Timeline for probability puzzle - selecting a person
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 14, 2010 at 0:23 | comment | added | Neil | Oh, I assumed that people get up from the table when they can no longer be head of the table... Sorry. | |
| Apr 13, 2010 at 22:48 | comment | added | Douglas Zare | @Neil If there are 4 people at the table, then when the coin is passed left and then right, the second pass does not eliminate a person. The coin is back in the original position, and there are two people who have not received the coin. The player to the right of the head has probability 1/3 of being the last not to hold the coin, and the player opposite the head now has a 2/3 chance to be the last. At the start, the probability distribution is uniform over all people except the head. | |
| Apr 13, 2010 at 22:48 | comment | added | HJRW | Neil, I don't think your last comment is right. For instance, HTHH selects person 0, it seems to me. | |
| Apr 13, 2010 at 22:13 | comment | added | Neil | In fact, this is clearly not right for n=4: heads-heads selects person 0, and tails-tails person 2, but person 1 is selected for either heads-tails or tails-heads. Thus, the probability is not uniform. | |
| Jan 16, 2010 at 14:54 | comment | added | Douglas Zare | For each person P, it's the same random walk problem: One neighbor is eliminated. Will the coin go around the table to the other neighbor before getting to P? If so, P is selected. If not, P is eliminated. It's also a standard exercise that the probability is 1/(n-1) as discussed <a href="mathoverflow.net/questions/7004/…>. | |
| Jan 16, 2010 at 14:47 | comment | added | Rohit Banga | not very clear. how is it the same for each person. | |
| Jan 16, 2010 at 14:29 | history | answered | Douglas Zare | CC BY-SA 2.5 |