0

# Probability - Gambler A and B

Fri 31 Jul, 2015 04:39 pm
Gambler A plays a game with Gambler B, in which A and B bet exactly one dollar in each play. The game ends when one of players loses all his money. We assume the probability of A winning any betting is 2/3.
i) Assume that gambler A starts wtih one dollar and B with two dollars. What is the probability of A losing?
ii) Assume that gambler A starts wtih one dollar and B with infinity money. What is the chance of A losing?
iii) Assume that gambler A starts with m dollars and B with n dollars. What is the chance of A winning?
• Topic Stats
• Top Replies
Type: Question • Score: 0 • Views: 617 • Replies: 9
No top replies

markr

1
Fri 31 Jul, 2015 09:56 pm
@pp99,
The first one is easy. There are four states (0,3), (1,2), (2,1), and (3,0). The first and last are terminal states, and the game starts in state (1,2). The probability of going from (1,2) to (0,3) is 1/3. The probability of going from (1,2) to (2,1) to (1,2) is 2/3 * 1/3 = 2/9. Therefore, the probability of A losing is 1/3 + 1/3*2/9 + 1/3*2/9*2/9 + ...

This is 1/3 * (1 + 2/9 + (2/9)^2 + (2/9)^3 + ...), which is 1/3 times the sum of a geometric series with r=2/9.
Therefore the probability is 1/3 * 1/(1-2/9) = 1/3 * 1/(7/9) = 1/3 * 9/7 = 3/7.
markr

1
Fri 31 Jul, 2015 10:03 pm
@markr,
The second one would seem to be 1 since B has an infinite amount of money and can't get to 0. No matter how much money A has, there is a non-zero probability that consecutive losses will wipe him out.
markr

1
Fri 31 Jul, 2015 10:44 pm
@markr,
The previous post is incorrect. When starting with (1,N), the probability of A losing is (2^N - 1) / (2^(N+1) - 1) which approaches 1/2 as N approaches infinity. Therefore, the second answer is 1/2.
markr

1
Fri 31 Jul, 2015 11:14 pm
@pp99,
For the third one, starting with (m,n), the probability of A losing is:
(2^n - 1) / (2^(m+n) - 1)

I don't have proofs for 2 and 3, but after modeling with Markov chains, it's pretty clear that the answers are correct.
0 Replies

Herald

1
Sat 1 Aug, 2015 02:55 am
@markr,
markr wrote:
The previous post is incorrect. When starting with (1,N), the probability of A losing is (2^N - 1) / (2^(N+1) - 1) which approaches 1/2 as N approaches infinity. Therefore, the second answer is 1/2.
I am not sure that it is 1/2. The probability of A to lose on the first round is 1/2. The probability to win is also 1/2, thus becoming with 2 USD. The probability to loose them both in the next two rounds is 1/2*1/2=1/4 and the probability to have that outcome is 1/2(win)*1/2(lose)*1/2(lose)=1/8, from where follows that the probability to lose up to the 3rd round is 1/2 + 1/8. As you are so knowledgeable what is the probability of A to lose everything after two successive wins in the beginning (1/2*1/2)? IMV it must be 1/4*1/8, and the probability to A to become absolute loser at any time is 1/2 + 1/2.4 + 1/4.8 + 1/8.16 + 1/16.32 + ...
markr

1
Sat 1 Aug, 2015 11:04 am
@Herald,
You're overlooking the fact that the probability of A winning any bet is 2/3 - not 1/2.
Herald

1
Sat 1 Aug, 2015 11:27 am
@markr,
markr wrote:
You're overlooking the fact that the probability of A winning any bet is 2/3 - not 1/2.
We are talking about case ii), aren't we. You don't have 'any bet' - A is putting 1 USD on the table and B is putting 1 USD on the table - gambling (toss a coin or throw dice or s.th.) - and the outcome is either A wins 1 USD, or B wins 1USD - where do you see three outcomes here?
markr

1
Sat 1 Aug, 2015 05:12 pm
@Herald,
I can think of one obvious way to interpret this sentence:
"We assume the probability of A winning any betting is 2/3."

I don't know how you're coming up with something other than that. There are two outcomes: A wins or B wins. The probability that A wins is 2/3.

If my interpretation of that sentence is incorrect, what interpretation do you propose?
Herald

1
Sat 1 Aug, 2015 10:04 pm
@markr,
markr wrote:
"We assume the probability of A winning any betting is 2/3."
O.K., it is in the assignment. Then let's see what we have after Awins on the 1st round, and then follows the worst case scenario ending up with losing everything: 1rst round - Awin = 2/3; Alose = 1/3; 2nd - Alose = 2/3.1/3; 3rd round: Alose=2/3.1/3.1/3. It must be something of the kind: 1/3 + 2/27 + 4/243 ... and further the polynomial starts acquiring negligibly small members of the series.
0 Replies

### Related Topics

Probability of 11 cards - Question by ptkaisen
Probability - Question by sean2
Exercise in Probability!!! - Question by evinda
Winning chance - Question by ghodhereek
Family Wine Tasting - Question by mrmac
what is the probability of... #4 - Question by w0lfshad3

1. Forums
2. » Probability - Gambler A and B