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        riddles >> medium >> Game probability.
        
(Message started by: WhattheWhat on Mar 24^th, 2009, 8:52am)

Title: Game probability.
Post by WhattheWhat on Mar 24^th, 2009, 8:52am

I don't know the answer to this one, I'd appreciate a guide on how to calculate it.

John and Tony play a game. (The actual game mechanics are irrelevant). The outcome of this game works out to a .653 chance that John will give Tony $1 and a .347 chance that Tony will give John $2. If after a few hands John is up $10 what is the probability that Tony will never ever break even, assuming both men have infinite resources and patience?

Title: Re: Game probability.
Post by towr on Mar 24^th, 2009, 9:49am

You can use a recursive function. At each step you're get either closer to breaking even, or further off.

Suppose P(D) is the probability of breaking even if John is D dollars up on Tony.
Then we have
P(D) = .653 P(D-2) + .347 P(D+4) [edited]
a .653 probability that the difference decreases, and a .347 probability it increases.
1- P(10) is the probability tony never breaks even.

The base case is P(0) = 1, and now the trick is to solve the recursive formula.

Title: Re: Game probability.
Post by WhattheWhat on Mar 24^th, 2009, 10:00am

on 03/24/09 at 09:49:23, towr wrote:

You can use a recursive function. At each step you're get either closer to breaking even, or further off.

Suppose P(D) is the probability of breaking even if John is D dollars up on Tony.
Then we have
P(D) = .653 P(D-1) + .347 P(D+2)
a .653 probability that the difference decreases, and a .347 probability it increases.
1- P(10) is the probability tony never breaks even.

The base case is P(0) = 1, and now the trick is to solve the recursive formula.

I think I chose an appropriate username....

Title: Re: Game probability.
Post by towr on Mar 24^th, 2009, 10:14am

on 03/24/09 at 10:00:12, WhattheWhat wrote:

I think I chose an appropriate username....

Ok, let's try it with an example.
Say John has $110 and Tony has $100.
Then after the next game, there is a probability of .653 That Tony won, and so the score will be $109 vs $101, but there is also a chance of .347 that John won and the score is $112 vs $98
In the first case Tony is closer to breaking even, in the second he is further off.

We can repeat these steps from both cases.
case1: .653 chance of $109 vs $101 --> .653*.653 chance of $108 vs $102 and .653*.347 chance of $111 vs $99
case2: .347 chance of $112 vs $98 --> .347*.653 chance of $111 vs $99 and .347*.347 chance of $114 vs $96
etc.

Title: Re: Game probability.
Post by WhattheWhat on Mar 24^th, 2009, 11:28am

If I understand you, you're saying that the probability is effectvely the same as

(probability Tony will ever get $1 ahead)^10

which makes sense and seems simpler. I'm not quite understanding how to find that probability though.

Title: Re: Game probability.
Post by towr on Mar 24^th, 2009, 12:14pm

That's good thinking.
Bear in mind though, if John gives Tony a dollar, the difference in their totals changes by 2. Unless they don't count the infinite supply of money they pay the other from(?) Not that it makes much of a difference for the process.

We have
P(2N) = .653 P(2N-2) + .347 P(2N+4)
P(0) = 1
And, P(2N) = P(2)^N

So, P(2) = .653 P(0) + .347 P(6) = .653 + .347 P(2)³
Which we can solve (http://72.3.253.76:8080/webMathematica3/quickmath/page.jsp?s1=equations&s2=solve&s3=basic) for P(2)
It has 3 possible solutions,
~ -1.96, which can't be a probability,
~ 0.96,
and 1
So it's one of the latter two. I'm not quite sure how to decide between the two. But there are people on this forum that are much better at these things than me, so I'm sure it'll get resolved sooner or later.

[edit]A quick simulation seems to suggest it is the ~0.96. Although a more rigorous approach would still be nice.
So we get P(10) = P(2)⁵ ~= 0.8157
[/edit]

Title: Re: Game probability.
Post by WhattheWhat on Mar 24^th, 2009, 12:52pm

Ah, I think I've just noticed a miscommunication. Tony doesn't care about John's total. In reality "Tony" is me. (My name's not Tony.) "John" is an online casino. I lost £10 playing a game, and was just idly curious as to what the chances of me recovering that money on a "do or die" raid would be. (I don't intend to do this in case it is not clear.)

Title: Re: Game probability.
Post by towr on Mar 24^th, 2009, 1:01pm

Ah, well, for the process it doesn't make a lot of difference.

P(N) = .653 P(N-1) + .347 P(N+2)
P(0) = 1
And, P(N) = P(1)^N

So, P(1) = .653 P(0) + .347 P(3) = .653 + .347 P(1)³
Same solutions as before, P(1) ~= 0.96
giving P(10) = P(1)¹⁰ ~= 0.665

Title: Re: Game probability.
Post by WhattheWhat on Mar 24^th, 2009, 1:13pm

Thanks.

(I don't follow your process but I trust your answer. ;D)

Title: Re: Game probability.
Post by towr on Mar 24^th, 2009, 1:16pm

In case you're interested, risking another $10 gives you a 0.386 chance of breaking even, but also 0.406 chance at losing an additional 9$ and 0.208 chance at losing an additional $10.

All in all though, $10 is not that much to learn not to gamble ;)