Some one commented to me that when looking for an item the one that you want is always the last one.
If I have 10 identical envelopes and one has £10 in it what is the probability of it being in the last one that I choose?
Does it make a difference if I simply pick each envelope from a pile or make a random selection each time?
It's the same probability as it being the first one or any other one, i.e. 1/10. How you select it doesn't make any difference, assuming of course the starting order is random, and not set by somebody wanting to prove it's always the last one.
Is it not the last one because you stop looking once you've found it?
It's always the last one because you stop looking when you find it
Read up on matrices.
Of course it's the last one, because once you've found it you stop looking.
Anyway, isn't the problem to open all ten envelopes, but the one with the tenner in it to be the last one opened? In which case, I think it's:
(9/10)*(8/9)*(7/8)*(6/7)*(5/6)*(4/5)*(3/4)*(2/3)*(1/2)
"The last one" as in the tenth one.
If you check each one in turn then the odds should be 1/10 that it will be the last one.
However, if you choose random envelopes then your first choice is 1/10 then 1/9, 1/8 etc. until you find the £10. Wouldn't this give a different result as at each choice the probabilty is against you? (In a Monty Hall type way)
Matrices I'll have a look.
It's more complicated than 1/10. But I might be wrong.
Ok it's the same (allways 1 in 10) as the mutiplied odds are 1/10.
Thanks all
the chance of it being in any envelope is 1/10. You could say that it is 1/10 that it is the last[deductive logic] * However if you were to do this it would require more than 10 attempts for it to be in the last envelope.... the probability of it being the last is the sum above of each odds multiplied [empirical evidence - what happens in the real world.
the sum shoulkd be 9x8x7x6x5x4x3x2x1 or 1 in 362880
Remember each time you pick if you get the money you need to start again.
Graham will be along with a spreadsheet soon
I hope my maths is correct...for a change
*1 in 10 if 10 people are picking the envelopes and one each is the only way it can occur in real world
Assuming the 10 envelopes are in front of you then your essentialy saying the odds of the £10 being in the 'last' envelope are 1in10.
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