Bret Maverick's solitaire

[James Dow Allen]

With all the Poker puzzles, let's mention a game shown on the Maverick TV
series about 40 years ago.

Select 25 cards at random from a deck of 52. What is the probability that
you can partition them into five hands of five cards each so that each hand is
a straight or better in Poker?

Bret Maverick was the holdout for acquittal in an 11-1 jury. He convinced
ten jurors of the defendant's innocence and soon it was 1-11. He got the
final juror to bet the man's life on whether Bret could find five "pat
hands" in 25 random cards.

James

[Courtenay Footman]

In article <3c09ba86...@news.mco.bellsouth.net>, Yore Nimlak wrote:

>I don't know if it was on Maverick, but I remember it from Alias Smith and
>Jones, the cowboy show about Kid Curry and Hannibal Hayes. The object was
>to create five "pat hands", which I guess amounts to the same thing as a
>straight or better. It's partly the inspiration for my Five by Five Poker
>game. (http://www.5x5poker.com)

No. A pat hand is one that uses all five cards, and would be broken up
if any where discarded. Thus four of a kind is not a pat hand. This
sucker bet has been around for a long time.

--
Courtenay Footman I have again gotten back on the net, and
c...@lightlink.com again I will never get anything done.
(All mail from non-valid addresses is automatically deleted by my system.)

[James Dreier]

c...@adore.lightlink.com (Courtenay Footman) wrote:

> No. A pat hand is one that uses all five cards, and would be broken up
> if any where discarded. Thus four of a kind is not a pat hand. This
> sucker bet has been around for a long time.

The term "pat hand" can be used either way: as you've defined it, or as
a hand which cannot be improved by drawing to it. In the latter case,
four-of-a-kind is, of course, a pat hand. See

http://www.vegascorner.com/howto/article_listing.cfm?rec_id=1081

This article defines "pat hand" in the second way, and also mentions
both "Maverick" and "Alias Smith and Jones" (and complains about the way
the 25 card bet was used in "Maverick").

-Jamie

[James Dow Allen]

On Thursday, November 29, 2001 at 2:51:06 PM UTC+7, James Dow Allen wrote:
> With all the Poker puzzles, let's mention a game shown on the Maverick TV
> series about 40 years ago.
>
> Select 25 cards at random from a deck of 52. What is the probability that
> you can partition them into five hands of five cards each so that each hand is
> a straight or better in Poker?

There are two threads in rec.puzzles on Maverick's solitaire; the other one is

https://groups.google.com/forum/?hl=en#!searchin/rec.puzzles/maverick$20solitaire/rec.puzzles/9Pk0UVnznes/-pgrq-cjwcoJ

My favorite solitaire game requires that the cards be shuffled well
each time, so I alternate it with a different solitaire to help shuffle.
Lately the solitaire to-help-me-shuffle has been Maverick's solitaire.

Partition the following 25 cards into five pat hands:
KC QC 8C 6C 3C
AD JD TD 9D 7D 5D 4D
QH TH 9H 8H 7H 6H 5H 4H 3H
AS KS 9S 3S

James

[Jonathan Dushoff]

On Wednesday, June 26, 2019 at 7:50:27 AM UTC-4, James Dow Allen wrote:

The problem is ambiguous, since four-of-a-kind is not usually defined as a pat hand, but you sort-of kind-of imply you mean to include it.

Jonathan

[James Dow Allen]

On Thursday, June 27, 2019 at 12:39:48 AM UTC+7, Jonathan Dushoff wrote:
> On Wednesday, June 26, 2019 at 7:50:27 AM UTC-4, James Dow Allen wrote:
>
> The problem is ambiguous, since four-of-a-kind is not usually defined as a pat hand, but you sort-of kind-of imply you mean to include it.
>
> Jonathan

I was quite aware of the ambiguity; I deliberately refrained
from disambiguating because

Gurer ner ab dhnqehcyrgf nzbat gur gjragl-svir pneqf.

