Question about Odds.

[Mango99]

If I put 3 players into a festival and we all get drawn to play the same day, what are the odds that we all get drawn on the same table?

Say for this example there are 155 players on the first day, tables of 10 (and some 9's obv).

[AndrewT]

My attempt at this.

I assume 155 players and a ten seat table.

The chances of you taking your seat and finding your three backees is.

(3/154 * 2/153 * 1/152) * 84 = 7105/1

84 being the number of arrangements of three people at a nine seat table

9x8x7
  3x2

[SuffolkPunch]

My attempt at this.

I assume 155 players and a ten seat table.

The chances of you taking your seat and finding your three backees is.

(3/154 * 2/153 * 1/152) * 84 = 7105/1

84 being the number of arrangements of three people at a nine seat table

9x8x7
  3x2

Exactly what I was about to say 

[Mango99]

My attempt at this.

I assume 155 players and a ten seat table.

The chances of you taking your seat and finding your three backees is.

(3/154 * 2/153 * 1/152) * 84 = 7105/1

84 being the number of arrangements of three people at a nine seat table

9x8x7
  3x2

Aha, thanks for the answer

I actually meant 3 of us in total. Please can you do the math again lol Or is that what your the above has worked out already?

[Sheriff Fatman]

There's a mathematical answer to this question and the 'as seen at the WSOP 2005' answer, where people were allocated seats as they registered for the event.  Consequently, groups of friends who registered together often found themselves at the same table.

I recall this from feedback made about the early events (as I was over there at the time).  I think they might have sorted things out by the time of the main event.

[tikay]

My attempt at this.

I assume 155 players and a ten seat table.
The chances of you taking your seat and finding your three backees is.

(3/154 * 2/153 * 1/152) * 84 = 7105/1

84 being the number of arrangements of three people at a nine seat table
9x8x7
  3x2

Surely, the multiplier is 120, as we assumed a 10 seat table at the head of the equation?

[RED-DOG]

Does anyone have anything heavy that they require lifting?

[AndrewT]

My attempt at this.

I assume 155 players and a ten seat table.
The chances of you taking your seat and finding your three backees is.

(3/154 * 2/153 * 1/152) * 84 = 7105/1

84 being the number of arrangements of three people at a nine seat table
9x8x7
  3x2

Surely, the multiplier is 120, as we assumed a 10 seat table at the head of the equation?

You're already sitting in one of the seats. You're trying to work out the odds that three of the nine other seats are filled by your guys.

EDIT: Thinking about it again, I'm not 100% sure that's a correct assumption to make. But it's Saturday and I'm not spending my day thinking about A-Level maths questions.

[ariston]

why do you guys have to make things so complicated. Whichever table you get drawn on your 2 friends have to be drawn on as well. There are either 9 or 8 seats left on your table (depending on a ten or 9 handed table) and they have to occupy 2 of these.
9/154 x 8/153 = 325.5-1 on a ten handed table
8/154 x 7/152 = 418-1 on a 9 handed table

If there are an equal number of 9 and ten handed table giving you an equal chance of being drawn on a 9 or ten handed table we can just work out the mean average of the 2 figures giving you odds of 371.75-1 of being drawn on the same table table as you 2 mates.

[KingPoker]

Does anyone have anything heavy that they require lifting?

LMFAO!!!

[ariston]

in fact thinking sensibly if there were 155 runners then that would be 11 ten handed and 5 nine handed tables so a mean average is no good.

(11x325.5)+(5x418)/16= 354.41 so the odds of being drawn on the same table as your 2 mates is 354.41-1.

[tikay]

in fact thinking sensibly if there were 155 runners then that would be 11 ten handed and 5 nine handed tables so a mean average is no good.

(11x325.5)+(5x418)/16= 354.41 so the odds of being drawn on the same table as your 2 mates is 354.41-1.

But he has 3 mates.......

[I KNOW IT]

Very high in the Grovesnor, as I believe there is a glitch in there random draw software. I told them this about 12 months ago as in about 12 tournys me and a mate were drawn on the same table every time. I think its if you enter at the same time

as for the true mathmatical equation I cant be arsed to work it out

[ariston]

in fact thinking sensibly if there were 155 runners then that would be 11 ten handed and 5 nine handed tables so a mean average is no good.

(11x325.5)+(5x418)/16= 354.41 so the odds of being drawn on the same table as your 2 mates is 354.41-1.

But he has 3 mates.......

read the thread granddad. A few posts down he says he meant himself and 2 mates. Another Tikay missread lol

[KingPoker]