i start with two tosses of a coin, for every head i get i get two extra tosses
could this game lasts for someone say as to get a trillion extra tosses?
how likely is to last at this game?

i start with two tosses of a coin, for every head i get i get two extra tosses
could this game lasts for someone say as to get a trillion extra tosses?
how likely is to last at this game?
Yes it could, obviously, you could just toss a trillion heads in a row (and claim the title of the worlds greatest tosser )Originally Posted by luxtpm
Anyway back to probability.
OK I should say I worked this out as I typed it in and changed my 'method
after the ============, I might have made some errors but I think I got there
in the end!!
If you can follow it perhaps you can tell me if you agree?
I might try a computer program later to verify the results (if I get the program right  lol)
For the first two tosses you can have
hh
th
ht
tt
Thus you have a 75% chance of making 3 tosses.
you will only exit if you toss a t after a t
chance you have a h is 66% t 33%
If you have t then you have a 50% chance of another= 16%
consider 3 tosses
hhh
thh
hth
tth
hht
tht
htt
ttt
3 contain tt so you have a 3/8 chance of making 4 tosses
hhhh
thhh
hthh
tthh
hhth
thth
htth
ttth
hhht
thht
htht
ttht
hhtt
thtt
httt
tttt
8/16
================
hh
th
ht
tt
third coin
hhh
thh
hth
hht
tht
htt
so 5/6 chance of surviving, leaving
hhhh
thhh
hthh
hhth
thth
hhht
thht
htht
hhtt
thtt
again 8/10 chance of surviving!
hhhhh
thhhh
hthhh
hhthh
ththh
hhhth
thhth
hthth
hhhht
thhht
hthht
hhtht
ththt
hhhtt
thhtt
hthtt
13/16
hhhhhh
thhhhh
hthhhh
hhthhh
ththhh
hhhthh
thhthh
hththh
hhhhth
thhhth
hthhth
hhthth
ththth
hhhhht
thhhht
hthhht
hhthht
ththht
hhhtht
thhtht
hththt
hhhhtt
thhhtt
hthhtt
hhthtt
ththtt
21/ 26
So let me recap to get a formula
5/6
8/10
13/16
21/ 26
Now I am in big rouble if I have screwed this up but.........
we have the terms x,y in the form x/y
to get the next y we go....y=2x' ( x'= previous x)
the next x is harder, I will go back a bit, we had 26 terms, 13 ended in t,
we removed 5 of them, leaving 8 ending in t. (out of 21)
We will double that 21 (add h to one half and t to the other)
that gives 42 terms, 8 of which will have tt, 428= 34
34/42
Adding that to the sequence
5/6
8/10
13/16
21/ 26
34/42
3 5 8 13
OK I think we know that sequnce add the preceeding numbers to get the next
3+5=8, 5+8=13 etc....so next term is 8+13=21!
34x2= 68, 13+21 =34
so next term = 47/68
5/6
8/10
13/16
21/ 26
34/42
47/68
WEll I hope you can see a pattern emerging (cos I'm f***ed if I can (lol).
I will stick the first term in at the start which is 2/3? or is it 3/4? lol
5/6
8/10
13/16
21/ 26
34/42
55/68
OK I wil come clean with you I am getting a bit tired and concused and error prone,
but there is a formula!!
I will stick my neck out and add a few more terms...
5/6
8/10
13/16
21/ 26
34/42
55/68
89/110
144/178
233/288
thats it the next 'x' is the sum of the previous 2 x's and the next y is 2x
So for the a sequence of 2 its 3/4
for a sequence of 3 its 3/4 x 5/6
for a sequence of 4 its 3/4 x 5/6 x 8/10
for a sequence of 5 its 3/4 x 5/6 x 8/10 x 13/16
etc.....
I may have made an error I admit but I am too tired to check now
But I am sure there is a fairly simple forumula similar to the above.
wow thanks a lot wasnt expecting this have to read it deeply
edit:
so i see chances to win at this game are pretty high, yeah should have supposed it,thanks ill keep taking a look at what you did
Originally Posted by luxtpm
It's not straight forward probability IMO because, it depends on the last coin you tossed as well as the first.
Anyway so I look at all the possibilities, there are four for the firsts two coins.
hh th ht and tt.
You eleiminate the tt and are left with 3 you have left.
These can be followed by a head or a tail, giving you 6 possibilites, you again eleimate the lines with tt and repeat.
Actually I think I may have gone wrong. I am not sure if I should eliminate
the ones which end in tt or not!!
in which case it would be
3/4
6/8
12/16
24/32
this seem more 'right'
So I think your chances would be 3 to the power of a trillion divided by 4 to the power of a trillion.
Yes I think that is correct.
So to do 10 coins (11 including the first coin) you have a 5.6% chance of doing it.
Yes i think that's right,what I did earlier seems like a load a garbage.
You have a 1 in 4 chance of spinning tt, it does not matter what the previous coin was it could have been a head or a tail in equal likely hood.
So it if was a head which will happen 50% of the time, you then have no chance of tt, but the other 50% of the time you will have a tail giving a 50% chance of another tail, 25% in total.
Well OK I got it wrong at first but it was a good exercise doing it I think!!
Acually it seems I may have been right the first time judging by the results of
a computer simulation I did.
I did a test on 30 spins and it came up with about 7961 to get a run of 30.
The computer results were
run 29 3524578 / 4356618 1.236068 r2 492.
run 30 5702887 / 7049156 1.236068 r2 609.
run 31 9227465 / 11405774 1.236068 r2 753. <
run 32 14930352 / 18454930 1.236068 r2 930.
That 753 is in the right 'ball park'.
I may have made a few 'fiddly errors' I will come back to this later
when I have sorted them out.
Seem I was right the first time (my computer simulation may have errors)
anyway the answer is also the (n+2th) Fibonacci number, divided by 2^n.
I seem to have arrived at that by a different method, but not realised it.
Anyway it seem another way of writing what I got was
Code:1 x F1 x F2 x F3 x F4 ...F(n1) X F(n)  2F1 x 2F2 x 2F3 x 2F4 ...2F(n1)
Then all the F1 F2 F3 all cancel apart from the last, leaving F(n)/2^n
Where 'F' is the Fibonacci number for n.
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