Thread: chances to last in this tossing the coin 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?

Originally Posted by luxtpm

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 )

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, 42-8= 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

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

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 796-1 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(n-1) X F(n)
-----------------------------------------------
      2F1 x 2F2 x 2F3 x 2F4 ...2F(n-1)

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.