odds question

[JimD82 0]

What are the odds of losing 10 hands within a span of only 100 hands played where you are 90% or better to win on the flop? I know an exact figure probably isn't possible, but I'm curious where this one falls.

[custom36 4]

1 in 5.

[JimD82 0]

should've phrased the question better. Within a 100 hands, there are 10 played where the player is in on the flop a 90% or better favorite to win. The other 90 hands are relevant only for the sake of frequency. What are the odds the player who is the favorite will lose all 10 of those hands.

[93transam 0]

i was never good at math.this post makes me want to cry.

[JimD82 0]

I know it's a complicated a equation and I haven't taken a math course in 4 years so that's why I'm asking. Thought maybe someone might be smart enough to figure it out. I was all reading and writing in school.

[Sushiman 0]

This is gonna be some extremely rough guesstimating math but:90%+ fav to win on the flop usually means your opponent is drawing to 2 outs or runner runner. In the case of 2 outs thier chance to catch out of 45 remaining cards in a deck is roughly 0.067. 10 times in a row would be 0.067^10 or 1.8E-12. This is assuming the flop situation is something like you have the opponent's hand dominated and you both catch a pair on the flop.

[Municipal Hare 0]

Let's say one in ten billion.

[ajs1281 0]

If pokerstars then: 75%reality: you should win 9 of 10 times each time, my math, probably bad but its late and im really lazy.You would lose.02% 10 times in a rowEDIT........ok that doesnt look right at all, I probably forgot a decimal somewhere, I always mess up some mundane detail."I cannot believe what a bunch of losers we are. We're looking up 'money laundering' in the dictionary!"

[Stylin_Fish 0]

Pokerroom=100%

[Dubey 1,035]

uhh .10^10, no?.0000000001%

[troyomac 0]

uhh .10^10, no?.0000000001%

agreed. for those of you that don't know what this means.....10*.10*.10*.10*.10*.10*.10*.10*.10*.10

[Municipal Hare 0]

uhh .10^10, no?.0000000001%

agreed. for those of you that don't know what this means.....10*.10*.10*.10*.10*.10*.10*.10*.10*.10

Let's say one in ten billion.

[Drwnded 0]

uhh .10^10, no?.0000000001%

This is the correct answer, because to calculate the probability of several events occuring, you just multiply the individual probabilities together. This ends up being useful to know in a lot of different holdem situations when figuring pot odds. For example, let say you're trying to figure whether to call an all-in bet when you're holding the nut four-flush with spades after the flop. Assuming you win if you make your flush, then the probability of not hitting a spade on the turn is (about)75%, and likewise the chance on not hitting a spade on the river is 75%. Thus the chance of not making your flush is .75 X .75 = .56, meaning your chance of hitting at least one spade by the river is 44%. Then you can calculate whether the pot is laying you enough odds to make the call.

[JimD82 0]

Thansk for the replies. This actually happened to me last might. I lost to 9 two outers and head a set lose to a backdoor straight within a half hour of 4 tabling. Not looking for sympathy, but it seemed like a pretty statistically significant string of beats. I thought I might have a better chance of floppign a royal.

[Municipal Hare 0]

Thus the chance of not making your flush is .75 X .75 = .56, meaning your chance of hitting at least one spade by the river is 44%. Then you can calculate whether the pot is laying you enough odds to make the call.

Your reasoning is perfect, but your guesstimate of .75 has thrown the result. You must count a four-flush as 9 outs on the turn, 9 outs again on the river.(38/47)*(37/46) = 65% to miss, so 35% to hit.Huge difference.

[Drwnded 0]

Thus the chance of not making your flush is .75 X .75 = .56, meaning your chance of hitting at least one spade by the river is 44%. Then you can calculate whether the pot is laying you enough odds to make the call.

Your reasoning is perfect, but your guesstimate of .75 has thrown the result. You must count a four-flush as 9 outs on the turn, 9 outs again on the river.(38/47)*(37/46) = 65% to miss, so 35% to hit.Huge difference.

No argument - I should've been more precise, but I was really just trying to illustrate that math behind calculating the probability of multiple events occuring. You're right, big difference.