I found this discussion by William Chen in the rec.gambling site, and some replies from David Skylansky and others. Here's a long excerpt of the discussion, which can also be found
here>
Tournament coin toss question, part deux
From: Bill chen - view profile
Date: Thurs, Jan 3 2002 4:46 am
Email: w...@cyra.com (Bill chen)
Groups: rec.gambling.poker
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dsklan...@aol.com (Dsklansky) wrote in message <news:20011231234324.25701.00004908@mb-fi.aol.com>...
> Neglecting hourly rate considerations, the answer should be pretty close to
> your assessment of your chances of doubling up vesus going broke.
Except that if you are a skilled player, you increase your winning
chances by doubling up immediately rather than later. This is because
you should have some postive expectation for your play. Suppose you
have 2000 and the next round is an hour of 50-100. Then if you are a
1 big bet/hr winner, having 2000 at the beginning of the round should
be (to borrow David's phrase) pretty close to having 2100 at the end
of the round. Also because of the same effect, P(doubling up) is
greater earlier and less later.
This further complicates the question. In David's model if you assume
you have twice the average chance of winning in an 128 player field,
your chances of doubling up are roughly 55% on average. It's a
reasonable assumption to say P(doubling) is 55% if you are an average
stack, less if less than average. So say you are average with 1000 at
the beginning of the 50-100 round, and a genie (or some other plot
device) gives you the chance to double up immediately, what odds do
you need? Well, if you figure it will take you a round on average to
double up, which seems reasonable since the variance per hour round is
10 big bets. If you win you will have 2000, which we have said to be
worth 2100 one round later, as opposed to taking a 55% chance to have
2000 one round later. Hence you should have a 52.5% chance of winning
the coin flip.
Note that you only need a 50.6:49.4 edge to risk half your chips.
Now, many assumptions are made here and many more left out that should
be considered, but I've tried a few different models and I have gotten
similar results. What does this mean to me in practice? Well in
general I think I am a pretty quantative guy, but I can't really say
"well base on my read of the range of hands my opponent(s) have
including implied odds, I think I have a 1% edge here but since this
is a tournament situation I can now fold!" "But wait a minute that
15-30 game there seems juicy. Let me see, based on the players there
my win rate is $30/hr. So like time value converted to tourney chips
is."
Seriously though, don't sweat it too much until it gets close to the
money. At the beginning of a tourney just put in your chips if you
have an edge, fold if it just seems like a toss up. Not bad advice for
a ring game either.
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From: Dsklansky - view profile
Date: Thurs, Jan 3 2002 11:37 am
Email: dsklan...@aol.com (Dsklansky)
Groups: rec.gambling.poker
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>> Neglecting hourly rate considerations, the answer should be pretty close to
>> your assessment of your chances of doubling up vesus going broke.
>Except that if you are a skilled player, you increase your winning
>chances by doubling up immediately rather than later.
Yes. That is why I said pretty close.
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From: jacksup - view profile
Date: Thurs, Jan 3 2002 12:28 pm
Email: mattmat...@hotmail.com (jacksup)
Groups: rec.gambling.poker
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Nice post, Bill. Just one question:
> In David's model if you assume
> you have twice the average chance of winning in an 128 player field,
> your chances of doubling up are roughly 55% on average.
Could you explain how you arrived at this figure? It's not obvious to
me, and the 55% is crucial to your conclusions so I want to be sure to
understand how you got it.
Thanks a lot,
Matt
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From: Bill chen - view profile
Date: Thurs, Jan 3 2002 2:33 pm
Email: w...@cyra.com (Bill chen)
Groups: rec.gambling.poker
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mattmat...@hotmail.com (jacksup) wrote in message <news:473b8d61.0201031128.4236138c@posting.google. com>...
> Nice post, Bill. Just one question:
> > In David's model if you assume
> > you have twice the average chance of winning in an 128 player field,
> > your chances of doubling up are roughly 55% on average.
> Could you explain how you arrived at this figure? It's not obvious to
> me, and the 55% is crucial to your conclusions so I want to be sure to
> understand how you got it.
> Thanks a lot,
> Matt
Well you need to double up 7 times to win. So say your chance of
winning is 1/64 instead of 1/128. LEt x be your chance of doubling
up. Then we have x^7 = 1/64. So we solve for x.
Bill
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Tournament coin toss question
From: NOSPAMdave - view profile
Date: Thurs, Jan 3 2002 8:00 pm
Email: NOSPAMd...@sebastian9.com
Groups: rec.gambling.poker
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According to JP Massar <mas...@alum.mit.edu>:
- Hide quoted text -
- Show quoted text -
> On Sat, 29 Dec 2001 07:33:27 -0800, "bobf"
> <robertfedun...@email.msn.com> wrote:
> >Does anyone know of a rigorous mathematical proof of the commonly-held
> >belief that each tournament chip won is worth slightly less than each one
> >lost? It seems sensible intuitively and qualitatively, but I've never seen
> >it proven quantitatively.
> All the chips, when possessed by the collection of players, are worth
> the prize pool.
> All the chips, when possessed by a single player, are only worth first
> prize, which is usually around 40% of the prize pool.
> Since you start with 1/N of the chips and 1/N of the equity (N = # of
> players entering the tournament), all else being equal, you cannot
> arrive at the limit of all the chips being worth only a fraction of
> the prize pool without the value of additional chips being worth less
> than value of the chips you currently possess.
> QED
But that's not a proof of the question as stated, and certainly not
rigorous. All that you've shown is that the accumulation of *all*
the remaining chips is worth less, on a per-chip basis, than your
original stack (and that only for place-paying tournaments, where
all players are equally skilled). In order to show that *each* chip
won is worth progressively less, you would at least need to show
that the marginal value of additional chips is monotonic with
respect to the number of chips in your stack at all stages of the
tournament, and you haven't come close to showing that yet. For example,
it might well be the case that going from 1/N to 2/N of the chips early
on more than doubles your chance of winning, but the difference between
70% and 90% of the chips is negligible if everyone else is still at 1/N.
Such a situation might arise if players are typically giving up
significant chip EV to avoid the risk of busting out. Then the first
few chips you win early might add significant value in letting you
exploit more marginal situations, but beyond that the utility might drop off.
So it might even be the case that it's right to push marginal opportunities
early if and only if most of your opponents think it's wrong.
--
Dave Wallace (Remove NOSPAM from my address to email me)
It is quite humbling to realize that the storage occupied by the longest
line from a typical Usenet posting is sufficient to provide a state space
so vast that all the computation power in the world can not conquer it.
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Tournament coin toss question, part deux
From: jacksup - view profile
Date: Mon, Jan 7 2002 4:35 pm
Email: mattmat...@hotmail.com (jacksup)
Groups: rec.gambling.poker
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Bill,
Thanks for explaining the 55%. But wait a minute, I have a new
question.
> Well, if you figure it will take you a round on average to
> double up, which seems reasonable since the variance per hour round is
> 10 big bets. If you win you will have 2000, which we have said to be
> worth 2100 one round later, as opposed to taking a 55% chance to have
> 2000 one round later. Hence you should have a 52.5% chance of winning
> the coin flip.
Except that if I lose the coin flip, I'm out of chips. This will
happen 47.5% of the time. In the case where I decline the coin flip,
55% of the time I will have 2000 one round later, but the other 45% of
the time, I will usually still have chips, right? I would think there
are many cases where a good player does not double up, but at least
maintains his chip count. And that has to increase his overall
tournament EV. Therefore I would think you would need significantly
better than a 52.5% chance to take the coin flip.
What do you think? Are there some flaws in this reasoning?
Thanks,
Matt