setup a formula -- EDIT: for this I'm thinking certainly the first player because they have one more opportunity to win by choosing a king first
probability
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user1029102912012
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up vote
5
down vote
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I am struggling with this interview prep question... SOS
Two players pick cards from standard 52 card deck without replacement: 1st player
picks a card, then 2nd, then again 1st, then 2nd etc. They stop once somebody picks a king (of any suit),
the player who picks the king wins. What is the probability that the first player wins? the second player?
Which player has more chances of winning: the first or the second? It is sufficient to set up the correct
formula for the probabilities, no need to evaluate it numerically. The last question can be answered without
numerical evaluation, by analysis of the formulas
What is the probability that the first player wins? the second player? couldn't be far of from 50% for both... right?
Which player has more chances of winning: the first or the second? Intuition tells me the first person but I'm not sure if that follows or how to setup a formula -- EDIT: for this I'm thinking certainly the first player because they have one more opportunity to win by choosing a king first
probability
asked 4 hours ago
user1029102912012
262
New contributor
user1029102912012 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
It will be near 50%, but exactly 50% of course. I would say that there is no need to evaluate the number explicitly. The first person wins with probability $frac452+frac4852cdotfrac4751cdotfrac450+frac48cdot 47cdot 46cdot 4552cdot 51cdot 50cdot 49cdot frac448+dots$, each term of which being added is the probability that the first player wins on the first draw, second draw, third draw, etc... wolfram results show it as $frac433833approx 51.98%$
– JMoravitz
4 hours ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I am struggling with this interview prep question... SOS
Two players pick cards from standard 52 card deck without replacement: 1st player
picks a card, then 2nd, then again 1st, then 2nd etc. They stop once somebody picks a king (of any suit),
the player who picks the king wins. What is the probability that the first player wins? the second player?
Which player has more chances of winning: the first or the second? It is sufficient to set up the correct
formula for the probabilities, no need to evaluate it numerically. The last question can be answered without
numerical evaluation, by analysis of the formulas
What is the probability that the first player wins? the second player? couldn't be far of from 50% for both... right?
Which player has more chances of winning: the first or the second? Intuition tells me the first person but I'm not sure if that follows or how to setup a formula -- EDIT: for this I'm thinking certainly the first player because they have one more opportunity to win by choosing a king first
probability
asked 4 hours ago
user1029102912012
262
New contributor
user1029102912012 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am struggling with this interview prep question... SOS
Two players pick cards from standard 52 card deck without replacement: 1st player
picks a card, then 2nd, then again 1st, then 2nd etc. They stop once somebody picks a king (of any suit),
the player who picks the king wins. What is the probability that the first player wins? the second player?
Which player has more chances of winning: the first or the second? It is sufficient to set up the correct
formula for the probabilities, no need to evaluate it numerically. The last question can be answered without
numerical evaluation, by analysis of the formulas
What is the probability that the first player wins? the second player? couldn't be far of from 50% for both... right?
Which player has more chances of winning: the first or the second? Intuition tells me the first person but I'm not sure if that follows or how to setup a formula -- EDIT: for this I'm thinking certainly the first player because they have one more opportunity to win by choosing a king first
probability
probability
asked 4 hours ago
user1029102912012
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user1029102912012 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago
user1029102912012
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edited 4 hours ago
asked 4 hours ago
user1029102912012
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asked 4 hours ago
user1029102912012
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asked 4 hours ago
user1029102912012
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It will be near 50%, but exactly 50% of course. I would say that there is no need to evaluate the number explicitly. The first person wins with probability $frac452+frac4852cdotfrac4751cdotfrac450+frac48cdot 47cdot 46cdot 4552cdot 51cdot 50cdot 49cdot frac448+dots$, each term of which being added is the probability that the first player wins on the first draw, second draw, third draw, etc... wolfram results show it as $frac433833approx 51.98%$
– JMoravitz
4 hours ago
add a comment |Â
It will be near 50%, but exactly 50% of course. I would say that there is no need to evaluate the number explicitly. The first person wins with probability $frac452+frac4852cdotfrac4751cdotfrac450+frac48cdot 47cdot 46cdot 4552cdot 51cdot 50cdot 49cdot frac448+dots$, each term of which being added is the probability that the first player wins on the first draw, second draw, third draw, etc... wolfram results show it as $frac433833approx 51.98%$
– JMoravitz
4 hours ago
It will be near 50%, but exactly 50% of course. I would say that there is no need to evaluate the number explicitly. The first person wins with probability $frac452+frac4852cdotfrac4751cdotfrac450+frac48cdot 47cdot 46cdot 4552cdot 51cdot 50cdot 49cdot frac448+dots$, each term of which being added is the probability that the first player wins on the first draw, second draw, third draw, etc... wolfram results show it as $frac433833approx 51.98%$
– JMoravitz
4 hours ago
It will be near 50%, but exactly 50% of course. I would say that there is no need to evaluate the number explicitly. The first person wins with probability $frac452+frac4852cdotfrac4751cdotfrac450+frac48cdot 47cdot 46cdot 4552cdot 51cdot 50cdot 49cdot frac448+dots$, each term of which being added is the probability that the first player wins on the first draw, second draw, third draw, etc... wolfram results show it as $frac433833approx 51.98%$
– JMoravitz
4 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
In a given round of two draws, you start with $n$ cards of which $4$ are kings. The first player wins with probability $frac 4n$. The second player wins with probability $frac n-4ncdot frac 4n-1=frac 4ncdotfrac n-4n-1$ because they need the first player not to draw a king and there is one less card in the pack for their draw. As the last factor is less than $1$, on each round the first player has a greater chance to win than the second, so the first player has a greater chance to win overall.
