Author Message
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. Jim Anderson (thejim2020)
Guest Posted: Thu Apr 21, 2005 6:00 am    Post subject: Re: OT: math question Yes to both. 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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Posted using RecPoker.com v2.2 - http://www.recpoker.com 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. _______________________________________________________________ Block Lists, Favorites, and more - http://www.recpoker.com 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 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 one-dimensional 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 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 well-known 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). 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. 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? 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 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/  Jim Anderson (thejim2020)
Guest Posted: Thu Apr 21, 2005 6:01 pm    Post subject: Re: OT: math question Thanks to everyone that responded. 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

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

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 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 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. --- New April Episode - Watch Now at http://PokerUpdates.com --- Posted via http://LiveActionPoker.com --- Where to play Online Poker: http://www.pokerupdates.com/html/op.html Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First 