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Old 2018-07-01, 09:09   #12
Citrix's Avatar
Jun 2003

5·307 Posts

Originally Posted by akea View Post
Can u help me with thi quesiton pls

Assume that a gambler plays a fair game where he can win or lose 1 dollar in each round . His initialstock is 200 dollar. He decides a priory to stop gambling at the moment when he either has 500 dollars or 0 dollars in his stock. Time is counted by the number of rounds played.
i) show that the probability that he will never stops gambling is zero
ii ) Compute the probability that at the time when he stops gambling he has 500 dollars , and theprobability that he has zero dollars

Starting with $200
Assuming that at n turns the game stops then n=x+200 losses and x wins (min n=200)
Or n=x looses and x + 300 wins (min n=300)
Also note that n must be even
Odd of this happening is 1 in n for reaching zero & 1 in n for reaching 500

Probability of loosing=1/200+1/202+....+1/infinity= infinity (divergent series)
Probability of winning=1/300+1/302+....+1/infinity = infinity (divergent series)
Probability of ending the game(winning or loosing)=1/200+1/202+....1/infinity+1/300+1/302+....1/infinity = infinity (divergent series)
Probability of not ending the game=1-infinity=0 (as probability cannot be negative)

Of note, if n was not very large (limited turns e.g. 500 turns) then the series would not diverge.
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Old 2018-07-01, 12:57   #13
wblipp's Avatar
May 2003
New Haven

2×32×131 Posts

Originally Posted by Citrix View Post
Probability of not ending the game=1-infinity=0 (as probability cannot be negative)
Something's wrong - the game will end at least some of the time.

I think the biggest problem is that you forgot to account for when the game ended previously. For example, 700 plays with 200 losses followed by 500 wins shows up as ending on both play 200 and play 700.
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