Fifth card makes last card of straight, 4C1 = 4
4 x 40 x 12 x 8 x 4 x 4 = 2048 = 0.0788 %
52 51 50 49 48 2598960
Fifth card makes an inside card of straight, 4 x 3 = 12 (4 places lead card of straight, 3 places end card)
12 x 40 x 12 x 8 x 4 x 4 = 6144 = 0.2364 %
52 51 50 49 48 2598960
Sums to 0.3940%, so is correct
Remove the fifth card (last 4/48 term) to see the picture with 4 cards dealt
= Chance of getting the first 4 cards of a straight (fifth card inside or outside)
0.3940% *48/4 = 4.728 %
For outside straights only
(0.0788% + 0.0788%) *48/4 = 1.8912 %
ANSWER
Chance of getting the first 4 cards of a straight deal, where a fifth card can make a straight on the start or end = 1.8912 %
If you get to throw out a card and get a fresh one, for the second chance straight probability
multiply the 0.3940 % by
(48/4 * 44/48 * 4/47) = 0.936
= 93.6%
almost same chance of getting a straight on the 6th card as on the 5th card
48/4 cancels the 5th card prob, because we no longer deal with a wanted card there
44/48 (= [48-4]/48) is the prob that the 5th card is unwanted for a straight
so you discard the incorrect 5th card and draw a 6th card, the chance the 6th card is what you want is 4/47
0.3940% (chance of straight on a 6 card deal, on the 5th card)
0.3940% x 0.936 = 0.3689 % (chance of straight on a 6 card deal, on the 6th card)
0.3940% x 1.936 = 0.7629 % (chance of straight on a 6 card deal, on the 5th or 6th card)
ANSWER:
Chance of a straight with a 1 card redraw = 0.7628 % (1 in 131.1)
(on a deal of either 5 or 6 cards)
Im happy to expand on the information in this post or the above post if anyone who wants to understand is still stuck
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Here is the probability of straight on a 6 card deal, on the 6th card in an expanded form
5 x 40 x 16 x 12 x 8 x 44 x 4 = 112640 x 4 = 0.3689 %
52 51 50 49 48 47 2598960 47
But the middle card doesnt have to be the first, it can be dealt in any of the 5 positions
either 1st 2nd 3rd 4th or 5th, so multiply by 5
The order of the other cards does matter
5 x 40 x 16 x 12 x 8 x 4 = 10240 = 0.3940 %
52 51 50 49 48 2598960
ANSWER
Chance of a straight in the first deal of 5 cards = 0.3940 % (1 in 253.8)
This is the figure given in wiki for a straight
en.wikipedia.org/wiki/Poker_probability
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Also notice that poker tables list Royal Flushs and Straight Flushs seperate to the other flushes
4 Royal Flush + 36 Straight Flush + 10200 Other Straights = 10240 All Straights
The guy at the first post made a similar mistake for For Flushes, its 1 in 504.8 for any flush in the first hand,
not 1 in 508 as he says, which is for other flushes but not the royal or straight flushes
I have this in my notes:
Hmmmm. Computer count yields an answer close to that guy's formula but slightly lower.
]]>Yes, I am playing draw, and drawing one card to the outside straight.
D
]]>Are you playing Draw Poker where you will discard one card trying to fill the straight?
]]>I am asking essentially the same question as started this thread, except I am looking for the probability of being dealt a STRAIGHT draw, rather than a flush draw (four spades, say).
In other words, what is the probability that I will be dealt 6-7-8-9 or T-J-Q-K, and so on. A hand that gives you a one card, outside, draw to one of those 10,200 straights in the deck.
Let`s consider only hands that offers an outside straight, (6-7-8-9 rather than 5-6-8-9).
Thanks very much.
D
]]>Please rephrase the question in an exact form. Are you asking the probability of being dealt a straight?
]]>I just found this old thread, which answered my question vis a vis flush draws. What about straights?
This is a more complicated question, given that once one id dealt an initial card the seond one (in most cases) must be one of the eight that occur within 4 on either side. (Eg say you are dealt an 8. The next card must be one of 4,5,6,7 (from the "low" side), or 9,T,J,Q on the high.
But what about when one is dealt a King? The second card must be one of the 4 from the low, or an ace.
I am more interested in the result, but am likewise interested in the underlying logic/math.
Thanks
D
]]>While having this argument on another forum. I found this, you might want to check this page out. It verifies Fruityloops's use of the hypergeometric distribution.
]]>Which still seems weird , because when you have four of the suit there are 9 more of that suit in the pack, and 39 others , so you would think the ratio would be 9 to 39 , or 1 in 4.3
But then I thought if you applied that to just six cards, five of which were clubs , then applying that logic you would get a flush 1 out of two goes, whereas in fact you would only get it 1 in 6 goes as the odd card would appear in the first five cards dealt five times out of six !
Thanks for your help on here now, just glad to have got it resolved in my head !!
]]>I believe you need to use the hypergeometric distribution formula. You are looking for 4 cards out of a group of 13 and 1 card from the other 39 so we have...
I am saying you could be dealt a non-club, then four clubs - or a club, then a non-club, then three clubs - etc - So how many times will you get five clubs off the deal , rather than four plus a card of another suit ( that could be the first, 2nd,3rd,4th, or 5th card dealt.)
maybe this is a better way of putting it -
For every one occasion that you 'flop' five clubs on the deal , on how many occasions will you flop four clubs plus one card of a different suit (in any order ) on the deal ?
]]>It could be:
"You are dealt four cards and all of them have the same suit. You are then dealt another card - what is the probability that this suit also matches?"
or
"You are dealt five cards and at least four of them have the same suit. What is the probability that five of them have the same suit?"
]]>