I made a spreadsheet to compute the probability. I find the first player wins about $51.98%$ of the time.
answered 4 hours ago
Ross Millikan
280k23189355
add a comment |Â
up vote
1
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Let $E_k$ be the event that the kings are all in the last $k$ cards. Then a win in turn $52-k$ is the event $E_k+1setminus E_k$. Since $E_ksubset E_k+1$, we have $mathsf P(E_k+1setminus E_k)=mathsf P(E_k+1)-mathsf P(E_k)$, so the probability for the first player to win is
$$
sum_j=2^26left(mathsf P(E_2j)-mathsf P(E_2j-1)right)=sum_k=4^52(-1)^kmathsf P(E_k)=sum_k=4^52(-1)^kfracbinom k4binom524=frac433833approx51.98%;.
$$
You can tell that this is greater than the second players win probability without computing the sum by noting that each summand $mathsf P(E_2j)-mathsf P(E_2j-1)$ is greater than the corresponding summand $mathsf P(E_2j-1)-mathsf P(E_2j-2)$ for the second player.
answered 2 hours ago
joriki
167k10180333
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
In a given round of two draws, you start with $n$ cards of which $4$ are kings. The first player wins with probability $frac 4n$. The second player wins with probability $frac n-4ncdot frac 4n-1=frac 4ncdotfrac n-4n-1$ because they need the first player not to draw a king and there is one less card in the pack for their draw. As the last factor is less than $1$, on each round the first player has a greater chance to win than the second, so the first player has a greater chance to win overall.
I made a spreadsheet to compute the probability. I find the first player wins about $51.98%$ of the time.
answered 4 hours ago
Ross Millikan
280k23189355
add a comment |Â
up vote
3
down vote
In a given round of two draws, you start with $n$ cards of which $4$ are kings. The first player wins with probability $frac 4n$. The second player wins with probability $frac n-4ncdot frac 4n-1=frac 4ncdotfrac n-4n-1$ because they need the first player not to draw a king and there is one less card in the pack for their draw. As the last factor is less than $1$, on each round the first player has a greater chance to win than the second, so the first player has a greater chance to win overall.
I made a spreadsheet to compute the probability. I find the first player wins about $51.98%$ of the time.
answered 4 hours ago
Ross Millikan
280k23189355
add a comment |Â
up vote
3
down vote
up vote
3
down vote
In a given round of two draws, you start with $n$ cards of which $4$ are kings. The first player wins with probability $frac 4n$. The second player wins with probability $frac n-4ncdot frac 4n-1=frac 4ncdotfrac n-4n-1$ because they need the first player not to draw a king and there is one less card in the pack for their draw. As the last factor is less than $1$, on each round the first player has a greater chance to win than the second, so the first player has a greater chance to win overall.
I made a spreadsheet to compute the probability. I find the first player wins about $51.98%$ of the time.
answered 4 hours ago
Ross Millikan
280k23189355
In a given round of two draws, you start with $n$ cards of which $4$ are kings. The first player wins with probability $frac 4n$. The second player wins with probability $frac n-4ncdot frac 4n-1=frac 4ncdotfrac n-4n-1$ because they need the first player not to draw a king and there is one less card in the pack for their draw. As the last factor is less than $1$, on each round the first player has a greater chance to win than the second, so the first player has a greater chance to win overall.
