hahahaha 
odds of this happening?
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odds of this happening?
Re: odds of this happening?
Simple odds assuming 2 hands playing , unknown hole cards.
the odds are 1/ (48 x 47 x 46 x 45 x 44) = 1/(205476480) =0.000000005 , 0000005 % the odds of such a thing happening indepedant of the suit = 4 / (48 x 47 x 46 x 45 x 44) = 4/(205476480) = an increased chance of 000002 %
Re: odds of this happening?
joke
Last edited by BLKH on Tue Jun 02, 2009 12:07 pm, edited 1 time in total.
Re: odds of this happening?
I bet u lose all the time online. considering the 1st card to be a 5 to a 10 48/48 * 4/47 * 3/46 * 2/45 * 1/44 its only 0.00056% don mislead others
Re: odds of this happening?
haha...well its good thing its a split pot
Re: odds of this happening?
its not 0.00056% 1) 1st card being 5 or 10 is not 48/48 its 2/48 2) any card can make a straight flush not only 5/10 with 1st card. assumptions: royal flush is encompassed, we are looking at specific suit (so as to allow for a valid comparision with 1st reply) if 1st card is 2/K, cards available on 2nd draw to make the hand will be 5 onlya if 1st card is 3/Q, cards available on 2nd draw to make the hand will be 6 onlyb if 1st card is 4/J, cards available on 2nd draw to make the hand will be 7 onlyc if 1st card is 5 to 10 or A, cards available on 2nd draw to make the hand will be 8d looking at scenario d, probability is 7/48 1st card 5 to 10 or A..... 2nd card probability is 8/47 if 2nd card is difference of 4 with 1st card (i.e. 1st card  2nd card value = absolute value of 4) then the 3rd card probability will be 3/46 the 4th = 2/45 n the 5th = 1/44 if the 2nd card is difference of 3 with 1st card then the 3rd card can have a probability of 4/46 which then brings different probabilities for the 4th card depending on wat the 3rd card is....if its a card dats in between the 1st n 2nd in value it will be 3/45 and if the 4th is in between again it will be 2/44 for the 5th implying an openended draw if the 3rd card is not within the value of the 1st n 2nd card, the draw will end up in a gut shot draw with lower probability...... im too lazy to calculate everything cos there are sites on the net afterall dat gives u exact probability of straight flush draw....im just giving the mechanics to it.... the easier way to see it will be....for example.... 1st card is K....i can only make a straight flush with A, Q J 10, 9 as 2nd card..... if 1st card is 6, i can make straight flush if 2nd card is 2,3,4,5,7,8,9,10 if 1st card is 6 n 2nd is 10 only running cards of any order of 7,8,9 will make straight flush if 1st card is 6 n 2nd card is 9, 3rd card can be 5,7,8,10 if 1st cards is 6 2nd card is 9 and 3rd card is 5 or 10, last 2 cards must be 7,8, if 1st card is 6 2nd card is 9 3rd card is 7, 4th card can be 5,8,10 if 4th card is 8 last card can be 5 or 10.....if 4th card is 5 or 10 last card have to be 8.....(this is the gut shot/open ended thing i was refering to)
Re: odds of this happening?
i sumhow dont tink that the previous poster whom u quoted had the intention of misleading anyone.... but.....if u were to tink that expressing one's views and attempting to share info dat is incorrect is in fact an intention to mislead....then u sir are in that category as well.... just becos u know that all straights have to encompass a 5 or 10 doesnt mean that u can sound intelligent by adding that fact into an equation that makes no sense at all.... pls at least be in a position where u are protected from rebuttals if u wish to be mean. i dunno how good u tink u are at poker......at poker i may be a fish/donkey....but maths is a different area....try me....
Re: odds of this happening?
My initial mention was the chance of any one hand , in a simple to understand manner i think. didn't expect it to generate such strong emotions.
lets begin with the chance of a straight flush http://en.wikipedia.org/wiki/Poker_probability Straight flush — Each straight flush is uniquely determined by its highest ranking card; and these ranks go from 5 (A2345) up to A (10JQKA) in each of the 4 suits. Thus, the total number of straight flushes is: {10 \choose 1}{4 \choose 1} = 40 with 40 being the total number of types of straight flushes available , 10 for each suit.(inclusive of royal) with out of action, ( due to them being on both sides's hands ) this take away a few of these straight flushes out there. the spade and club series are unaffected , with 10 sets of straight flushes shared among them. the diamond series has 2 sets of straight flushes taken out. to and to Diamond has 8 sets of straight flush left. the heart series , a couple is taken out of action As the result needs to be shared among both players , the only remaining sets available are  ,  Heart has 2 sets of straight flush left. so instead of a total of 40 straight flush , there are 10 from club , 10 from spade , 8 from diamond , 2 from hearts. so its a total of 30 possible selection out of 2,598,960 hands. = 0.000011543 or 0.0011543 % chance assuming both set of hole cards are known. This is assuming order is not taken into consideration. if order is taken into consideration , then you can multiply the odds by 1 / ( 5 x 4 x 3 x 2 x 1) Disclaimer : i am thinking from permutation and combinations point of view from statistics. i am not here to challenge as this thread i view as just a mind teaser. i am practically a fish in poker , do invite me over to donate $$ to you
Re: odds of this happening?
nice angle...
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