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| | | Subject: On the other hand
Posted Feb 25, 2003 by Ed Morg
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1
 | | People also have a weakness in wanting to stay with a gambling decision - they're unwilling to change even when the odds change. For instance the classic example of teh game where you have 10 doors - behind one there is a prize. You choose a door (odds 10:1) The host will then open another door to show you that there is no prize there. Do you change your guess? Most people won't.
In fact you should as teh odds of the first door are still 10:1 but the other 8 doors now have better odds. (about 8:1)
Aleks
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| | | Subject: On the other hand
Posted Feb 25, 2003 by Recumbentman
This is a reply to this Posting
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2
 | | But your door's odds have also improved equally! So you shouldn't change (or at least it won't improve your odds to do so).
I read once that your best chance of winning in a night's gambling is to bet all you intend to on red or black at Roulette, all at once. It's not a plan for an evening's gambling, but no strategy will give you better odds -- almost 49-51 (there are 18 red squares, 18 black and one white, where the house takes its profit).
If you lose, you waste the least time finding out; if you win, other diversions will suggest themselves for the evening.
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| | | Subject: On the other hand
Posted Feb 25, 2003 by born dippy
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4
| | No, the doors thing is correct. It's often misunderstood as it is counter intuitive.
Say there's 3 doors, behind one £10000 and behind the other 2 nothing. You choose a door:
probability of money 1/3
probability of nothing 2/3.
One of the other doors is then opened, and there's nothing there.
If you then choose the other door:
Proability you chose money first time, you change, and you lose, 1/3
Probabilty you chose nothing frist time, you change, so you win, must be 2/3
i.e. every time you guess wrong first time, you change, and win. So you have more chance of getting the money if you switch.
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| | | Subject: On the other hand
Posted Feb 25, 2003 by born dippy
This is a reply to this Posting
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5
| | No, the doors thing is correct. It's often misunderstood as it is counter intuitive.
Say there's 3 doors, behind one £10000 and behind the other 2 nothing. You choose a door:
probability of money 1/3
probability of nothing 2/3.
One of the other doors is then opened, and there's nothing there.
If you then choose the other door:
Proability you chose money first time, you change, and you lose, 1/3
Probabilty you chose nothing first time, you change, so you win, must be 2/3
i.e. every time you guess wrong first time, you change, and win. So you have more chance of getting the money if you switch.
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| | | Subject: On the other hand
Posted Feb 25, 2003 by Spoadface
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6
| | I don't understand how changing the door you want can improve your actual chance of success.
Taking 3 doors ABC
Choose A
if it's right or wrong, the guy's still going to open another door, bringing the odds down to 1/2.
That means that essentially, your first choice is irrelevant.
Since both remaining doors have a 1 in 2 chance of being the right one, whether you choose another one or not makes no difference.
Or have I missed something?
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| | | Subject: On the other hand
Posted Feb 26, 2003 by silverygibbon
This is a reply to this Posting
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7
| | Spoadface is right.
You select a door. This is an EVENT. With 3 doors you have a 1/3 chance of chosing correctly.
If someone opens another door and shows you there is no prize, you choose again (either to keep or change your original selection). This choice is a NEW EVENT, and is statistically totally unrelated to the previous one.
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| | | Subject: On the other hand
Posted Aug 2, 2003 by The Noble Robot
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11
| | This discussion is far more interesting then the article itself. Both conclusions seem logical to a point. But the answer is simple if you increase the scale.
Say you have a million doors, and you choose one, number 42.
Monty reveals to you 999,998 doors that are not yours, all are empty storage closets. What is left is your door and door number 563,876.
What would you do?
Because the door you picked always remains an option in the second choice, the second pick is NOT an independant event, as the first choice has direct influence on the options in the second choice. The chance of your door being correct during the second choice remains 1 in a 1,000,000, not 1 in 2.
In this rare case, it is very unlikly that 42 is the answer. Switch doors.
nr.
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