Tom Seewald's ranking
| Ranking by Earnings | ||
| Last 12 months worldwide | n/a | pos. n/a |
| Last 12 months national | n/a | pos. n/a |
| All time worlwide | 25147th | = |
| All time national | 14003rd | = |
| All time WSOP | 16985th | N/A |
| All time WPT | n/a | pos. n/a |
| All time EPT | n/a | pos. n/a |
The biggest cashes of Tom Seewald
$19,333
Rank: 3 / 286$540 Limit Texas Holdem
Jan 4, 1998
| Tournament | Results |
|---|---|
|
Event 15: $2500 Pot Limit Texas Holdem 24th Annual World Series of Poker 1993 Binion's May 4, 1993 |
$9,975 7 / 114 |
|
$500 Texas Holdem 1993 World Poker Finals Foxwoods Resort Casino Dec 3, 1993 |
$7,280 2 / 68 |
|
$1000 Limit Texas Holdem Carnivale of Poker Rio Jan 10, 1998 |
$3,520 8 / 176 |
|
$535 Pot Limit Texas Holdem 1997 Four Queens Poker Classic Four Queens Casino Jan 11, 1997 |
$3,211 8 / 162 |
|
$2000 Limit Texas Holdem 1993 World Poker Finals Foxwoods Resort Casino Dec 18, 1993 |
$3,080 6 |
|
$300 Texas Holdem 1993 World Poker Finals Foxwoods Resort Casino Dec 2, 1993 |
$2,623 4 / 159 |
|
$1060 Limit Texas Holdem Hall of Fame Poker Classic 1997 Binion's Aug 11, 1997 |
$1,560 16 / 195 |
|
$1560 Pot Limit Texas Holdem Hall of Fame Poker Classic 1992 Binion's Nov 3, 1992 |
$1,380 16 / 115 |
Statistics of Tom Seewald
Compare with other players:
All player's results
| Tournament | Results | |
|---|---|---|
|
Carnivale of PokerSaturday 10 January 1998 |
$3,520 8th / 176 |
|
Carnivale of PokerSunday 4 January 1998 |
$19,333 3rd / 286 |
|
Hall of Fame Poker Classic 1997Monday 11 August 1997 |
$1,560 16th / 195 |
|
1997 Four Queens Poker ClassicSaturday 11 January 1997 |
$3,211 8th / 162 |
|
1993 World Poker FinalsSaturday 18 December 1993 |
$3,080 6th |
|
1993 World Poker FinalsFriday 3 December 1993 |
$7,280 2nd / 68 |
|
1993 World Poker FinalsThursday 2 December 1993 |
$2,623 4th / 159 |
|
24th Annual World Series of Poker 1993Event 15: $2500 Pot Limit Texas Holdem Tuesday 4 May 1993 |
$9,975 7th / 114 |
|
Hall of Fame Poker Classic 1992Tuesday 3 November 1992 |
$1,380 16th / 115 |
