Monty Hall problem with coin flip












1












$begingroup$



Before each show, Monty secretly flips a coin with probability p of Heads. If the coin
lands Heads, Monty resolves to open a goat door (with equal probabilities if there is
a choice). Otherwise, Monty resolves to open a random unopened door, with equal probabilities. The contestant knows p but does not know the outcome of the coin flip.
When the show starts, the contestant chooses a door. Monty (who knows where the car is) then opens a door. If the car is revealed, the game is over; if a goat is revealed, the contestant is offered the option of switching. Now suppose it turns out that the contestant opens door 1 and then Monty opens door 2, revealing a goat. What is the contestant’s probability of success if he or she switches to door 3?




This is from "Introduction to Probability" By Joseph K. Blitzstein



The solution that I came up with is the following:



Ci - event that car is behind door i



H - event that coin landed head



Xi - event that participant picks door i initially
Oi - event that Monty opens door i



Given this we have:



$P(C3|X1, O2) = frac{P(O2|C3, X1)P(C3|X1)}{P(O2|C3, X1)}$



And:



$P(O2|C3, X1) = P(O2|C3, X1, H)P(H|C3, X1) + P(O2|C3, X1, Hc)P(Hc|C3, X1)$



I'd appreciate some thoughts on this.






combinatorics conditional-probability monty-hall







share|cite|improve this question









$endgroup$












  • $begingroup$
    Well.... did you try plugging in some numbers? It seems like a straightforward Baye's theorem.
    $endgroup$
    – fleablood
    Nov 26 '18 at 23:18










  • $begingroup$
    Don't think you need to worry about probability that contestant picks door i. We are given that he picked door 1. That's a given.
    $endgroup$
    – fleablood
    Nov 26 '18 at 23:20
















1












$begingroup$



Before each show, Monty secretly flips a coin with probability p of Heads. If the coin
lands Heads, Monty resolves to open a goat door (with equal probabilities if there is
a choice). Otherwise, Monty resolves to open a random unopened door, with equal probabilities. The contestant knows p but does not know the outcome of the coin flip.
When the show starts, the contestant chooses a door. Monty (who knows where the car is) then opens a door. If the car is revealed, the game is over; if a goat is revealed, the contestant is offered the option of switching. Now suppose it turns out that the contestant opens door 1 and then Monty opens door 2, revealing a goat. What is the contestant’s probability of success if he or she switches to door 3?




This is from "Introduction to Probability" By Joseph K. Blitzstein



The solution that I came up with is the following:



Ci - event that car is behind door i



H - event that coin landed head



Xi - event that participant picks door i initially
Oi - event that Monty opens door i



Given this we have:



$P(C3|X1, O2) = frac{P(O2|C3, X1)P(C3|X1)}{P(O2|C3, X1)}$



And:



$P(O2|C3, X1) = P(O2|C3, X1, H)P(H|C3, X1) + P(O2|C3, X1, Hc)P(Hc|C3, X1)$



I'd appreciate some thoughts on this.






combinatorics conditional-probability monty-hall







share|cite|improve this question









$endgroup$












  • $begingroup$
    Well.... did you try plugging in some numbers? It seems like a straightforward Baye's theorem.
    $endgroup$
    – fleablood
    Nov 26 '18 at 23:18










  • $begingroup$
    Don't think you need to worry about probability that contestant picks door i. We are given that he picked door 1. That's a given.
    $endgroup$
    – fleablood
    Nov 26 '18 at 23:20














1












1








1


2



$begingroup$



Before each show, Monty secretly flips a coin with probability p of Heads. If the coin
lands Heads, Monty resolves to open a goat door (with equal probabilities if there is
a choice). Otherwise, Monty resolves to open a random unopened door, with equal probabilities. The contestant knows p but does not know the outcome of the coin flip.
When the show starts, the contestant chooses a door. Monty (who knows where the car is) then opens a door. If the car is revealed, the game is over; if a goat is revealed, the contestant is offered the option of switching. Now suppose it turns out that the contestant opens door 1 and then Monty opens door 2, revealing a goat. What is the contestant’s probability of success if he or she switches to door 3?




