linksjunkie
linksjunkie
Joined: Feb 28, 2014
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August 2nd, 2019 at 10:22:32 AM permalink
Need some help please.

Trying to determine.fair odds on a bet using a standard deck of cards.

What are the odds of not matching ranks of two consecutive cards at least once while flipping thru a full deck of 52?

I have found some different answers online.

15:1 seems likely but canít do the math myself
Son you ainít paying attention Iím cutting you but you ainít bleeding - Foghorn Leghorn
Ayecarumba
Ayecarumba
Joined: Nov 17, 2009
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August 2nd, 2019 at 11:10:50 AM permalink
Quote: linksjunkie

Need some help please.

Trying to determine.fair odds on a bet using a standard deck of cards.

What are the odds of not matching ranks of two consecutive cards at least once while flipping thru a full deck of 52?

I have found some different answers online.

15:1 seems likely but canít do the math myself



I believe you will only run into this condition ~5 times out of every 100 trials.

Quote: ask.metafilter.com

Given a card, the odds that the next card doesn't match is 48/51. There are 51 pairs. Each pair can be considered independently -- the odds don't change as the deck is dealt. So, the probability that all 51 pairs don't match are (48/51)^51, or 0.0454. The odds that at least one pair matches is 1-(48/51)^51, or 0.9546.

Simplicity is the ultimate sophistication - Leonardo da Vinci
7craps
7craps
Joined: Jan 23, 2010
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August 2nd, 2019 at 11:21:38 AM permalink
Quote: linksjunkie

What are the odds of not matching ranks of two consecutive cards at least once while flipping thru a full deck of 52?

a solution to this type of question has been around for centuries
and there can be found many correct answers online with the math shown
https://math.stackexchange.com/questions/310971/no-two-identical-ranks-together-in-a-standard-deck-of-cards

without showing the math, I get
the probability is exactly
672058204939482014438623912695190927357/14778213400262135041705388361938994140625
as a decimal = 0.045476282331094304489685910971641185908
(1 chance in 21.989484380438290073317173171574198242)

for fair odds, that would be exactly
14106155195322653027266764449243803213268/672058204939482014438623912695190927357
as a decimal
20.989484380438290073317173171574198242 to 1
in agreement with a very fast simulation
winsome johnny (not Win some johnny)
linksjunkie
linksjunkie
Joined: Feb 28, 2014
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August 2nd, 2019 at 11:55:02 AM permalink
Thank you both very much for you quick responses.

Been hosing my gambling buddies more than I thought. They usually are happy at 8 to 1. Lol

Funny thing happened last night. One of the guys offered the 15 to 1 for one run only. Thought he would pick up a quick ten spot after seeing 8 or 9 different people loose. You can guess what happened....
Son you ainít paying attention Iím cutting you but you ainít bleeding - Foghorn Leghorn
onenickelmiracle
onenickelmiracle
Joined: Jan 26, 2012
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August 2nd, 2019 at 6:34:22 PM permalink
Quote: linksjunkie

Thank you both very much for you quick responses.

Been hosing my gambling buddies more than I thought. They usually are happy at 8 to 1. Lol

Funny thing happened last night. One of the guys offered the 15 to 1 for one run only. Thought he would pick up a quick ten spot after seeing 8 or 9 different people loose. You can guess what happened....

Sounds like you can tempt them with better odds on a second shot. Could scare them off though.
In the land of the blind, the man with one eye is the care taker. Hold my beer.
RS
RS
Joined: Feb 11, 2014
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August 3rd, 2019 at 1:22:37 AM permalink
Quote: 7craps

a solution to this type of question has been around for centuries
and there can be found many correct answers online with the math shown
https://math.stackexchange.com/questions/310971/no-two-identical-ranks-together-in-a-standard-deck-of-cards

without showing the math, I get
the probability is exactly
672058204939482014438623912695190927357/14778213400262135041705388361938994140625
as a decimal = 0.045476282331094304489685910971641185908
(1 chance in 21.989484380438290073317173171574198242)

for fair odds, that would be exactly
14106155195322653027266764449243803213268/672058204939482014438623912695190927357
as a decimal
20.989484380438290073317173171574198242 to 1
in agreement with a very fast simulation


Howíd you do the math?
7craps
7craps
Joined: Jan 23, 2010
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  • Posts: 1977
August 5th, 2019 at 9:46:29 AM permalink
Quote: RS

Howíd you do the math?

first, I did this type of problem years ago and started with half a deck of playing cards to make it easier (math) and less time to go thru the deck.I think I offered 1.5 to 1 that there would be NO 2 ranks together.
using pari/gp calculator
nWays=vector(13, n, sum(k=0, n, binomial(n, k)*(-1)^(n-k)*(n+k)!/2^k));
\\n=13;k=2;p=nWays/((n*k)!/(k!)^n)
p=17752366094818747392000/(26!/(2!)^13)
1.*p
q=1-p;
x=q/p;
xDec=1.*x;
x
xDec
that resulted in this
gp > p=17752366094818747392000/(26!/(2!)^13)
%1 = 63352450072/175685635125
gp > 1.*p
%2 = 0.36060119557825459408060411848654834636
gp > q=1-p;
gp > x=q/p;
gp > xDec=1.*x;
gp > x
%6 = 112333185053/63352450072
gp > xDec
%7 = 1.7731466569222412187942002597660797853


basic inclusion-exclusion formula at work for nWays

for k=4 (4 of each rank) and 13 kinds (52 card deck), I have only seen an integral formula (Laguerre polynomial) and the Mathematica code used to work in Wolfram Alpha (no longer does)
https://math.stackexchange.com/questions/129451/find-the-number-of-arrangements-of-k-mbox-1s-k-mbox-2s-cdots/129802#129802

so for nWays I looked it up as there are a few sources for it.
gp > p=4184920420968817245135211427730337964623328025600/(52!/(4!)^13)
%1 = 672058204939482014438623912695190927357/14778213400262135041705388361938994140625
672058204939482014438623912695190927357 can be found by itself too)

gp > q=1-p;
gp > x=q/p;
gp > xDec=1.*x;
gp > x
%5 = 14106155195322653027266764449243803213268/672058204939482014438623912695190927357
gp > xDec
%6 = 20.989484380438290073317173171574198242

that is how I did it.
remembering for a simple binomial type wager, the 'fair' X to 1 payoff is simply X = q/p
fair coin Heads, fair payoff = 1/2 / 1/2 = 1/2 * 2/1 = 1 = X
fair die (face 6 for example), fair payoff = 5/6 / 1/6 = 5/6 * 6/1 = 5 = X

added: classic example of
"IF it was that easy to do, everyone would do it."
also, this is very easily simulated as an exact answer really is not required.
winsome johnny (not Win some johnny)

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