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Jim Anderson (thejim2020) Guest

Posted: Thu Apr 21, 2005 6:00 am Post subject: OT: math question 


I have $10. My buddy has $4. We're going to draw high card for $1
until one of us is broke. What are my odds of winning? What's the EV
for each of us?
Also, how much does my advantage go up if the stakes go down (.50, .25,
et cetera)?
Additionally, please explain your answer to me so that I can learn how
you came to your conclusion.
Thanks in advance for those of you who respond. I appreciate your time
and effort. 

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Jim Anderson (thejim2020) Guest

Posted: Thu Apr 21, 2005 6:00 am Post subject: Re: OT: math question 


Yes to both. 

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Kincaid Guest

Posted: Thu Apr 21, 2005 6:00 am Post subject: Re: OT: math question 


Are you going to reshuffle the deck after each draw? And are you going to take
turns drawing first?
On Apr 20 2005 10:02 PM, Jim Anderson (thejim2020) wrote:
Quote:  I have $10. My buddy has $4. We're going to draw high card for $1
until one of us is broke. What are my odds of winning? What's the EV
for each of us?
Also, how much does my advantage go up if the stakes go down (.50, .25,
et cetera)?
Additionally, please explain your answer to me so that I can learn how
you came to your conclusion.
Thanks in advance for those of you who respond. I appreciate your time
and effort.

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KookieMonstr Guest

Posted: Thu Apr 21, 2005 7:00 am Post subject: Re: OT: math question 


The equation you want is x*0.5^n where x is the value bet ($1) and n is the
number of dollars you have (4 or 10 respectively).
Your buddy's chance of going broke when he holds $1 is: (1)(0.5)^1 = 0.50 or
50%. Chance of going broke becuase of 4 consecutive losses ($4): (1)(0.5)^4 =
0.125 or 12.5%.
Both of your expectation values are 0 becasue you are taking 50/50 shots, but
YOU have a significantly higher chance of outlasting him becasue of your deeper
stack.
KookieMonstr.
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Guest

Posted: Thu Apr 21, 2005 7:01 am Post subject: Re: OT: math question 


You could just say you are holding 10 Tournament Chips, and he has 4
Tournament Chips, and you are both risking 1 tournament chip per hand.
Your chance of winning is equal to your proportion of chips. So you
will win 10 times out of 14 and lose the other 4.
Your question about EV is interesting. The way you phrase the
question, you both have 0 EV. He will go broke 10 times in 14, but
will only lose $4 each time. You will go broke less often, but will
lose $10 each time. Neither one of you has an advantage.
Now if you meant you both started with $7, and you've taken the lead at
104, then you have a postitive EV.
Playing Heads up, the "stakes" don't matter. The EV would be the same
whether you "ante" $1, $0.25, or $20 (essentially meaning one or both
of you are allin every hand). Your win percentage will remain the
same (10/14). If you were playing with three or more people, this will
not hold. So if you have $10, he has $4, and I have $20, our chance of
winning the freezeout will still be our proportion of the chips no
matter how large the ante is, but the probabilities of finishing second
and third will change as you change the "ante." 

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Randy Hudson Guest

Posted: Thu Apr 21, 2005 7:38 am Post subject: Re: OT: math question 


In article <[email protected]>,
Jim Anderson (thejim2020) <[email protected]> wrote:
Quote:  I have $10. My buddy has $4. We're going to draw high card for $1
until one of us is broke. What are my odds of winning? What's the EV
for each of us?

This is mathematically modeled as onedimensional Brownian motion, or
"random walk." With no advantage for either side, your EV is zero each;
since you lose $10 when he wins, and win only $4 when you win, you must win
2.5 times as often as he.

Randy Hudson 

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Stephen Jacobs Guest

Posted: Thu Apr 21, 2005 3:00 pm Post subject: Re: math question 


"Jim Anderson (thejim2020)" <[email protected]> wrote in message
news:[email protected]...
Quote:  I have $10. My buddy has $4. We're going to draw high card for $1
until one of us is broke. What are my odds of winning? What's the EV
for each of us?

Others have answered the question. I'd just like to mention that the
modified version: "We're going to do this until one of us goes broke or
something happens to distract us" is a big money maker for the casinos. It
takes a lot longer for the short stack to bust the big one than vice versa.
So in addition to the wellknown fact that house games favor the house a bit
(the probability distribution of the results is shifted in their favor), the
disparity in resources between the typical player and the house makes it
impractical for the typical player to win amounts that really hurt the house
(the tail of the results distribution that is bad for them is cut off). 

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allfyre Guest

Posted: Thu Apr 21, 2005 5:02 pm Post subject: Re: OT: math question 


Mathematically you're dead even. He's going to go broke more often
than you will, but when you go broke you lose more money. From a
gamblers perspective this isn't necessarily a bad proposition, it's
just a waste of time. 

