Fair odds on bar bet
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.
