Statistics riddle problem continued.

[Xanthul]

250 its not 2/n. Its not picking 2 doors between N doors, its picking a door from N doors, then leaving only 2 doors, and choosing between them. Read the problem carefully, its not the same.

For example, with 100 doors: The host knows that prize is in door number 55, for example. So now, you pick a door randomly, for example #34. Then the host opens all doors except door #55 because HE KNOWS that the prize is there and he cant show it. He has to leave only 2 doors closed: one is your pick, and the other one most of the times will be the prized one (unless you picked the prized one in your first choice, in which case he will have to choose a door randomly).

Now, the discussion is whether door 34 and 55 have 50% chances each, or if door 34 has only 1% chance and door 55 has 99% chance. Im sure its the second one, because 99% of the times you will have MISSED your first choice (even if you still DONT KNOW IT), and the host will have to leave closed the prized door, therefore POINTING you which one is the prized one.

With N doors, if you ALWAYS change your first choice, you will win the prize with a probability of N-1/N.

[LennonCook]

Let me explain Again: Well use 3 doors Here., although the number only matters for the values not the result:
If it is Presented to us as a LOGIC problem: If on the first choice u pick the wrong door (66% Chance) U will only be right if u change.
If u pick the right door (33%) U will win only if u stay with this door. So u win 66% of the time if u Change, 33% If u dont. Therefore U should change.

If Presented as a PROBABLITY OR STATS PROBLEM:
Chance= NUmber of Useful Choices/ Total Possible Options.
So first Choice: 1 Useful, 3 POssible 1/3 or 33% Chance of Getting it right.
Second Choice: 1 Useful, 2 Possible- 50% Chance.

This was presented to us as a Probability puzzle, so therefore we must say that the second answer is the right one, Because it uses the Statistical Method.

[250]

ohhh, I see, so when you first pick a door, you DON'T look at it.
then the host ask if you want to switch

in that case, definitely switch
the first one is 1/n

the second is 1/(n-1)

[Xanthul]

Ehmmmm not exactly 250 [img]smile.gif[/img] . You pick a door, then the host DISCARDS the rest of the doors except one, so you have 2 choices: stick with your first choice or switch to the OTHER door (there is only ONE door more).

Thats why people said it was a 50/50 thing, but as i reasoned above, im sure its 1/n and (n-1)/n.

[250]

nope. not exactly Ertai

I think you understand the question wrong. in that case, switch is always way much better to the state of pointless

who would make such a stupid contest?

nope, I dont think so

[Xanthul]

Of course its better to switch 250. The clear example is with a high number of doors, lets say 1000. In that case, your first choice would be wrong 99,9 % of the times, so the host would have to leave closed ONLY THE PRIZED DOOR, because he cant open that one, so the chances of winning if you switch doors is 99,9 % in this case, (n-1)/n in general.

[Legolas]

No Ertai, if you want to use that approach, you have to remember the odds of that particular door with the chicken remaining is also 0.1%. So you have two doors which originally had a 0.1% chance of containing the prize, and the chance the prize is behind one over the other is 0.1 / 0.1 which equals 50%

[Xanthul]

Well Legolas, look at the numbers and forget about statistics: the truth is that if there are 1000 doors and you switch your first choice, you will win 99,9 % of the times. You cant argue that. The thing that changes the probability is that the host KNOWS where is the prize, and he cant open that door. The 50/50 thing would be if you DIDNT know which door is your RANDOM choice and which door is the host´s FORCED choice.

[250]

no ertai, that is NOT the case. you understand mistakingly. as I said, no one will give this kind term in a game, it is just plain stupid

[Sir Kenyth]

I think defining the perspective is the key here.

Subsequent tries always have a higher probability for a given task. The straight probability is only for a one try deal. If you try a 50/50 task four times, there's a better chance of you winning it at least once. Statistical concepts are somewhat abstract and fairly malleable. That's why statisticians can pretty much come up with whatever numbers the person who hires them wants.

The first try of ten doors is 1 in 10. The second try can be seen as 1 in 9 or 1 in 5 depending on what logic you use. If it is a subsequent try by the same person, it would be 1 in 5. If a different person made their first try at the remaining doors, it would be 1 in 9. If you look at it from the perspective of both people as a group effort, it again becomes 1 in 5 for the group. Even though the individual chance is 1 in 9. It's strange to think that the exact same situation can produce different probabilities depending on where you look at it from.

Even if a person participated in two entirely different game shows with ten doors each. Since they are separate problems, the chance is 1 in 10 for each respectively. When you take into perspective the history of there being two games played, the chance becomes higher the second game. A total of 20 doors, 2 choices, and 2 prizes. If memory serves, I think this is the situation in which 250s equation is pertinent. A third game would bring the total to 30 doors, 3 choices, and three prize doors. The stats change slightly in favor of the participant each subsequent game, but only if you take into account the historical data. The individual task still has a 1 in 10 chance.