[Jonathan Dushoff]

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There are no possible four-of-a-kinds. JD is a the only jack, so it has to be in a straight or a flush.

If JD is in a flush, it "cuts off" the two high spades, since they can't be in any flush and now can't be in a straight. Thus we need two full houses for these two cards. But the king-high house completely cuts off the QC: it can no longer be in a straight or a flush.

So we put JD in a straight. This rules out any high houses and leaves us with a T, Q, K and A that need to be in flushes. Since we can't have a spade flush (only four spades ever) we need KC and AD in flushes (and AS and KS in our high straight).

From there, it's not hard, and there's a pleasant surprise at the end.

TH JD QH KS AS
4D 7D AD 5D TD
KC 3C 6C 8C QC
9S 9H 9D 3S 3H
4H 5H 6H 7H 8H

Jonathan

[James Dow Allen]

On Friday, June 28, 2019 at 3:39:15 AM UTC+7, Jonathan Dushoff wrote:
> On Thursday, November 29, 2001 at 2:51:06 AM UTC-5, James Dow Allen wrote:
> > With all the Poker puzzles, let's mention a game shown on the Maverick TV
> > series about 40 years ago.
>
> > Select 25 cards at random from a deck of 52. What is the probability that
> > you can partition them into five hands of five cards each so that each hand is
> > a straight or better in Poker?

> From there, it's not hard, and there's a pleasant surprise at the end.

Well done, Jonathan!

In case anyone else finds these fun, here are a dozen more
starts for Maverick solitaire, hand-selected to be somewhat
more difficult than typical starts. For the purpose of these
puzzles:
(a) Four-of-a-kind IS considered a compliant "pat hand."
(b) Proper puzzles have a UNIQUE solution, but some of these
will have two very similar solutions.

1. KC JC TC 8C 6C 5C JD TD 9D 8D 7D 5D 4D 2D KH TH 8H 7H 6H 5H 4H AS 9S 5S 4S

2. AC TC 9C 8C 7C 6C 5C 3C 2C AD KD TD 9D 8D 4D 3D AH QH 6H 3H KS TS 5S 4S 3S

3. AC 8C 6C 5C 4C 2C AD QD JD 2D AH 8H 7H 6H 5H 4H 3H 2H KS JS 9S 5S 4S 3S 2S

4. QC TC 9C 8C 6C 5C 3C KD 9D 7D 2D KH QH JH 8H 7H 6H 4H 3H KS QS JS 6S 4S 2S

5. KC TC 6C 5C 3C AD KD 6D 3D KH QH JH 6H 4H 2H AS KS QS JS TS 9S 8S 7S 4S 2S

6. AC QC 7C 5C 4C AD KD 9D 8D 7D 6D 4D 3D 2D QH 9H 8H 4H 2H QS JS 9S 8S 7S 3S

7. AC KC QC TC 7C 4C KD QD JD TD 9D 8D 6D AH KH QH JH 9H 6H 4H 2H KS 6S 3S 2S

8. JC TC 9C 6C 4C AD KD JD TD 9D 8D 7D KH 9H 7H 2H AS QS JS TS 9S 8S 7S 6S 5S

9. AC JC TC 7C 5C 4C 3C AD JD 8D 7D 4D 2D KH JH 8H 2H AS JS TS 9S 7S 6S 3S 2S

10. JC TC 9C 8C 7C 5C 3C KD JD TD 2D AH JH 8H 7H 6H 5H 3H AS JS TS 7S 6S 4S 2S

11. AC KC JC TC 9C 7C 5C 2C AD QD TD 9D 8D 4D AH QH 9H 6H 4H 3H 2H KS 8S 4S 3S

12. KC 8C 7C 4C AD KD QD JD TD 8D 5D 4D 2D AH QH 9H 8H 5H 3H AS 9S 8S 5S 4S 2S


James

[Eric Sosman]

On 6/26/2019 7:50 AM, James Dow Allen wrote:
> On Thursday, November 29, 2001 at 2:51:06 PM UTC+7, James Dow Allen wrote:
>> With all the Poker puzzles, let's mention a game shown on the Maverick TV
>> series about 40 years ago.
>>
>> Select 25 cards at random from a deck of 52. What is the probability that
>> you can partition them into five hands of five cards each so that each hand is
>> a straight or better in Poker?

