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.