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Paul2432
10-05-2004, 12:39 AM
Party Poker advertises their bad beat distribution as 70% to the table (with the most to the player badly beaten), 20% to seed the next jackpot, and 10% withheld as Party's fee.

In fact, in the long run party will rake more than 10% of the money taken in by the bad beat rake. Why is this and what is the true long run percentage?

I'll post the solution in a few days if no one comes up with an answer.

Paul

pzhon
10-05-2004, 01:52 AM
[ QUOTE ]
Party Poker advertises their bad beat distribution as 70% to the table (with the most to the player badly beaten), 20% to seed the next jackpot, and 10% withheld as Party's fee.

In fact, in the long run party will rake more than 10% of the money taken in by the bad beat rake. Why is this and what is the true long run percentage?

[/ QUOTE ]
Good point. Party gets 10% out of 80% every time the jackpot is won, so Party should get 1/8 of the money coming out. Another way of looking at it is that for every dollar put in, Party will take .10 as an administrative fee from the first jackpot, and .20 is added to the second jackpot. Of this, .02 goes to Party when the second jackpot is won, and .04 goes into the third jackpot. .004 goes to Paty when the third jackpot is won, and 0.008 is returned, etc. 1/8 = .125 = .10 + .02 + .004 + .0008 + .00016 + ...

Even though 7/8 of the money is returned to the players, people often treat a jackpot drop as a rake. Perhaps those who play well get a lower share of the jackpot by folding suited trash like 74s that could be part of a straight-flush, or causing players to fold by raising and reraising preflop, or by folding 76s on a 999 flop.

Paul2432
10-05-2004, 09:31 PM
Your answer is the same as mine. At first I thought they did the bad beat the way they do so they could post higher jackpots. Then I realized they sneekily collect 25% more rake this way.

Paul

LondonBroil
10-06-2004, 12:24 AM
[ QUOTE ]
Party Poker advertises their bad beat distribution as 70% to the table (with the most to the player badly beaten), 20% to seed the next jackpot, and 10% withheld as Party's fee.

In fact, in the long run party will rake more than 10% of the money taken in by the bad beat rake. Why is this and what is the true long run percentage?

I'll post the solution in a few days if no one comes up with an answer.

Paul

[/ QUOTE ]

Imagine a JP of \$100,000. Party takes \$10K for fees, and \$20K gets seeded into the next JP. Someone else hits the \$20K right away, Party gets \$2K from that, then \$200, then \$20, then \$2. So Party made \$12,222, or 12.2%.

Is this right?

Paul2432
10-06-2004, 10:36 AM
You're on the right track. To arrive at the precise solution you need to sum the infinite geometric series:

1 + 0.2 + 0.02^2 + 0.02^3 + . . . + 0.02^n + . . .

The sum of an infinite geometric series can be found using the formula

S = (first term) / (1 - multplier) = 1 / (1-0.2 ) = 1.25

10% x 1.25 = 12.5%

Paul