Fifty50 Payout Distributed on Chip Count %

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Fifty50 Payout Distributed on Chip Count %

Postby Tuck Fheman » Sun Jan 07, 2018 2:44 pm

I don't think there's a way to accomplish this in the current system with a Fifty50 tourney, so unless I'm mistaken I suggest this addition if deemed necessary and in demand by others.

PokerStars has their Fifty50 work like this and I like it better than an even split of rewards among the remaining players.

Half of the prize pool will be distributed equally among the 5 winners, and the other half of the prize pool will be distributed among those same 5 players based on the chip count percentages at the end of the event. The 6th-10th place finishers win nothing. So in a 10-player $10+$1 Fifty50, the prize pool is $100. Each of the final 5 finishers would receive $10 each, plus a percentage of the remaining $50 in the prize pool.


The only way I've seen the current Fifty50 tourney pay rewards to the last 5 (in a 10 player game) based on chip count is if I set the timer to remove inactive players and a player(s) is removed creating an uneven amount of players.

If I'm wrong, please correct me, thanks!
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Re: Fifty50 Payout Distributed on Chip Count %

Postby Kent Briggs » Sun Jan 07, 2018 4:13 pm

The only way to do this in the current system is if you coded the proportional payout part yourself. For example, instead of a 10+1 buyin, you would make a 5+6 buyin with stop-on-chop enabled. Then write some API code (triggered by the tournament finished callback) that distributed 5 of the 6 house chips back to player's individual accounts based on their final stacks that you pulled from the handhistory. That's likely more trouble than it's worth, though.
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