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If he sticks with his original choice his chance of success is 1/3. If he changes his mind his chances of success are 2/3.
His second choice offers him more information that wasn’t available with his first.
spot on echo , I’m impressed
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If he sticks with his original choice his chance of success is 1/3. If he changes his mind his chances of success are 2/3.
His second choice offers him more information that wasn’t available with his first.
spot on echo , I’m impressed
This is actually wrong.
The Monty Hall problem needs 4 or more choices before the probability of winning increases from switching. With three choices the probability is 1/3 in both cases.
I feel like we’ve discussed this before.
If he sticks with his original choice his chance of success is 1/3. If he changes his mind his chances of success are 2/3.
His second choice offers him more information that wasn’t available with his first.
spot on echo , I’m impressed
This is actually wrong.
The Monty Hall problem needs 4 or more choices before the probability of winning increases from switching. With three choices the probability is 1/3 in both cases.
I feel like we’ve discussed this before.
so much for you being educated
I’ll endeavour to educate you drac, 3 counters ( 2 white and one black) means there is a statistical chance of someone randomly picking a black counter of 33.3%. If the person picks the black counter there are two white counters left, if the person picks a white counter there is one black and one white counter left. In any scenario, removing a white counter from the two counters not picked doesn’t affect the 33.3% chance of the original choice of counter being black as the odds on there being a white counter left is 100% in the two remaining counters so removing one has no impact whatsoever.
I’ll endeavour to educate you drac, 3 counters ( 2 white and one black) means there is a statistical chance of someone randomly picking a black counter of 33.3%. If the person picks the black counter there are two white counters left, if the person picks a white counter there is one black and one white counter left. In any scenario, removing a white counter from the two counters not picked doesn’t affect the 33.3% chance of the original choice of counter being black as the odds on there being a white counter left is 100% in the two remaining counters so removing one has no impact whatsoever.
Nope
Lets play out all scenarios – remember there are 3 hidden counters ( 2 white and 1 black). The question asks , after person A picks one random counter concealed under a beer mat and a white counter is removed from one of the two left ,does his probability of picking a black counter increase by switching to the other remaining beer mat. Here are all the possibilities of what lies under each beer mat if he picks the first beer mat and has a white counter removed from one of the two left
Scenario 1) Beer mat 1 has a black counter , 2 and 3 have white counters. one white counter is removed leaving 1 black and 1 white. He switches from his original choice of black to white = white counter
Scenario 2) Beer mat 1 has white counter , 2 has black , 3 has white. White counter is removed leaving his white counter and one black – he switches from his initial choice of white to end up with a black counter
Scenario 3) Beer mat 1 has a white counter, 2 has white , 3 has black- one of the white counters not picked by him in mats 2 and 3 is removed leaving one white ( his initial choice and one black) he switches to leave a black counter.
His probability of ending with a black counter is 2/3 by switching. Play out all scenarios by not switching and it’s 1/3.
Stay off the drugs drac, they’re frazzling your brain
You mention monty hall, allow me to educate you drac .. replace the counters with goats and it confirms you’ve lost the plot once again
Insulting me doesn’t convince me that you’re right
Whatever counter the original one is , is bound to leave either one white counter left or two. Removing one white counter left doesn’t alter the statistical probability of the original choice which was 1 in 3 or 33% of the counter being black. The chances of a counter left in the two remaining is 100%, a computer or in this case “barman” removing one doesn’t tell us anything new and has no effect on the original odds.
If you have 50 counters – 49 white and one black. The chances of one being picked randomly turning out black is 2% ( 1 in 50). After that counter is picked , there are guaranteed to be either 48 or 49 white counters left… if 48 white counters are removed from the other side, leaving 2 left your first choice and the one remaining counter then your chance of your original choice being a black counter is still one in fifty not one in two. The same principle applies to 3 counters or 3 million.. so much for the French education system. I’m embarrassed for you Drac
Insulting me doesn’t convince me that you’re right
but I am right drac, as well you know
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