stud/8: chance of making a low starting with 4 babies

[Nick_Foxx]

i was reading stern's stud/8 section in "championship stud" and he states the probability of making a low when starting with 4 babies (consideration of upcards aside) is approx. 80%

i decided to calculate some #s myself and the answer i'm getting is closer to 70% than 80%:

there are 32 cards 8 or below in the deck... if u have 4 babies by 4th street, there are 4 remaining ranks you can catch to make a low... so 16 "outs" total...

prob. of not hitting one of these 16 on 5th st is 32/48 (2/3)
prob. of catching a brick on 6th st is 31/47
prob. of catching a brick on 7th st is 30/46

prob. of not making a low: 32/48*31/47*30/46=29%
approx. prob of making a low: 100%-29%=71%

what did i do wrong?

mike

[well]

[ QUOTE ]
there are 32 cards 8 or below in the deck... if u have 4 babies by 4th street, there are 4 remaining ranks you can catch to make a low... so 16 "outs" total...

[/ QUOTE ]

...and then there were Aces!

[TommyTutone]

I think you cannot count out the fact that you know more than the value of 4 cards at that point. I'm just shooting from the hip here but 8/13 of the deck is 8 or less, or 32 cards.

At a full table of 8, you will have information on 10 of the cards at the start of the hand, and based on the above information, you can guess that 6 of them will fall into the 8 or less category and 4 will not (and you have to assume three of the 6 are yours for your presumption to be true).

That means on the deal, you have 42 cards unaccounted for, and of those, usually 26 of them are low.

Let's continue hypothetically and say your 4th card is a baby and everyone else's is face down. You now know there are 41 cards left unknown and of them 25 should be low, but only 16 of them will give you the low.

Using those numbers in your equation, your chances of not hitting are now: (25/41)*(24/40)*(23/39)= 21.5%
or you have a 78.5% chance you will hit your low.

[Nick_Foxx]

huh?

[Nick_Foxx]

thanks, makes a lot of sense

[TommyTutone]

[ QUOTE ]
thanks, makes a lot of sense

[/ QUOTE ]

I can't tell if you are serious based on your previous post, but I can try to spell it out a bit more. Like I said, I was kind of improvising as I went. I think the numbers are solid, I just went about it in a halfbaked way.

[Nick_Foxx]

no i was actually genuinely saying "thanks" /images/graemlins/smile.gif

i wonder what the #s would be if you use your 3rd street assumption about the down cards (that approx 8/13 % of them would be babies) on 4th street as well... that is, given that you catch a baby and a certain # of other individuals do as well, what is your adjusted probability of making a low?

i wonder about the 3rd street assumption as well... i know the probability of the other players receiving babies are slightly reduced by the fact that you have 3 of them, but overall their base rate for receiving them on each card should be approx 8/13 %... given this is the case, there would be more than 3 additional babies other than your 3 dealt out on 3rd st..

i'd love to hear your thoughts

mike

[BruceZ]

[ QUOTE ]
i was reading stern's stud/8 section in "championship stud" and he states the probability of making a low when starting with 4 babies (consideration of upcards aside) is approx. 80%

i decided to calculate some #s myself and the answer i'm getting is closer to 70% than 80%:

there are 32 cards 8 or below in the deck... if u have 4 babies by 4th street, there are 4 remaining ranks you can catch to make a low... so 16 "outs" total...

prob. of not hitting one of these 16 on 5th st is 32/48 (2/3)
prob. of catching a brick on 6th st is 31/47
prob. of catching a brick on 7th st is 30/46

prob. of not making a low: 32/48*31/47*30/46=29%
approx. prob of making a low: 100%-29%=71%

what did i do wrong?

mike

[/ QUOTE ]

Nothing. You're completely correct, and the author is wrong. Your answer also agrees with the appendix of Super System p. 604 for imporoving to an 8 low or better starting with A234 (71.32%). Of course in a real situation you would take into account the exposed cards, and this would change your answer, but to get the average probability of making your hand we consider that there are only 4 seen cards, with 32 high and 16 low.

[TommyTutone]

[ QUOTE ]
no i was actually genuinely saying "thanks" /images/graemlins/smile.gif

i wonder what the #s would be if you use your 3rd street assumption about the down cards (that approx 8/13 % of them would be babies) on 4th street as well... that is, given that you catch a baby and a certain # of other individuals do as well, what is your adjusted probability of making a low?

i wonder about the 3rd street assumption as well... i know the probability of the other players receiving babies are slightly reduced by the fact that you have 3 of them, but overall their base rate for receiving them on each card should be approx 8/13 %... given this is the case, there would be more than 3 additional babies other than your 3 dealt out on 3rd st..

i'd love to hear your thoughts

mike

[/ QUOTE ]

Well, I'm a little hungover, but I'm willing to give it a shot. For the sake of argument, let's assume that all 8 players stay in to see 4th street. This is unlikely to happen, and this is part of why I wanted to keep the 4th street cards face down, because this may introduce a larger variance.

That said, based on that assumption, at 4th street, you have information on exactly 18 cards (two for everybody plus two more for your down cards). Continuing with the fact that on average 8/13 of those will be low cards, 11.07 cards (11 is good enough) will be low and 7 will be high, AND you know that 4 of those low cards are yours.

Another assumption that I want to make right now is that you will have fewer than 16 outs to finish your low hand. This is because some of those 7 low cards that don't belong to are outs that you have lost. Let's say for arguments sake that you have lost 4 outs because of this. This means you now have 12 outs to finish your low hand. On the flip side, there are only 34 cards in the deck that you do not have information on.

This means that for you to miss your low, the math becomes (22/34)*(21*33)*(20*32)= 25.7%

So to summarize, just so I have my head on straight:
- you know 18 cards
- you know that 11 of them are low and 4 of those are yours
- of the 7 that are not yours, 4 of them were outs you could have used, but have lost
- all 8 players have gone on to see 4th street

I think this formula is less useful than the original because the original works regardless of how many players stay in for 4th street. In this case, the assumption about all 8 players staying around is a lot less likely to be true.

[AJo Go All In]

all of this is nonsense. the math in the original post is sound.

[TommyTutone]

[ QUOTE ]
all of this is nonsense. the math in the original post is sound.

[/ QUOTE ]

I'm not knocking the original post, but can you explain, at least in my first post, what the problem is? You are not playing in a closed environment and you need to take other factors into consideration in your equation.

I just used averages to try to work the problem out to a practical solution, and the number just happened to come out to what the original poster had read as accurate.

I'm not saying there are not mistakes somewhere along the lines in my thought process, but something beyond 'this is nonsense' would be a bit more constructive and appreciated.

[Nick_Foxx]

using the same logic, i went ahead and calculated your probability of making a low when starting out with 4 unpaired babies given that you know what has appeared on 3rd street...

if none of your outs showed up on 3rd st, you have a 78.4% chance of making your low
1 out - 75.6%
2 outs - 72.6%
3 outs - 69.3%
4 outs - 65.7%
5 outs - 61.9%
6 outs - 57.8%
7 outs - 53.5% (that rare occasion when every doorcard is a card that would've helped make your low)

the most frequently occurring situation is 2-3 of the cards you need are dead... so i think 70% would be a good benchmark number

mike

[AJo Go All In]

[ QUOTE ]
and you need to take other factors into consideration in your equation.

[/ QUOTE ]

no, you don't.