Razz - Working out the percentages - seventh street

[Mathew82]

Instead of using a simulator, I thought I would learn how to do this manually.

Unfortunately I do not come up with the exact same answer as the simulator.

I created this hand

Player 1 - KT9832
Player 2 - QJ872A

So right now player 1 has the best hand...

If player 2 makes an 8 then player 1 can't win and since there are 4 sixes, 4 fives, 4 fours and 3 threes left in the deck totaling 15 of the forty remaining cards. Therefore 15/40 x 100 = 37.5% of the time player 2 wins.

I then took the chance of hitting a nine or a ten with the remaining three tens and three nines making 6 outs in the 40 remaining cards. Now I had to work out what the chance player 1 would have of missing making his 9 to win. To hit his nine, he would have to hit one of the 3 sevens, 4 sixes, 4 fives, 4 fours, or 3 Aces equaling 18 out of the 39 remaining cards therefore the chances of him missing is 18-38= 21/39

So therefore 6/40 x 21/39 x 100 = 8.077%

So adding these percentages together, 37.5 + 8.077 = 45.577%

So the chances of player 2 winning is 45.577%

Except when I run a simulation I get 46.15%


Edited title to reflect that we're talking about razz.

[Andy B]

I think that you didn't account for this scenario:

Player 1 - KT98327
Player 2 - QJ872A9

[Mathew82]

Ahhhh... no I didn't...

So let me split this up

If player 2 hits a ten 3/40 and him then missing his nine is 21/39 the chance of this occuring is 3/40 x 21/39 = 63/1560 x 100 = 4.03846%

If player 2 hits a nine 3/40 and him then missing a better nine is 24/39 - the chance of this occuring is 3/40 x 24/39 = 72/1560 x 100 = 4.61538%

So 37.5% + 4.03846% + 4.61538% = 46.15384%

Cool !

[Andy B]

Y'know, the simulator is a lot easier and you're less likely to make a mistake. [img]/images/graemlins/grin.gif[/img]

[RustyBrooks]

And now you know why we all like plug-n-chug.

I think it's great that you're going for the math though. I've been working on short-hand for razz, things like... if I can catch up to his current hand in one street X% of the time, but can not get a lock on him (he can draw a card to improve that will beat my improvement), how often will I win (see Prax's recent tournament hand post for an example). Obviously this is complicated by exposed cards and stuff but I'm interested in the "average" solution because honestly I don't think I can calculate it every time. A short hand for this partiucular problem is, how ofen do I draw a card beneath N, which is lower than the card my opponent draws.

[Mathew82]

[ QUOTE ]
Y'know, the simulator is a lot easier and you're less likely to make a mistake. [img]/images/graemlins/grin.gif[/img]

[/ QUOTE ]

True, but I like to know I can do it without one too [img]/images/graemlins/smile.gif[/img]

[Praxising]

"Oh, Professors? Yoo-hooooooo!"

prax waves arm in air at back of room - teachers exchange a long-suffering look

"How useful is all this when all those cards aren't actually in the deck any longer? I mean, don't you have to toss in a few dead cards just to make it realistic?"

[Mathew82]

All I want to do is to find out how to work these things out. It might not be that useful other than by solving alot of these puzzles that you have a greater intuition about the percentages. I plan to do alot of this type of maths.

This hand could very conceivably be a heads up hand if the order of the cards appearing was right...

[RustyBrooks]

It's a little tough to wrap your head around, and it's a very common mistake. I once played poker (holdem) with someone who felt that if he had 4 outs on the turn, that his odds of making his hand were 4/28, because there were already 20 cards dealt face down and 4 dealt face up. I did not correct him.

When there is a card you are looking for, you never know if it's in the deck or in the cards that have already been dealt. All you know is that it is one of the cards that is "unknown" - this means already dealt or in the deck. But the thing is, there is basically no difference between a card being already dealt, and it being at the bottom of the deck, where it can never be dealt. So think of the unknown cards, dealt and undealt, as being one big deck where the already dealt cards are at the bottom. Sometimes the card you're looking for is on top of this deck, sometimes it's #2, sometimes it's #2, sometimes it's #10, etc, on average it's likelihood of coming up next is K/N where K is the nubmer of cards you're looking for, and N is the number of unknown cards.

[RustyBrooks]

I feel the same way, by the way... I always want to know how to do something, even if, in practice, I won't be doing it myself. Sometimes it really does help.