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Jim Anderson (thejim2020) · Guest

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 Anderson (thejim2020) · Guest

Yes to both.

Kincaid · Guest

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:

KookieMonstr · Guest

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
10-4, 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 all-in 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."

Randy Hudson · Guest

In article <[email protected]>,
Jim Anderson (thejim2020) <[email protected]> wrote:

Stephen Jacobs · Guest

"Jim Anderson (thejim2020)" <[email protected]> wrote in message
news:[email protected]...

allfyre · Guest

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.

Jim Anderson (thejim2020) · Guest

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

Jim Anderson (thejim2020) · Guest

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 10-sided
die and he gives me 6 sides to his four)? Then what's the equation to
figure out my chance at busting him?

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
6-2.

Now that it's 6-2, 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 4-4.

Now that's it 4-4, 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/ Cool

Jim Anderson (thejim2020) · Guest

Thanks to everyone that responded.

jeff · Guest

Jim Anderson (thejim2020) wrote:

wamplerr · Guest

"What if I somehow had a slight advantage (say we're rolling a 10-sided
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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