I made a spreadsheet to compute the probability. I find the first player wins about $51.98%$ of the time.
answered 4 hours ago
Ross Millikan
280k23189355
answered 4 hours ago
Ross Millikan
280k23189355
answered 4 hours ago
Ross Millikan
280k23189355
answered 4 hours ago
Ross Millikan
280k23189355
280k23189355
add a comment |Â
add a comment |Â
up vote
1
down vote
Let $E_k$ be the event that the kings are all in the last $k$ cards. Then a win in turn $52-k$ is the event $E_k+1setminus E_k$. Since $E_ksubset E_k+1$, we have $mathsf P(E_k+1setminus E_k)=mathsf P(E_k+1)-mathsf P(E_k)$, so the probability for the first player to win is
$$
sum_j=2^26left(mathsf P(E_2j)-mathsf P(E_2j-1)right)=sum_k=4^52(-1)^kmathsf P(E_k)=sum_k=4^52(-1)^kfracbinom k4binom524=frac433833approx51.98%;.
$$
You can tell that this is greater than the second players win probability without computing the sum by noting that each summand $mathsf P(E_2j)-mathsf P(E_2j-1)$ is greater than the corresponding summand $mathsf P(E_2j-1)-mathsf P(E_2j-2)$ for the second player.
answered 2 hours ago
joriki
167k10180333
add a comment |Â
up vote
1
down vote
Let $E_k$ be the event that the kings are all in the last $k$ cards. Then a win in turn $52-k$ is the event $E_k+1setminus E_k$. Since $E_ksubset E_k+1$, we have $mathsf P(E_k+1setminus E_k)=mathsf P(E_k+1)-mathsf P(E_k)$, so the probability for the first player to win is
$$
sum_j=2^26left(mathsf P(E_2j)-mathsf P(E_2j-1)right)=sum_k=4^52(-1)^kmathsf P(E_k)=sum_k=4^52(-1)^kfracbinom k4binom524=frac433833approx51.98%;.
$$
You can tell that this is greater than the second players win probability without computing the sum by noting that each summand $mathsf P(E_2j)-mathsf P(E_2j-1)$ is greater than the corresponding summand $mathsf P(E_2j-1)-mathsf P(E_2j-2)$ for the second player.
answered 2 hours ago
joriki
167k10180333
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $E_k$ be the event that the kings are all in the last $k$ cards. Then a win in turn $52-k$ is the event $E_k+1setminus E_k$. Since $E_ksubset E_k+1$, we have $mathsf P(E_k+1setminus E_k)=mathsf P(E_k+1)-mathsf P(E_k)$, so the probability for the first player to win is
$$
sum_j=2^26left(mathsf P(E_2j)-mathsf P(E_2j-1)right)=sum_k=4^52(-1)^kmathsf P(E_k)=sum_k=4^52(-1)^kfracbinom k4binom524=frac433833approx51.98%;.
$$
You can tell that this is greater than the second players win probability without computing the sum by noting that each summand $mathsf P(E_2j)-mathsf P(E_2j-1)$ is greater than the corresponding summand $mathsf P(E_2j-1)-mathsf P(E_2j-2)$ for the second player.
answered 2 hours ago
joriki
167k10180333
Let $E_k$ be the event that the kings are all in the last $k$ cards. Then a win in turn $52-k$ is the event $E_k+1setminus E_k$. Since $E_ksubset E_k+1$, we have $mathsf P(E_k+1setminus E_k)=mathsf P(E_k+1)-mathsf P(E_k)$, so the probability for the first player to win is
$$
sum_j=2^26left(mathsf P(E_2j)-mathsf P(E_2j-1)right)=sum_k=4^52(-1)^kmathsf P(E_k)=sum_k=4^52(-1)^kfracbinom k4binom524=frac433833approx51.98%;.
$$
You can tell that this is greater than the second players win probability without computing the sum by noting that each summand $mathsf P(E_2j)-mathsf P(E_2j-1)$ is greater than the corresponding summand $mathsf P(E_2j-1)-mathsf P(E_2j-2)$ for the second player.
answered 2 hours ago
joriki
167k10180333
answered 2 hours ago
joriki
167k10180333
answered 2 hours ago
joriki
167k10180333
answered 2 hours ago
joriki
167k10180333
167k10180333
add a comment |Â
add a comment |Â
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It will be near 50%, but exactly 50% of course. I would say that there is no need to evaluate the number explicitly. The first person wins with probability $frac452+frac4852cdotfrac4751cdotfrac450+frac48cdot 47cdot 46cdot 4552cdot 51cdot 50cdot 49cdot frac448+dots$, each term of which being added is the probability that the first player wins on the first draw, second draw, third draw, etc... wolfram results show it as $frac433833approx 51.98%$
– JMoravitz
4 hours ago