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3rd May 2009, 11:05 AM | #1 |
Philosopher
Join Date: Jul 2004
Posts: 5,169
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Monty Hall scenario revisited.
Perhaps this very example has been done before on this forum, but I haven't discovered it ... so let me give it a try.
I actually saw this play out as described, which is why it started me thinking if the contestant actually went with the best course of action. It was on the game show Who Wants to be a Millionaire, where a question was given and four (4) possible choices are given --- one being the correct answer. The contestant had already used the "Ask the Audience" lifeline on a previous question, so he had only two lifelines left. He had no idea as to what the correct answer was, so he chose to "Phone a Friend". His friend gave an answer (choice A), but said it was merely a guess. Now even more frustrated, the contestant used his last lifeline, the "50/50", where two wrong answers are removed (randomly), leaving one wrong answer and the correct one. Choice A was still present as a possibility. After thinking for a moment, he went with the other choice that was left --- and he got it right. Now, did he win that question by mere luck with a 50/50 probability, or did he actually have the 75% chance of having it correct because the choice made by his friend originally still only had a 25% chance of being correct --- even after two wrong answers were removed? |
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Our greatest challenge is not just to ask the important questions, but to recognize the meaningless ones. |
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3rd May 2009, 11:17 AM | #2 |
Philosopher
Join Date: Oct 2007
Posts: 8,613
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First, a question on the rules: could the 50/50 lifeline have removed his friend's incorrect answer (and just didn't happen to in this instance), or are the rules such that the friend's answer always remains on the board in that situation, even if it's incorrect?
If it's the latter and we assume his friend was truly guessing, he was right to switch (and would then have a 75% chance of winning). If it's the former and the friend was guessing, it makes no difference - he had a 50% chance either way. |
3rd May 2009, 11:22 AM | #3 |
New Blood
Join Date: Apr 2009
Posts: 12
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If the 50/50 is random, then it doesn't matter. The friend doesn't have inside knowledge of the answer, and their choice did not influence what answers remained.
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3rd May 2009, 11:23 AM | #4 |
Guest
Join Date: Jul 2006
Posts: 6,387
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In the Monte Hall problem, the source of the surprising probability-change is that the host selectively rules out a door which is (a) not the one you initially picked and also (b) not the prizewinning door. This conveys information about both remaining doors (the one you picked and the one Monte doesn't pick).
In the example you just gave, the lifelines conveyed information that (say) C and D were wrong. Lacking the Monte-Hall-like constraint, this says nothing whatsoever about any differences between A and B. So, no: your contestant originally had a 25% chance; by getting two useful lifelines he reduced it to a 50% chance; then he got lucky. |
3rd May 2009, 12:20 PM | #5 |
Philosopher
Join Date: Jul 2004
Posts: 5,169
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I guess that one could also look at the reverse order of lifelines. If the 50/50 was chosen first it's obvious that he only has a 50% chance. Then phoning someone that picks one of the remaining two, yet still doesn't have a clue helps in no way whatsoever.
Thanks all. |
__________________
Our greatest challenge is not just to ask the important questions, but to recognize the meaningless ones. |
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