My friend was having an argument relating to statistics. Basically, there are five people and four tickets to the Astros game. The group decided to put five names into a hat. The question they had was, should they pick out one name and that person would be the one not to go to the game, or should they have four different drawings (each person whose name is drawn will go to the game)? The group members tried calculating probabilities they could go to the game in both scenarios, but everyone stumbled. Common sense says the probabilities are the same. Is that the case? Usually I'm pretty good with math/stats, but my mind's a little numb right now...
In the first case, your chance of going is 4/5. However, in the second case, with each successive drawing, your chance of going decreases from 4/5 to 3/4 to 2/3 to 1/2. they all kinda end up the same...but second is more suspenseful I guess. for all intents and purposes though, you're just prolonging the process and not changing the odds fundamentally.
Yeah, it's the same. For the picking of four names, there are 5 ways of doing this (5 combination 4), and in one of these ways does person A (assuming the five people are A,B, etc.) not get picked. So that's a 1 in 5 probability. For the picking of one name not to go, it's a simple 1 out of 5 probability that his name gets picked to not go.
They are precisely equal. In the first case, your chance of not going is 1/5. In the second case, your chance of not going is (4/5)*(3/4)*(2/3)*(1/2) = 1/5.
With one drawing you have a 20% chance of being drawn and not going. With 4 drawings it's: (1) 20% of going (2) 25% "" (3) 33% "" (4) 50% "" And one lucky man. (1-t) (.8)*(3/4)*(2/3)*(1/2) = 20% They are the same, but I would add the extra 30 seconds of suspense.
Second one is funner though, your odds start to "shift" as new information comes in, even if they were fundamentally the same before and are the same if you consider the overall picture. But with the second, you can slowly glimpse your passage into "not getting reward hell". then again, astros game. meh.
this seems like it ALMOST qualifies as a monty hall problem. http://en.wikipedia.org/wiki/Monty_Hall_problem HOWEVER - there is no omniscient third party that is forced to reveal a "wrong door".... which means that i agree - either way you choose, you always have a 20% chance of not going.
