Probabilty of board being all one suit

[Toro]

My buddy is shift manager at Caesars in LV and the topic came up of the probablity of the board in HE being all one suit. He asked me and this is what I came up with. Is this the correct way to calculate that? (copy of my email to him)

I couldn't find the answer to how often all one suit comes on the board in Holdem in any of my odds books but I think this analysis works. You have to take it card by card. When doing any calcualtions in holdem you do it based on unknown cards. So all player cards are unknown except your own and so in this case you said the player didn't have that suit so he assumes all 13 (lets pick a suit) spades are live.

In reality they all won't be live with a full table but the other suits won't be live either so it cancels out over many trials. So . . .

The chances of the first card being a spade are 13/52 = 25% or .25
2nd card 12/52 = .2307692
3rd card 11/52 = .2115385
4th card 10/52 = .1923076
5th card 9/52 = .1730769

So the chances of all 5 being spades are all those fractions multiplied together:

.25 x .2307692 x .2115385 x .1923076 x .1730769 = .0004062

To find how many deals would have to occur for that to happen, take the inverse of that number so:

1 divided by .0004062 = 2461.8414 (let's round it up) so statistically:

Once in every 2462 deals all of one suit will hit the board. Not sure if my logic is correct but that number sounds reasonable.

[djames]

There are 50 choose 3 ways to flop a board of 3 cards from the remaining 50 cards in the deck. If you have i cards of the suit in question, there are 13-i left. From those you must choose 3 for success.

The probability of success is then (13-i choose 3) / (50 choose 3).

This is approximately 1.46%, 1.12% and 0.84% if you have 0, 1 or 2 of the suit, respectively.

[AkuJoe]

It's calculated this way: (13/52)*(12/51)*(11/50)
Which is 0.25*235294117647*0.22
It makes 0.012941176471. So, about 1.3% of the time.

[Mellor]

djames and AkuJoe, you only calculated that three of a suit would be on the board. And you forgot to account the remain too any way as it affects the odds.


Toros original post was close, but you forgot to reduce the deck size each time. There is a quicker way.

13[c]5 (number of all suit flops)
divided by
52[c]5 (total number of flops)

if you dont understand x[c]y, or rather do it just in numbers the above breaks down into

(13x12x11x10x9)/(52x51x50x49x48)

154440/311875200

=.05% (slightly rounded up)

[djames]

Yes, I read the post as the flop being one suit as this is a common sidebet, and not the whole board. He did say the whole board, so I didn't enumeratehis question. My mistake.

[Siegmund]

If the flop is all one suit, whether or not there will BE a turn or river depends quite a lot on how many cards of that suit are held by the players still involved in the hand.

This is not a prop bet to touch with a ten-foot pole.

The chance of five consecutive cards in an otherwise random deck being all the same suit is 4 times the number in OP's post since any of four suits might come up.

[AkuJoe]

Oops, I calculated it wrong. And thought that it's flop only.
Flop percentage being only one suit is of course 1*(12/51)*(11/50)=0.0517= 5% of the time.
The whole board: 1*(12/51)*(11/50)*(10/49)*(9/48)

[jogsxyz]

[ QUOTE ]
Oops, I calculated it wrong. And thought that it's flop only.
Flop percentage being only one suit is of course 1*(12/51)*(11/50)=0.0517= 5% of the time.
The whole board: 1*(12/51)*(11/50)*(10/49)*(9/48)

[/ QUOTE ]

I second this answer. Only it's ~ 5% of the time.
The first card is any of the 52 cards. That's 100%. Then the second card is 12 of the remaining 51 cards, etc.