GdB - 21 November 2012 11:35 AM
Clearer as here I cannot explain it, ...
Well, maybe I can. Don’t ever forget (if you do not agree with this then you are completely lost): the chance of an event is the number of times the event occurs, divided by the total number of events. Take the coin: there are two possible events, heads or tails. Say I throw 100 times, then we expect about 50 times heads. So the chance of heads is 50/100 = 0.5. And the die: the chance for a 3 is 1/6 means that if I throw the die 600 times, about 100 times it will be a 3. In both cases, based on symmetry I can say that the chances are effective 1/2 and 1/6 respectively. Right? If you do not agree with this definition of chance we are done.
Now the TEP: we must find the events we look for (maximise gain). Now it is a bit difficult to do this for all possible TEP pairs (these are infinite). But it suffices that the player does not know with what TEP pair he is playing. But we can probe the different possibilities for just one pair, and then state that this is true then for every TEP pair.
So what we must do is run through all possible scenarios for one TEP pair, with the addition that the player forgets immediate with what pair he is playing (or we play everytime with somebody else who does not know what has happened before).
Now we must compare the strategy ‘always switch’ with ‘not switching’.
As example we take the pair (10,20).
So the possible events are:
- You see you have chosen 10: you suppose the pair could have been (5,10) or (10,20). You switch, and you gain 10.
- You see you have chosen 20: you suppose the pair could have been (10,20) or (20,40). You switch, and you loose 10.
And not switching:
- You see you have chosen 10, you keep it, you loose 10.
- You see you have chosen 20, you keep it, you gained 10.
In both case the nett gain is 0.
Your rephrasing of the problem just gives you the wrong suggestion. Let’s compare it with the other situation, you choose an envelope, you look at the amount, and then the other envelope is filled with half or twice the amount you have, decided by the tossing of a fair coin (i.e. the chance of half or twice is 50%)
- You choose 10, you switch, the other contains 5, you loose 5
- You choose 10, you switch, the other contains 20, you gain 10 (from here I could stop, but for completeness I’ll do 20 too)
- You choose 20, you switch, the other contains 10, you loose 10
- You choose 20, you switch, the other contains 40, you gain 20
Obviously, you gain by switching. And obviously both situations are completely different. Clear now?
Corrolarium: the same argumentation holds if you do not look, so also in TEP without looking you do not gain by switching.