Wrong probabilities on the World Series of Poker...
 

I was watching a final round of one of the World Series of Poker tournaments (I don't know which one, at the moment. I can come back later with that information...) and I couldn't help but notice that the displayed probabilies was egregiously incorrect. This was the hand:

<table border=1><tr><th>player</th><th>hand</th><th>to win</th></tr><tr><th>1</th><th>4&spades; 4&clubs;</th><th align=right>98%</th></tr><tr><th>2</th><th>K&spades; Q&spades;</th><th align=right>2%</th></tr></table>
flop

2&diams; 6&diams; 5&spades; 3&diams; ?

As you can see, player 1 has already made his straight, while player 2 currently has nothing. Now, the river card has yet to be revealed, hence the probabilites.
I was watching this game, thinking "man, 98% are pretty good odds! There's very little that player 2 could get to win..." and that's when it occured to me that that 2% was rather generous! In fact, there is no card that can allow player 2 to win. His chances of winning aren't 2%, they're 0%!

So, either there are roll-over rules in the WSOP that I'm unaware of or this is an error in the show.
I have a few questions concerning this. First, has anyone else seen blatant errors, such as this, on the show? Secondly, what could be the cause of this error? Surely, they use a computer program to determine these probabilities so perhaps there's a bug in their software?

Thanks for reading...


Oh, and I'm tempted to start a new thread on this subject but let me just congratulate Annie Duke on winning the WSOP Tournament of Champions! I believe she's the first woman to ever have won any of these tournaments...

Not having the winning hands memorized, I googled winning poker hands and came up with this answer:

http://govegas.about.com/cs/gambling...kerhandstp.htm

If another 4 was drawn, then player 2 wins. Three of a kind does not beat a straight. Therefore, since both players would have a straight, the high card holder would win. In this case, King high would win.

The 2% issue is a little more difficult. Since there are at least two 4's already out, the chance of pulling another 4 would probably be around 2 in 32 (assuming 6 players at the start - 12 dealt to players, four shown, four wasted). This would give about 6% chance to draw a four. From the 2% figure, I am going to assume they "knew" another four had already been wasted and then also rounded down. I am not 100% on this theory though.

Yeah you're right another 4 comming up would have split the pot. the 2% does not come from the High card then winning the hand as the pot would be split. The 2% is derived from the fact that a 4 comming on the river would split the pot between the KQ and the pocket 4's, not win it for the KQ. Their is no real way to show this on a graph like the one that they use on TV so what they do is take the percentage of a 4 comming on the river and split it between the two people in the pot. so while the KQ has no chance of winning the pot he has about a 4% chance of splitting the pot and this is represented in the box by a 2% chance of winning. Hope this is clear

THe K high would not win the pot as the K would not play as both players best hand is on the table. In texas hold'em iti s the best 5 card hand that plays in cases in which that is the same the pot is split the 6th or 7th card are never reverted to as tie breackers.

If another 4 comes out the pot is split. The first player has a 98% chance of winning the entire pot, while the second player has a 2% chance of getting half the pot. It's not very clear but they don't want to show 100% when the player still has a chance of having the best hand.

I stand corrected, multiple times.

I'm completely missing the logic of this somehow...

Player 1 has: 4♠ 4♣

Player 2 has: K♠ Q♠

Flop: 2♦ 6♦ 5♠ 3♦

How would the river card being a 4 help player 2 in any way whatsoever...?
Do BonesCPA, yoyoyobro and dbc know about a rule that I don't..?

I watch and play Texas Holdem rules quite regularly but I don't get what I'm missing on this one...

the best five cards would be common to both players as they are on the table.
The pot would split between them.

The only card that would garauntee a victorty for player 2 would be a 4. He would win the pot, albiet with another player, but he still is technically winning the the hand. Then, two people would win, player 1 and player 2.

Another thing, the high card does not come into play. You try and make the best 5 card hand possible, and both hand would be the 2-6 straight.

Ahaaaa! Thanks Wayne21 & SoccerChamp76, I missed the glaringly obvious there, maybe that's why I'm still no good at Texas Holdem... Or was I bluffing..? lol

If that were the metric, then why isn't the chances of player 1 winning a 100%. After all, there is no card that could allow him to lose, right?

I TiVo all these WSOP shows and watch them later and I remember this hand coming up. I had to rewind it and do a double take.

What I remember was the announcer (Chad) commenting about the statistic that although it said 98%, player 1 could not lose, only split.

And I always have to laugh at the witty comments from the 'color' guy. Dude, we know that the shows are taped. You saying something like it is off-the-cuff is actually pretty condescending.

Anyway, I love watching them. The guys are good. They know the odds. Because hey, it's not what you have it's what you think the other guy has.

I agree with DDDDave, these shows are great fun. Yes, the commentary is not "live", but unlike many sporting events (like the ones with Dick Vitale or Vin Scully), the announcers actually add to the event. I like the duo on WPT (trio, if you count Shana Hiatt!)

JcS

I don't know the specifics of the odds-making program the WOSP uses so I don't know if either of the following theories are actually the reason, but...

Theory One
As was mentioned earlier, a 4 on the river would have resulted in a split pot. Because only five cards are used to make up a hand, it doesn't matter that player two's 6th card is higher. But in the event that a pot is split and there is an extra chip left over, the player with the highest 6th card would get that extra chip. This is the only time a 6th card has any impact on winnings. So, since player two had the highest 6th card, he would have recieved a slightly larger portion of the pot. The odds-making program may have taken this into account and adjusted the percentages accordingly.

Theory Two
The odds-making program may not be able to handle percentage totals over 100%. Depending on the programing, it might not make sense to the computer for one player to have a 100% chance of winning and another player still have a 2% chance. To deal with this conflict,the program simply adjusted player one's percentage to accomadate player two's chances. I have seen odds-making programs that, in a situation like that, will allow for percentage totals over 100%, but not all programs are created equal.

So either the program was overly sophisticated or not sophisticated enough. Who knows?

This is an awful lot of fuss for something that's inconsequential. When *I* see an error on TV, I yell out, "HA! They messed up!" and then move on.

I got a question.

player1 Q 8
Player2 J 8

Then the rest of the cards

9 10 Q Q 8

Is the pot split? I kept telling my friend it was but he said since player 1 had the higher card in the straight he won. Can anybody find me information(website) to back up either claim?

Player 1 has a full house, that beats a straight by far.

Player 1 doesn't have a straight. He has a full house which beats Player 2's straight. Since you didn't show the suits, I assume Player 2 did not get a straight flush (which would beat the full house).

The cards
 

I messed up what happened but really happened was one person had a queen adn the other person had a 9 and they both had straights. DOes the person with the higher card in his hand even though they have the same straights win or is it a split pot.

From http://www.online-poker-rules.net/handValues.html:

This rule seems to be pretty consistent, so I would say that your friend is mistaken.