Probability/deal problem

[David Zanetti]

This is an old problem which took over a century to solve: many claimed
success before Blaise Pascal finally cracked it. In a match to sixty
points, with each game worth one point, play has to cease when the scores
are 50 to 30: what is the correct division of the prize money?

_________________________________________________________________
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[Dubious]

David Zanetti wrote:
> This is an old problem which took over a century to solve: many claimed
> success before Blaise Pascal finally cracked it. In a match to sixty
> points, with each game worth one point, play has to cease when the scores
> are 50 to 30: what is the correct division of the prize money?
>

OK, here's my take on it. We can assume that player 1 has a 50/80 chance
(.625) of winning each game and want to figure out the chance that he
will win at least 10 of the next 39 games (because after 38 games with
no winner, the score will be 59 to 59 and the next game will win it).

So, the probability he will win k games (10)in n attempts (39) with a
probability of p per attempt (.625) is
P(k)=n-choose-k*(p^k)*[(1-p)^(n-k)]. I must admit, I don't know an
algebraic formula for an "at least" answer and I'm too lazy to find one
so I'll just do this for all values less than k (chances we wins less
than 10 games) and subtract the sum from 1.

1-[P(1)+P(2)+...+P(9)]= 1-0.00000318=0.99999682 => 99.9997% chance
player 1 will win, so don't argue about fractions of a cent and just
give him the prize money.

On the other hand, if you assume they have the same chance at winning
each game, then p=0.5, same calculations, P(player 1 wins)=0.9983 so he
gets 99.83% (or basically all) of the prize money.

What did Pascal say?

--Jeff

[Hugh Candlin]

David Zanetti <anonymous@yahoo.com> wrote in message news:3f8c09bf$0$69333$9a6e19ea@news.newshosting.com...
> This is an old problem which took over a century to solve: many claimed
> success before Blaise Pascal finally cracked it. In a match to sixty
> points, with each game worth one point, play has to cease when the scores
> are 50 to 30: what is the correct division of the prize money?

============================================
Off the top of my head, in my humble opinion, feel free to disagree.
============================================

A is entitled to 50/80ths of 50/60ths of 100% of the prize money,
since A led B in the ratio of 50 to 30 at the time of play stoppage,
but only attained 50/60th of the final prize goal.

A is also entitled to 50% of the remaining unallocated prize fund,
as there is no reliable predictor of future results.

Therefore, based on performance, A gets
50/80 * 50/60 * 100% = 0.625 * 0.833 * 100% = 0.52 * 100% = 52%

B is entitled to 30/80ths of 30/60ths of 100% of the prize money,
since B trailed A in the ratio of 30 to 50 at the time of play stoppage,
and only attained 30/60th of the final goal.

B is also entitled to 50% of the remaining unallocated prize fund,
as there is no reliable predictor of future results.

Therefore, based on performance, B gets
30/80 * 30/60 * 100% = 0.375 * 0.5 * 100% = 0.1875 * 100% = 18.75%

52% + 18.75% = 70.75% is their total performance-based entitlement,
leaving 29.25% in the unallocated prize fund.

As the remaining games cannot be forecast with 100% accuracy,
I must divide the remainder of the prize money equally,
with each getting 50% of 29.25% = 14.625%

A's prize share total = 52.00% + 14.625% = 66.625%

B's prize share total = 18.75% + 14.625% = 33.375%

The final distribution ratio of approximately 6 to 3 does not mirror
the final score ratio of 5 to 3, but, as we all know,
past performance is not a guaranteed foretoken of future events.

My formula works for any number of contestants.

That was fun.

[Michael Lemkin]

On Oct 14 2003 7:35AM, David Zanetti wrote:

> This is an old problem which took over a century to solve: many claimed
> success before Blaise Pascal finally cracked it. In a match to sixty
> points, with each game worth one point, play has to cease when the scores
> are 50 to 30: what is the correct division of the prize money?

Well here is what you need to know. Was the 50-30 score the result of
superior play or was it simply luck. I believe in the original example we
have either a complete chance mechanism or a game where two players are
identically skilled.
Clearly if the lopsided score is a result of one player having superior
skills he should probably claim almost all the money - if not all.
However, let us assume that the two players are equally skilled and the
score is merely a random event. We know that after 39 events the game
would be over- either the 50 would get his 10pts or the 30 would get his
30. Taking all possible combinations of these 39events we would then find
the number where player B scored thirty points before Player A score 10.
Since the total is a huge number(2 to the 39th power), why not scale it
back to 6, 10, 12 - Then we have 2 to the fifth possible combos(32)- THIS
IS MORE MANAGEABLE.
tHEN THERE ARE ONLY LLLLW,LLLLL,LLWLL,WLLLL,LWLLLL,LLLWL OR SIX WINNING
POSSIBILITIES FOR THE LOWER PLAYER . Following this logic, the lower
player may be entitled to 6/32 of the prize pool or 18.75% -the higher
81.25
This should be roughly the same as the 30 50 60 problem.

