Simple Probability Question

[LuckyLucky]

Hello,

You have a 3 table, 30 person tournament and the top 6 people get paid. The tables are numbered 1, 2 and 3. Assuming that there is no skill difference, what are the chances of at least one person from table 1 making it to the money.

Can you show your work as well?

[WhiteWolf]

The chance than no person from Table 1 makes the money is:

20/30 * 19/29 * 18/28 * 17/27 * 16/26 * 15/25 = 0.065

The chance that at least one player from table 1 makes the money is 1.0 minus the above figure:

1.0 - 0.065 = .935, or 93.5%

[LuckyLucky]

Can you explain your work?

[elitegimp]

[ QUOTE ]
Can you explain your work?

[/ QUOTE ]

You have 30 players, all with a 1/30 chance of finishing in any specific place (since they are equally skilled). Therefore there is a 20/30 chance that the first place spot is occupied by one of the 20 players starting at table 2 or 3.

Given that the first place player came from table 2 or 3, there are now 29 players vying for second place, and 19 of them started on table 2 or 3, so it's a 19/29 chance that given first place from table 2 or 3, second place is from there as well.

Doing this for third - sixth place yields the other four terms, and you just multiply them all together to get the probability that all 6 things are true.

This gives the probability that nobody from table 1 makes the money, so to find the probability that at least one person from table one makes the money, you subtract this probability from one. This comes from the fact that the probability of an event happening plus the probability of the event not happening must sum to 1 ("nobody from table 1 monies" and "at least one person from table one monies" are opposites, so exactly one of the two events must happen).

[ncray]

You can think of it this way. There are 30 spots
|__________|__________|_________|
----table 1-----table 2-----table 3

How many permutations are there where people from tables 2 and 3 end up in the first six spots?

______|_______________________
first six

For the first spot, you have 20 ppl to choose from (anyone from tables two and three). For the next spot, you have 19, then 18, 17, 16, 15. So there are 20*19*18*17*16*15 permutations of people from tables 2 and 3 filling the first six spots (in the money) out of a total 30*29*28*27*26*25. Spots 7-30 don't matter. Since 20*19*18*17*16*15/(30*29*28*27*26*25) gives you the probability of the complement (nobody from table 1 makes money), subtract it from one to get the probability that at least one person from table one makes money.

[LuckyLucky]

Thanks that made perfect sense to me.

[pzhon]

[ QUOTE ]

You have a 3 table, 30 person tournament and the top 6 people get paid. The tables are numbered 1, 2 and 3. Assuming that there is no skill difference, what are the chances of at least one person from table 1 making it to the money.

[/ QUOTE ]
This is not a simple problem unless you make unrealistic assumptions. There isn't enough information.

Others have assumed that equal skill means that the players are not only equally likely to place first, but that other places do not depend on the locations of the players who finish first. This is unrealistic. If a player right next to you is accumulating chips en route to winning, this decreases your chance to place second. It should be more likely that at least one player from table 1 finishes in the money than that at least one player cashes from a random list of 6 players.

[Vertical Taco]

[ QUOTE ]
[ QUOTE ]

You have a 3 table, 30 person tournament and the top 6 people get paid. The tables are numbered 1, 2 and 3. Assuming that there is no skill difference, what are the chances of at least one person from table 1 making it to the money.

[/ QUOTE ]
This is not a simple problem unless you make unrealistic assumptions. There isn't enough information.

Others have assumed that equal skill means that the players are not only equally likely to place first, but that other places do not depend on the locations of the players who finish first. This is unrealistic. If a player right next to you is accumulating chips en route to winning, this decreases your chance to place second. It should be more likely that at least one player from table 1 finishes in the money than that at least one player cashes from a random list of 6 players.

[/ QUOTE ]

I believe that Lucky is simply trying to determine the probability of one player from table 1 making the money. Lets assume 6 players out of the 30 are selected at random without any poker being played - what are the odds of a player from table one being 'selected'?

I believe that might be a logical way of removing any extenuating circumstance such as chip accumulation.

[LuckyLucky]

Yeah, that's what I wanted to know. The real life question I'm trying to answer is a 30 person tournament with 10 women playing. I'm betting that one woman will make the money and I wanted to know the odds of that happening. And now I do. Thanks. Thread over.

[AaronBrown]

Although LuckyLucky got his answer, I think the original question is interesting for the reason pzhon mentions. I think the actual probability of having no money finishers from table 1 is less than 2% rather than 6.5%.

Think about it this way. The tables play for some period of time until there are 20 or fewer players left and the organizers break one table up. At that time, there could be as few as one or as many as ten survivors from table 1, but they will share 1/3 of the chips between them.

At this point, the chance that the winner comes from original table 2 or 3 is 2/3. The chance that the second place finisher comes from original table 2 or 3, given that the winner did, is roughly 2/3 minus the expected proportion of chips held at time of breakup by the eventual winner. In effect WhiteWolf's calculation assumes the amount to subtract is 1/30, but it has to be greater than that. WhiteWolf is computing the probability as of the beginning of the tournament, which ignores the fact that money must stay within table 1 until at least ten players are eliminated from the tournament.

The chance that the third place finisher comes from the original tables 2 or 3, given that the first two finishers did, is roughly 2/3 minus the expected combined proportion of the chips held by the first two finishers. And so on through the first six.

Even this calculation overestimates the chance because 2 times in 3 table 1 is preserved after the breakup (assuming the tournament organizers choose to break up the smallest table).

Making some reasonable guesses for the chip stack distribution at time of break up, I get about a 1.3% chance that no original table 1 player will finish among the top six.