Wikipedia:Reference desk/Archives/Mathematics/2017 December 27

= December 27 =

What is the probability that a prisoner will be set free?
Once upon a time, there were three jewish prisoners in prison await execution. As there is a important jewish festival coming up, the governor of the area would choose one of the prisoner to be set free and the other two would be executed at the date of the festival.

“Woe is me!” thought one of the prisoner, “for my chances for being set free is 1/3”. As he sat there in his cell, the warden of the prison walked by. “Warden! Tell me! Would I be set free?” the prisoner demanded. “Ho! Ho! Not so fast! The Governor insisted that none of the prisoners shall know if they are to be set free or to be executed until the day of the festival. Sorry!”

“Wait!” said the prisoner, “If prisoner Barnabas is to be set free then tell me that prisoner Jesus is to be executed. And if prisoner Jesus is to be set free then tell me that prisoner Barnabas is to be executed. And if I am to be set free then flip a coin and depending on the result of the coin flip, tell me that either prisoner Barnabas or prisoner Jesus is to be executed.”

“Do you take me to be a fool? If I tell you prisoner Jesus is to be executed then you know prisoner Barnabas is to be set free. If I tell you prisoner Barnabas is to be executed then you know prisoner Jesus is to be set free. And if you see me take out a coin and flip the coin then you know that you are to be set free.”

“Don’t do anything now. Go home tonight and do what you have to do at home. Come back tomorrow morning and tell me.”

“Good idea!” said the warden.

The next morning, the warden returned and told the prisoner that “prisoner Barnabas is to be executed.”

“Oh good! My chances of being set free has increased from 1/3 to 1/2 because there is no reason to believe that the governor will choose the prisoner Jesus over me.” said the prisoner

It was at this point that the prisoner woke up from his dream. All the talking to the warden is just a dream. The prisoner never had a chance to speak to the warden.

But wait, thought the prisoner. No matter what answer the the warden gave me in the dream, my chances of being set free will increased from 1/3 to 1/2.

My question is this: can the prisoner increase his probability of being set free by just having a thought experiment? Ohanian (talk) 12:55, 27 December 2017 (UTC)


 * If he can, I'd like to take him to Las Vegas with me. In other words, no, of course not.  The question should actually be: What's the flaw in the prisoner's reasoning?  –Deacon Vorbis (carbon &bull; videos) 14:32, 27 December 2017 (UTC)
 * The other prisoner was probably Barabbas. Bo Jacoby (talk) 15:14, 27 December 2017 (UTC).
 * The other prisoner was probably Barabbas. Bo Jacoby (talk) 15:14, 27 December 2017 (UTC).


 * Our article: Three Prisoners problem. (Though it does not address the twist of the dream.) -- ToE 16:09, 27 December 2017 (UTC)
 * As that article says, this is isomorphic to the Monty Hall problem. AndrewWTaylor (talk) 17:19, 27 December 2017 (UTC)
 * The situation is isomorphic, but the question this time is different. If we'd been asked about the chance that prisoner Jesus will survive, then (ignoring the dream aspect) the conclusion of the Monty Hall Problem would apply, i.e. it is now 2/3. --76.69.117.217 (talk) 06:02, 28 December 2017 (UTC)


 * The Monty Hall link above will hopefully explain the situation, but just to drive the point home: the crucial mistake is in My chances of being set free has increased from 1/3 to 1/2 because there is no reason to believe that the governor will choose the prisoner Jesus over me. Actually, there is such a reason, which is that the information was derived from a biaised game. The mistaken belief in the quoted sentence is one of conditional probability: P(J dies|B dies) = 1/2, and P(J dies|warden said B dies) = 1/2, but that is a mere coincidence and in particular P(unnamed prisoner dies|B dies) = 1/2 ≠ P(unnamed prisoner dies|warden said B dies) = 1/3, because the rules of the "warden says" game are assymetrical with respect to the unnamed prisoner and the other two.  Tigraan Click here to contact me 13:18, 29 December 2017 (UTC)

homotopy books
I want learn homotopy theory in isolation but I should begin from an elementary level so please introduce some elementary books to me. Alireza Badali (talk) 19:18, 27 December 2017 (UTC)
 * Basic topology. Ruslik_ Zero 20:45, 28 December 2017 (UTC)
 * Thank you so much, you resolved my problem, this book is elementary and intelligible for me and then can I read the book Algebraic Topology author Allen Hatcher? Alireza Badali (talk) 22:23, 28 December 2017 (UTC)
 * Perhaps! Alireza Badali (talk) 17:35, 1 January 2018 (UTC)

Five-suit poker probabilities
Suppose I'm playing a poker game with a deck of 65 cards, 13 each in 5 suits. What's the probability that, given 7 cards, I can build a hand of 5 that contains all 5 suits (which I'll hereinafter call a "shulf" as a phonemic reversal of "flush", since it's the opposite of a flush)? A shulf with a pair? A shulf with 3 of a kind? 4 of a kind? 5 of a kind? Two pair? A full house? A straight? (Keep in mind that 10-J-Q-K-A and A-2-3-4-5 both count as straights, but that J-Q-K-A-2, Q-K-A-2-3 and K-A-2-3-4 do not.) I know that if I had only 5 cards to use, the probability of a shulf would be 135/65C5 and of a 5-of-a-kind would be 13/65C5, but poker complicates the problem with its 2 hole cards and 5 community cards. Neon Merlin  23:09, 27 December 2017 (UTC)


 * For just any shulf, you can have 3 cards of one suit, and 1 card each of the remaining four suits, or you can have 2 cards of two suits, and 1 card of each of the remaining three suits. So then you get
 * $$\frac{5\binom{13}{3}\binom{13}{1}^4 + \binom{5}{2}\binom{13}{2}^2\binom{13}{1}^3}{\binom{65}{7}}.$$
 * The others are more complicated, but that should hopefully give you some ideas on how to proceed. That of course shouldn't have been $$\tbinom{65}{5}$$ in the denominator.  It's now correct; thanks ToE.  –Deacon Vorbis (carbon &bull; videos) 23:29, 27 December 2017 (UTC)