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 can’t do the math myself

Quote:linksjunkieNeed 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 can’t do the math myself

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

Quote:ask.metafilter.comGiven 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.

a solution to this type of question has been around for centuriesQuote:linksjunkieWhat are the odds of not matching ranks of two consecutive cards at least once while flipping thru a full deck of 52?

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

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.Quote:linksjunkieThank 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....

Quote:7crapsa 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

How’d 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.Quote:RSHow’d you do the math?

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.