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[Michael Lemkin]

On Oct 14 2003 6:01AM, Michael Lemkin wrote:

> On Oct 14 2003 7:35AM, David Zanetti wrote:
>
> > This is an old problem which took over a century to solve: many claimed
> > success before Blaise Pascal finally cracked it. In a match to sixty
> > points, with each game worth one point, play has to cease when the scores
> > are 50 to 30: what is the correct division of the prize money?
>
> Well here is what you need to know. Was the 50-30 score the result of
> superior play or was it simply luck. I believe in the original example we
> have either a complete chance mechanism or a game where two players are
> identically skilled.
> Clearly if the lopsided score is a result of one player having superior
> skills he should probably claim almost all the money - if not all.
> However, let us assume that the two players are equally skilled and the
> score is merely a random event. We know that after 39 events the game
> would be over- either the 50 would get his 10pts or the 30 would get his
> 30. Taking all possible combinations of these 39events we would then find
> the number where player B scored thirty points before Player A score 10.
> Since the total is a huge number(2 to the 39th power), why not scale it
> back to 6, 10, 12 - Then we have 2 to the fifth possible combos(32)- THIS
> IS MORE MANAGEABLE. OOPS!!! I got confused here - it would actually be 2 to
the seventh power or 128 possibilities.
We find all combinations where Bgets six before A gets 2 - they are
WWWWWWW,WWWWWWL,WWWWWLW,wwwwlww,wwlwwww,WWWLWWW,WLWWWWW,LWWWWWW -this is
8/128 so in this situation player be should get 6.25, player a 93.75
SORRY!

> tHEN THERE ARE ONLY LLLLW,LLLLL,LLWLL,WLLLL,LWLLLL,LLLWL OR SIX WINNING
> POSSIBILITIES FOR THE LOWER PLAYER . Following this logic, the lower
> player may be entitled to 6/32 of the prize pool or 18.75% -the higher
> 81.25
> This should be roughly the same as the 30 50 60 problem.

_________________________________________________________________
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[Kevin Cline]

"David Zanetti" <anonymous@yahoo.com> wrote in message news:<3f8c09bf$0$69333$9a6e19ea@news.newshosting.com>...
> This is an old problem which took over a century to solve: many claimed
> success before Blaise Pascal finally cracked it. In a match to sixty
> points, with each game worth one point, play has to cease when the scores
> are 50 to 30: what is the correct division of the prize money?

Assuming that player A with 50 points has a probability p of winning
each point, we can calculate the probability that A reaches 60 first
thusly:

Let f(m,n) be the probability that A wins when A is m points away
from victory and B is n points away.

Then:
f(0,n) = 1.0
f(n,0) = 0.0

f(m,n) = p * f(m-1,n) + (1-p) * f(m,n-1)

With p = 0.5, m = 10, n = 30, A's probability of winning is 99.95%.
Of course the fact that the score has reached 50-30 is a good
indication
that p is somewhat greater than 0.5.

#!perl -w
use strict;

my ($m,$n,$p) = @ARGV;
my $q = 1 - $p;

my @F;

foreach my $j (1..$n) {
$F[0][$j] = 1;
}

foreach my $i (1..$m) {
$F[$i][0] = 0;
}

foreach my $i (1..$m) {
foreach my $j (1..$n) {
$F[$i][$j] = $p * $F[$i-1][$j] + $q * $F[$i][$j-1];
}
}

print $F[$m][$n], "\n";

[stacy_friedman]

"David Zanetti" <anonymous@yahoo.com> wrote:
> This is an old problem which took over a century to solve: many claimed
> success before Blaise Pascal finally cracked it. In a match to sixty
> points, with each game worth one point, play has to cease when the scores
> are 50 to 30: what is the correct division of the prize money?

To be equitable, the prize money should be divided based on each
player's probability of winning. I assume that this is a fair game.

There can be no more than 39 plays remaining. There are 2^39 ways for
these plays to occur, and because this is a fair game these 2^39 are
equiprobable. Of those arrangements, if Player 1 wins 10 or more, he
wins. If Player 2 wins 30 or more, she wins.

Using the combinations function and the notation C(x,y) the
probability of Player 1 winning is therefore
p(player 1 wins) = SUM (C(39, i)), i = 10..39 / 2^39
= 99.9467%.
p(player 2 wins) = SUM (C(39, i)), i = 30..39 / 2^39
= 0.0533%.

Player 1 therefore gets 99.9467% of the money.