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Jim Anderson (thejim2020) Guest

Posted: Thu Apr 21, 2005 5:02 pm Post subject: Re: math question 


Is the win percentage really the same as the money ratio? I have a 10
in 14 chance? 

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Jim Anderson (thejim2020) Guest

Posted: Thu Apr 21, 2005 6:00 pm Post subject: Re: math question 


But if we're always betting in increments of 1 instead of whatever I
have in front of me, it seems to me the equation should be more complex
than this.
With so many people (whom I respect) saying it's a 0 EV play, I believe
and accept you're right. I just like thinking about it.
What if I somehow had a slight advantage (say we're rolling a 10sided
die and he gives me 6 sides to his four)? Then what's the equation to
figure out my chance at busting him? 

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Guest

Posted: Thu Apr 21, 2005 6:00 pm Post subject: Re: math question 


Here's one way to think about it.
I have 7 chips, you have 1 chip, and it's a fair game where we ante one
chip at a time. You have a 50% chance of doubling up to make the score
62.
Now that it's 62, you need to win twice before I win twice (and break
you). But it's a fair game, so again there's a 50% chance of you
doubling up to make the score 44.
Now that's it 44, it makes sense that you have a 50% chance of
winning.
So with 1 chip out of 8, your chance of winning is:
(1/2)*(1/2)*(1/2) = (1/ 

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Jim Anderson (thejim2020) Guest

Posted: Thu Apr 21, 2005 6:01 pm Post subject: Re: OT: math question 


Thanks to everyone that responded. 

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jeff Guest

Posted: Thu Apr 21, 2005 7:01 pm Post subject: Re: OT: math question 


Jim Anderson (thejim2020) wrote:
Quote:  I have $10. My buddy has $4. We're going to draw high card for $1
until one of us is broke. What are my odds of winning? What's the
EV
for each of us?
Also, how much does my advantage go up if the stakes go down (.50,
..25,
et cetera)?
Additionally, please explain your answer to me so that I can learn
how
you came to your conclusion.
Thanks in advance for those of you who respond. I appreciate your
time
and effort.

Jim,
First some notation. Let P(i) mean the probability of winning given
that you currently have i dollars. So, for example, P(0) = 0 and P(14)
= 1 since the game stops when you have either all the money or lost all
your money.
I will assume at each draw of the card you have a 0.5 probability of
winning; i.e., even chance of winning or losing. Under that
assumption, if you have i dollars, where 0 < i < 14, after a draw of
the card there is a 0.5 probability you will have i  1 dollars and a
0.5 probability that you will have i + 1 dollars. Note that this
assumtion means that P(7) = 0.5, since if you and your opponent each
have 7 dollars then (by symmetry) each of you has an equal chance of
ending up with all the money.
Now:
P(1) = 0.5*P(0) + 0.5*P(2) = 0.5*P(2), since if you start with 1 dollar
after a draw of the cards it is equally likely that you end up with 0
(in which case you lose) or 2 dollars. This gives us the relationship
P(2) = 2*P(1); i.e., if you start out with 2 dollars you are twice as
likely to win as when you start with one dollar
Now, proceed for the other probabilities:
P(2) = 0.5*P(1) + 0.5*P(3), or
2*P(1) = 0.5*P(1) + 0.5*P(3), which yields
P(3) = 3*P(1)
The claim is that, in general, P(i) = i*P(1), which if true will give
us the answers we want. Since P(14) = 1, P(14) = 14*P(1) would mean
that P(1) = 1/14, and in general P(i) = i/14. So, for example, if you
started with 10 then you would have a 10/14 probability of winning all
the money.
I will prove the claim using induction. Suppose that the claim is true
for all i <= j. In other words, if i <= j then P(i) = i*P(1). Then
P(j) = 0.5*P(j  1) + 0.5*P(j + 1) = 0.5*(j 1)*P(1) + 0.5*P(j + 1).
By the induction hypothesis we also have P(j) = j*P(1), so substituting
this into the left hand side of the previous expression gives us
j*P(1) = 0.5*(j  1)P(1) + 0.5*P(j + 1), and solving for P(j + 1)
yields
P(j + 1) = (j + 1)*P(1)
Make sense?
Jeff 

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wamplerr Guest

Posted: Thu Apr 21, 2005 9:00 pm Post subject: Re: math question 


"What if I somehow had a slight advantage (say we're rolling a 10sided
die and he gives me 6 sides to his four)? Then what's the equation to
figure out my chance at busting him?"
In this case, a computer simulation is probably the easiest way out.
You might also do a google search for the "risk of ruin" to see similar
problems.

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