Timeline for Google question: In a country in which people only want boys
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Apr 21, 2021 at 20:44 | comment | added | Marco Disce | @Alexandros ok, let's say the probability is 0.5 whatever I do, the question is: how is this implied by considering the alternative case when the cards are turned from the bottom instead of the top? | |
| Dec 19, 2020 at 3:43 | comment | added | BlueRaja | Peter Winkler also wrote a paper on this and similar problems: Games Pople Don't Play | |
| Sep 1, 2020 at 13:23 | comment | added | hb20007 | @SandeepSilwal My first thought was the same as yours. I even considered the extreme scenario where the first 26 cards all happened to be black, so there are no more black cards in the deck. Surely now the probability of winning is 100%? Actually, the key is to realize that you could have stopped at any point before this. At the point where there was still 1 black card left in the deck, you chose to play on instead of saying "now". So that's the point where you took a decision which had a 50/50 success chance. When you see the last black card, you simply know that you have won. | |
| Dec 2, 2016 at 1:10 | comment | added | Alexandros | @Sandeep: I think you are biased toward the favourable scenario where you've just seen X (say 5) black cards in a row and are thinking you should yell 'now'. You are dicounting the equallly probable scenario where you start by seeing X red cards in a row which is equally likely. For every favourable scenario there is an equally unfavourable one to counter-balance the average. So when you start playing, you might as well take your 50-50 chances and yell 'now'. | |
| Dec 13, 2015 at 4:22 | comment | added | Sandeep Silwal | It is not so clear to me why the answer is 0.5. So I understand that the dealer could have pulled out any card form the remaining deck and the game is equivalent. However, if a lot of black cards had been pulled already from the deck, then the number of black cards remaining in the deck would be low so it would be sensible to say any card pulled from the deck would have a high probability of being red. | |
| Apr 5, 2010 at 17:58 | comment | added | Timothy Chow | For the history of this problem, you could try asking Peter Winkler (at Dartmouth), who calls the bottom card of the deck the "Predestination Card." | |
| Mar 12, 2010 at 14:19 | comment | added | Douglas Zare | I also posted a variant in the ProjectEuler forums. forum.projecteuler.net/viewtopic.php?f=4&t=1445 Besides the symmetry argument, the probability of success in that puzzle is a martingale. | |
| Mar 12, 2010 at 13:57 | comment | added | Douglas Zare | It's funny that you mention that. I just discussed that puzzle in the StoxPoker.com forums (private), and was thinking of posting here to ask for the source. I learned of it on the TwoPlusTwo.com poker forums. | |
| Mar 12, 2010 at 11:45 | history | answered | Tom Leinster | CC BY-SA 2.5 |