20030906, 15:44  #1 
Sep 2003
2 Posts 
Infinite CoinToss?
Hello everyone! :)
I have a probability problem which i found very amusing to solve.. You have a game that consists of tossing a coin which can land with "heads" up with probability (P) and land with "tails" up with probability (1P). The game stops whenever the amount of "heads" equals the amount of "tails". Except for when both are zero. What is the probability of the game ending ever, given the probability P? Happy solving! 
20030916, 18:17  #2 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
I got as far as this sum:
Sum{n=1:infinity} p^{n}(1p)^{n}(2n)!/(n!)^{2} But I haven't spent too much time trying to evaluate it 
20030916, 22:51  #3 
"William"
May 2003
New Haven
3×787 Posts 
Keep score as "net heads", so every head scores +1 and every tail scores 1. Suppose the first toss is a head. What is the probability that your score will ever get down one level from here? Well, you might get down one level immediately by throwing a tail (probability 1p). But if you throw a head, ...
Saying any more gives it away completely. 
20030917, 12:35  #4  
Sep 2002
Vienna, Austria
3·73 Posts 
Quote:
Code:
(1/12p)1 

20030917, 16:36  #5 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
cool.
but what about when it's just a regular 50/50 coin? 
20030917, 21:24  #6 
Feb 2003
174_{10} Posts 
Travis 
I don't think the combinatorics are correct. You need to exclude the permutations which would result in equal heads/tails at a lower value of n. For n = 2, HTTH would not be valid because the game would have stopped at HT Unless I misunderstood the problem... 
20030917, 23:00  #7 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Curses! you're right .. arg.

20030917, 23:02  #8  
"William"
May 2003
New Haven
3·787 Posts 
Quote:


20030918, 00:36  #9 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
Let me suggest a slightly different way to count.
If the # Heads equal # Tails, the the total number of tosses is always even. So lets count pairs of tosses. If we count each head as +1/2 and each tail as 1/2, the outcome of a pair of tosses is either +1 (with probability p^2), 0 (with probability 2*p*(1p)), or 1 (with probability (1p)^2) Now look at the cumulative count after N pairs of tosses and call it V(N). Also define P(V) to be the probability of termination at V=0, from some initial point V. Clearly, V(N) = 0 and we are looking for P(0). More hints later ... 
20030918, 16:57  #10 
Sep 2003
2 Posts 
Im glad finally some people found my problem interesting
Here is a little hint Try to look at the problem after 2, 4, 6, 8.. tosses. Can you see any interesting relations between ((2n)!)/((n!)^2) (n = number of pair of tosses) and amount of combinations that lead to a termination of the game that do NOT include a shorter string of combinations that would end the game. Analyze both of these combinations for each step and try to find a relation. PS. It can be good to write down the combinations for each step (HH, HT, TH, TT), (HHHH, HHHT...). to get a good visualization of the problem. DS 
20030918, 21:01  #11 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
okay, I've done that, maybe too hastily .. now I get the sum
Sum{n=1:infinity} p^{n}(1p)^{n}(2n)!/(n!^{2}(2n1)) 
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