This is from "Introduction to Probability" By Joseph K. Blitzstein



The solution that I came up with is the following:



Ci - event that car is behind door i



H - event that coin landed head



Xi - event that participant picks door i initially
Oi - event that Monty opens door i



Given this we have:



$P(C3|X1, O2) = frac{P(O2|C3, X1)P(C3|X1)}{P(O2|C3, X1)}$



And:



$P(O2|C3, X1) = P(O2|C3, X1, H)P(H|C3, X1) + P(O2|C3, X1, Hc)P(Hc|C3, X1)$



I'd appreciate some thoughts on this.






combinatorics conditional-probability monty-hall







share|cite|improve this question









$endgroup$





Before each show, Monty secretly flips a coin with probability p of Heads. If the coin
lands Heads, Monty resolves to open a goat door (with equal probabilities if there is
a choice). Otherwise, Monty resolves to open a random unopened door, with equal probabilities. The contestant knows p but does not know the outcome of the coin flip.
When the show starts, the contestant chooses a door. Monty (who knows where the car is) then opens a door. If the car is revealed, the game is over; if a goat is revealed, the contestant is offered the option of switching. Now suppose it turns out that the contestant opens door 1 and then Monty opens door 2, revealing a goat. What is the contestant’s probability of success if he or she switches to door 3?




This is from "Introduction to Probability" By Joseph K. Blitzstein



The solution that I came up with is the following:



Ci - event that car is behind door i



H - event that coin landed head



Xi - event that participant picks door i initially
Oi - event that Monty opens door i



Given this we have:



$P(C3|X1, O2) = frac{P(O2|C3, X1)P(C3|X1)}{P(O2|C3, X1)}$



And:



$P(O2|C3, X1) = P(O2|C3, X1, H)P(H|C3, X1) + P(O2|C3, X1, Hc)P(Hc|C3, X1)$



I'd appreciate some thoughts on this.






combinatorics conditional-probability monty-hall




combinatorics conditional-probability monty-hall






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Nov 26 '18 at 22:02









Erik Cristian SeuleanErik Cristian Seulean

556




556












  • $begingroup$
    Well.... did you try plugging in some numbers? It seems like a straightforward Baye's theorem.
    $endgroup$
    – fleablood
    Nov 26 '18 at 23:18










  • $begingroup$
    Don't think you need to worry about probability that contestant picks door i. We are given that he picked door 1. That's a given.
    $endgroup$
    – fleablood
    Nov 26 '18 at 23:20


















  • $begingroup$
    Well.... did you try plugging in some numbers? It seems like a straightforward Baye's theorem.
    $endgroup$
    – fleablood
    Nov 26 '18 at 23:18










  • $begingroup$
    Don't think you need to worry about probability that contestant picks door i. We are given that he picked door 1. That's a given.
    $endgroup$
    – fleablood
    Nov 26 '18 at 23:20
















$begingroup$
Well.... did you try plugging in some numbers? It seems like a straightforward Baye's theorem.
$endgroup$
– fleablood
Nov 26 '18 at 23:18




$begingroup$
Well.... did you try plugging in some numbers? It seems like a straightforward Baye's theorem.
$endgroup$
– fleablood
Nov 26 '18 at 23:18












$begingroup$
Don't think you need to worry about probability that contestant picks door i. We are given that he picked door 1. That's a given.
$endgroup$
– fleablood
Nov 26 '18 at 23:20




$begingroup$
Don't think you need to worry about probability that contestant picks door i. We are given that he picked door 1. That's a given.
$endgroup$
– fleablood
Nov 26 '18 at 23:20










1 Answer
1






active

oldest

votes


















1












$begingroup$

The coin flip, the door that the contestant picks, and the placement of the car will be mutually independent .



However, the door that Monty picked depended on their result.   When the contestant and car indicate different doors, then if the coin landed heads, Monty certainly revealed the only goat he could, but otherwise he had unbiasedly choosen a door from those the contestant did not choose reguardless of the car placement.



$$begin{split}mathsf P(O_2mid X_1,C_3) &= mathsf P(H)~mathsf P(O_2mid H,X_1,C_3) + mathsf P(H^complement)~mathsf P(O_2mid H^complement,X_1)end{split}$$



Of course, if both the contestant and car choose the same door, then Monty's choiseof goats was unbiased whatever the coin may have said.




$$begin{split}mathsf P(C_3mid X_1,O_2) &= dfrac{mathsf P(C_3)~mathsf P(O_2mid X_1,C_3)}{mathsf P(C_1)~mathsf P(O_2mid X_1,C_1)+mathsf P(C_3)~mathsf P(O_2mid X_1,C_3)}end{split}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for this, yeah, I thought about being independent events, although didn't simplify.
    $endgroup$
    – Erik Cristian Seulean
    Nov 28 '18 at 20:57











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$begingroup$

The coin flip, the door that the contestant picks, and the placement of the car will be mutually independent .