By simulation, I get a win probability of

98.27% +- 0.12% if "straight or better" is a win
97.76% +- 0.15% if four-of-a-kind is not a "pat hand"

(Details: 10000 deals per trial, and the "+-" is the population
standard deviation estimated from 20 such trials, 200000 deals
per configuration all told.)

I have absolutely no idea whatsoever about how to tackle
the probability-of-winning problem analytically.

--
eso...@comcast-dot-net.invalid
Five hundred seventy-two days to go.

[James Dow Allen]

On Saturday, June 29, 2019 at 3:04:27 AM UTC+7, Eric Sosman wrote:
> On 6/26/2019 7:50 AM, James Dow Allen wrote:
> >> Select 25 cards at random from a deck of 52. What is the probability that
> >> you can partition them into five hands of five cards each so that each hand is
> >> a straight or better in Poker?
>
> By simulation, I get a win probability of
>
> 98.27% +- 0.12% if "straight or better" is a win
>

> (Details: 10000 deals per trial, and the "+-" is the population
> standard deviation estimated from 20 such trials, 200000 deals
> per configuration all told.)

> eso...@comcast-dot-net.invalid
> Five hundred seventy-two days to go.

Thanks, Eric! I see that in 18 years the best answer is
just mine from the other thread, where I wrote

]] After a random shuffle, the chance of success in
]] this solitaire is a bit less than 98% (if Quads are
]] not considered "pat"), or a bit more than 98%
]] (if Quads are considered "pat.")

Yesterday I generated 500,000 deals, looking for those with
only 1 or 2 solutions. I posted an exemplar of each 1-solution
position encountered. Almost by chance(!) my program printed
the count for each of the 500,000 trials and I see that 8837 of
them had no solution. I think this translates for comparison
with your simulated win probability (98.27 +- .12) above, to
98.23% +- .037%
where .037 = 1.96 * sqrt(.9823*(1- .9823)/500000) * 100
(Is that the correct formula for 95% confidence?)

This doesn't PROVE that either of our simulations
is correct, but the closeness of our results is reassuring.

James

[Eric Sosman]

On 6/28/2019 5:28 PM, James Dow Allen wrote:
> [...]

> Thanks, Eric! I see that in 18 years the best answer is
> just mine from the other thread, where I wrote
>
> ]] After a random shuffle, the chance of success in
> ]] this solitaire is a bit less than 98% (if Quads are
> ]] not considered "pat"), or a bit more than 98%
> ]] (if Quads are considered "pat.")

Just for kicks I simulated versions of the game with different
restrictions on what hands could be used: Straight or better, same
but without quads, flush or better, and so on. I got

Straight or better
19,716 eligible hands
98,195 wins in 100,000 deals

Straight or better, no quads
19,092 eligible hands
97,686 wins in 100,000 deals

Flush or better
9,516 eligible hands
39,897 wins in 100,000 deals

Flush or better, no quads
8,892 eligible hands
37,414 wins in 100,000 deals

Full house or better
4,408 eligible hands
1,162 wins in 100,000 deals

Full house or straight flush
3,784 eligible hands
147 wins in 100,000 deals

Quads or better
664 eligible hands
1 wins in 100,000 deals

Straight flush only
40 eligible hands
0 wins in 100,000 deals

This is clearly not The Ultimate Word on the matter, as the game
is obviously winnable even when only the forty straight flushes are
allowed. (Might take a *lot* of attempts, though ...)

I also tried fitting various curves to see if "probability of
winning" might be a tractable function of "eligible hands," but nothing
looked promising. Of the fits I tried, a quadratic was the least bad
but wasn't very good.

--
eso...@comcast-dot-net.invalid
Five hundred seventy days to go.