Choice of 4, 100 times. How often do you pick right the 1st, 2nd, 3rd and 4th times?

Moonsire

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The are four items in front of you to choose from, only one of which is the correct choice. If you choose wrongly, you must keep trying the other choices until you choose the right one. You must choose the correct item 100 times in total. What percentage of the time will you have chosen correctly the first, second, third and fourth times by the end?


This has been bothering me for a while because both instinct and logic tell me that the answer is 25% for each of the four percentages. However I find myself caught in a loop that I can't quite untangle when I start thinking like so;
If on attempt 1, the odds of picking the correct item first time are 25%, then it is 3 times more likely that you will pick incorrectly first (25% / 75%). If you do pick incorrectly first, then the chances of each of the other items being the right choice become 33% (of 75%, three of four remaining choices). 33% x 75% is still 25%, so we're still fine on the maths and logic front. It is now twice as likely within this sub-choice of three, that the correct item is either the third of fourth choice (33% / 66%) but once again, 66% x 75% = 50% and since we grouped two possibilities together, we must divide by two to get 25% for each.


I do understand that this logic is completely and utterly circular and therefore, seems pointless to contemplate; each option will be chosen 25% of the time by the end no matter what way you twist the maths.
In spite of myself though, I just want to confirm that there is no statistical significance to grouping, for example, a third and fourth choice together other than having to do back multiplication and division to get to the inevitable figure of 25%, and that grouping does not somehow make them more likely outcomes than earlier ones.


Apologies if this question has been irritating, but I have no background in statistics or probability and I just want to be absolutely sure that there isn't some strange rule or calculation that might change the maths, because I know probability can be weird at times. It just feels to me like the answers would somehow be closer to 20%, 25%, 35% and 20%, though I cannot possibly tell you why. Thanks in advance.
 
We need more information. When you choose incorrectly, do you now know that this item is incorrect for the next choices?
 
The are four items in front of you to choose from, only one of which is the correct choice. If you choose wrongly, you must keep trying the other choices until you choose the right one. You must choose the correct item 100 times in total. What percentage of the time will you have chosen correctly the first, second, third and fourth times by the end?


This has been bothering me for a while because both instinct and logic tell me that the answer is 25% for each of the four percentages. However I find myself caught in a loop that I can't quite untangle when I start thinking like so;
If on attempt 1, the odds of picking the correct item first time are 25%, then it is 3 times more likely that you will pick incorrectly first (25% / 75%). If you do pick incorrectly first, then the chances of each of the other items being the right choice become 33% (of 75%, three of four remaining choices). 33% x 75% is still 25%, so we're still fine on the maths and logic front. It is now twice as likely within this sub-choice of three, that the correct item is either the third of fourth choice (33% / 66%) but once again, 66% x 75% = 50% and since we grouped two possibilities together, we must divide by two to get 25% for each.


I do understand that this logic is completely and utterly circular and therefore, seems pointless to contemplate; each option will be chosen 25% of the time by the end no matter what way you twist the maths.
In spite of myself though, I just want to confirm that there is no statistical significance to grouping, for example, a third and fourth choice together other than having to do back multiplication and division to get to the inevitable figure of 25%, and that grouping does not somehow make them more likely outcomes than earlier ones.


Apologies if this question has been irritating, but I have no background in statistics or probability and I just want to be absolutely sure that there isn't some strange rule or calculation that might change the maths, because I know probability can be weird at times. It just feels to me like the answers would somehow be closer to 20%, 25%, 35% and 20%, though I cannot possibly tell you why. Thanks in advance.

1) "Odds" and "Probabilities" are NOT the same thing. Please don't use them interchangeably.
2) Assuming you can answer HoI's question in the affirmative, it's a simple survival model.
----- 21) 1 -- (1/4)
----- 22) 2 -- (3/4)*(1/3)
----- 23) 3 -- (3/4)*(2/3)*(1/2)
----- 24) 4 -- (3/4)*(2/3)*(1/2)*(1)
You seem to have talked yourself into the right idea.
 
Yes sorry, I failed to address that. You are aware that each of your previous choices have been incorrect. I struggled to find a suitable analogy for this, perhaps my initial idea would have been more palatable.
Consider it more like 4 paths that you must choose from. 3 of them lead you to a dead-end and the last is the correct one. You know once you have taken a path whether it was the right choice or not.
 
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