However, the door that Monty picked depended on their result.   When the contestant and car indicate different doors, then if the coin landed heads, Monty certainly revealed the only goat he could, but otherwise he had unbiasedly choosen a door from those the contestant did not choose reguardless of the car placement.



$$begin{split}mathsf P(O_2mid X_1,C_3) &= mathsf P(H)~mathsf P(O_2mid H,X_1,C_3) + mathsf P(H^complement)~mathsf P(O_2mid H^complement,X_1)end{split}$$



Of course, if both the contestant and car choose the same door, then Monty's choiseof goats was unbiased whatever the coin may have said.




$$begin{split}mathsf P(C_3mid X_1,O_2) &= dfrac{mathsf P(C_3)~mathsf P(O_2mid X_1,C_3)}{mathsf P(C_1)~mathsf P(O_2mid X_1,C_1)+mathsf P(C_3)~mathsf P(O_2mid X_1,C_3)}end{split}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for this, yeah, I thought about being independent events, although didn't simplify.
    $endgroup$
    – Erik Cristian Seulean
    Nov 28 '18 at 20:57
















1












$begingroup$

The coin flip, the door that the contestant picks, and the placement of the car will be mutually independent .



However, the door that Monty picked depended on their result.   When the contestant and car indicate different doors, then if the coin landed heads, Monty certainly revealed the only goat he could, but otherwise he had unbiasedly choosen a door from those the contestant did not choose reguardless of the car placement.



$$begin{split}mathsf P(O_2mid X_1,C_3) &= mathsf P(H)~mathsf P(O_2mid H,X_1,C_3) + mathsf P(H^complement)~mathsf P(O_2mid H^complement,X_1)end{split}$$



Of course, if both the contestant and car choose the same door, then Monty's choiseof goats was unbiased whatever the coin may have said.




$$begin{split}mathsf P(C_3mid X_1,O_2) &= dfrac{mathsf P(C_3)~mathsf P(O_2mid X_1,C_3)}{mathsf P(C_1)~mathsf P(O_2mid X_1,C_1)+mathsf P(C_3)~mathsf P(O_2mid X_1,C_3)}end{split}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for this, yeah, I thought about being independent events, although didn't simplify.
    $endgroup$
    – Erik Cristian Seulean
    Nov 28 '18 at 20:57














1












1








1





$begingroup$

The coin flip, the door that the contestant picks, and the placement of the car will be mutually independent .



However, the door that Monty picked depended on their result.   When the contestant and car indicate different doors, then if the coin landed heads, Monty certainly revealed the only goat he could, but otherwise he had unbiasedly choosen a door from those the contestant did not choose reguardless of the car placement.



$$begin{split}mathsf P(O_2mid X_1,C_3) &= mathsf P(H)~mathsf P(O_2mid H,X_1,C_3) + mathsf P(H^complement)~mathsf P(O_2mid H^complement,X_1)end{split}$$



Of course, if both the contestant and car choose the same door, then Monty's choiseof goats was unbiased whatever the coin may have said.




$$begin{split}mathsf P(C_3mid X_1,O_2) &= dfrac{mathsf P(C_3)~mathsf P(O_2mid X_1,C_3)}{mathsf P(C_1)~mathsf P(O_2mid X_1,C_1)+mathsf P(C_3)~mathsf P(O_2mid X_1,C_3)}end{split}$$






share|cite|improve this answer









$endgroup$



The coin flip, the door that the contestant picks, and the placement of the car will be mutually independent .



However, the door that Monty picked depended on their result.   When the contestant and car indicate different doors, then if the coin landed heads, Monty certainly revealed the only goat he could, but otherwise he had unbiasedly choosen a door from those the contestant did not choose reguardless of the car placement.



$$begin{split}mathsf P(O_2mid X_1,C_3) &= mathsf P(H)~mathsf P(O_2mid H,X_1,C_3) + mathsf P(H^complement)~mathsf P(O_2mid H^complement,X_1)end{split}$$



Of course, if both the contestant and car choose the same door, then Monty's choiseof goats was unbiased whatever the coin may have said.




$$begin{split}mathsf P(C_3mid X_1,O_2) &= dfrac{mathsf P(C_3)~mathsf P(O_2mid X_1,C_3)}{mathsf P(C_1)~mathsf P(O_2mid X_1,C_1)+mathsf P(C_3)~mathsf P(O_2mid X_1,C_3)}end{split}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 '18 at 23:20









Graham KempGraham Kemp

85.1k43378




85.1k43378












  • $begingroup$
    Thanks for this, yeah, I thought about being independent events, although didn't simplify.
    $endgroup$
    – Erik Cristian Seulean
    Nov 28 '18 at 20:57


















  • $begingroup$
    Thanks for this, yeah, I thought about being independent events, although didn't simplify.
    $endgroup$
    – Erik Cristian Seulean
    Nov 28 '18 at 20:57
















$begingroup$
Thanks for this, yeah, I thought about being independent events, although didn't simplify.
$endgroup$
– Erik Cristian Seulean
Nov 28 '18 at 20:57




$begingroup$
Thanks for this, yeah, I thought about being independent events, although didn't simplify.
$endgroup$
– Erik Cristian Seulean
Nov 28 '18 at 20:57


















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