A Conversation for The Gambler's Fallacy
On the other hand
Ed Morg Started conversation Feb 25, 2003
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
On the other hand
Recumbentman Posted Feb 25, 2003
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.
On the other hand
Decaf Silicon Posted Feb 25, 2003
Marilyn Vos Savant wrote a piece on this; whilst searching for it I followed a bunny trail, which led me to some Wordsworth but did nothing for this conversation. Methinks I've been perusing the "Serendipity" article too often.
On the other hand
born dippy Posted Feb 25, 2003
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.
On the other hand
born dippy Posted Feb 25, 2003
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.
On the other hand
Spoadface Posted Feb 25, 2003
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?
On the other hand
silverygibbon Posted Feb 26, 2003
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.
On the other hand
Recumbentman Posted Mar 3, 2003
OK here you go:-
Say there's 3 dollars, behind one £10000 and behind the other 2 nothing. You choose a dollar:
probability of money 1/3
probability of nothing 2/3.
All klear now?
On the other hand
Decaf Silicon Posted Mar 7, 2003
Definitely. I have no idea what these "door" things are. Crazy British.
Don't let the schmegglelippy hit your butt on the way out.
On the other hand
The Noble Robot Posted Aug 2, 2003
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.
On the other hand
Recumbentman Posted Aug 2, 2003
We've had a conversation on this very question in the Monty Hall thread F106080?thread=282209, and I have come out corrected. You *do* improve your chances by switching, as described above; if you are right and switch you lose, but if you are wrong and switch you win. The cases where you are wrong first time outnumber the cases where you are right, therefore the figures favour switching every time.
Key: Complain about this post
On the other hand
- 1: Ed Morg (Feb 25, 2003)
- 2: Recumbentman (Feb 25, 2003)
- 3: Decaf Silicon (Feb 25, 2003)
- 4: born dippy (Feb 25, 2003)
- 5: born dippy (Feb 25, 2003)
- 6: Spoadface (Feb 25, 2003)
- 7: silverygibbon (Feb 26, 2003)
- 8: Decaf Silicon (Mar 3, 2003)
- 9: Recumbentman (Mar 3, 2003)
- 10: Decaf Silicon (Mar 7, 2003)
- 11: The Noble Robot (Aug 2, 2003)
- 12: Recumbentman (Aug 2, 2003)
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