Anyone tell me the odds of.....

SCHOF78

losing 5 consecutive flips with a pair vs two overs???

55 v a9
55 v aq
66 v aq
55 v aj
88 v j10

ive had the pair every time, its not wonder people think there is something dodgy about sky somtimes.

Pysterman

In Response to Anyone tell me the odds of.....:

losing 5 consecutive flips with a pair vs two overs??? 55 v a9 55 v aq 66 v aq 55 v aj 88 v j10 ive had the pair every time, its not wonder people think there is something dodgy about sky somtimes.
Posted by SCHOF78

It depends on the exact suit configurations in each hand.

But, the hands above have winning probabilities of roughly... 55%, 55%, 55%, 54% and 53% respectively.

So the odds of all of these losing in a row are...

0.45³ x 0.46 x 0.47 = 0.0197...    about 2%

Goethe

In Response to Anyone tell me the odds of.....:

losing 5 consecutive flips with a pair vs two overs??? 55 v a9 55 v aq 66 v aq 55 v aj 88 v j10 ive had the pair every time, its not wonder people think there is something dodgy about sky somtimes.
Posted by SCHOF78

Depends at what point you committed. Assuming you both committed pre-flop . . .

The chances of the overcards pairing are 6/50 (number that can pair over the unseens) multiplied by 5/1 (number of opportunities to draw said cards) - so 30/50 or 60% chance of your opponent pairing. The chances of you pulling a set, 2/50 x 5/1 = 10/50 or 10%. The coincidence of these two events is 10% x 60% = 6%, or about 16-1. So the chances of your opponent hitting a pair and you not drawing to a set, so they end up with the better hand, are 54% (60% - 6%, see above).

Therefore the chances of you having a winning hand are 46% - the diffence between 100% and 54%.
This is the sum of:
(a) hitting your set (10%)
(b) hitting a set when oppo hits an overpair (6%)
(c) neither hitting in which case the pocket pair carries (40%)

There are also other potential draws such as straights or flushes etc and the figures above will differ depending on what the two overcards were, ie J,10 has a greater chance of drawing to a flush than A9. But let's stick with just the potential overpair/set contest for this.

So virtually a coin flip really. If you round the rough win probability above up to 50%, or 1/2, then 1/2 to the power of five (1/(2^5) = 1/32, or 31-1. High, but everyone playing at Sky will have experienced losing out to longer shots.

Moral of the story? The smaller the pair, the greater the chance that the oppo(s) will have two overcards, which makes going All-In a coin flip gamble. Profound question I suppose is whether you need to go All-In pre-flop (raising or calling?) - the answer is, of course, it depends.

Good cards.

Goethe

Two answers. . .  that's always a worry.

I don't think the possibility of flipping a coin and getting five heads in a row is 2/100 though.

Pysterman

In Response to Re: Anyone tell me the odds of.....:

Two answers. . .  that's always a worry. I don't think the possibility of flipping a coin and getting five heads in a row is 2/100 though.
Posted by Goethe

Several points...

I used computer simulated estimates (fairly accurate nonetheless) to find the probability of 55% for 55 vs A9. The reason I did this is that it is extremely complicated to consider all of the ways that each hand can win and not practical to work out by hand.

I believe there are flaws in your maths.

For example, the way you calculated the probability of your opponent making a pair...

Yes, he has 6 outs and 5 chances to hit, but the probability is not simply 6/50 x 5. Firstly, there are 48 cards left, not 50. Secondly, simply multiplying by 5 does not take into account all the ways in which he may hit 2 pairs, trips or quads with one or both cards.

Think of it this way... If he had 10 outs instead of 6, what would be the probability that he makes a pair, by your reasoning? Your maths would say (10/50) x 5, which equals 50/50 - certainty! Clearly, even with 10 outs he is not certain to hit one.

The way to work this out is to find the probability that he misses with all 5 cards...

42/48 x 41/47 x 40/46 x 39/45 x 38/44 = 0.4968

So, the probability that he pairs at least one card is 0.5032 - roughly 50%.

However, even if he hits, it doen't mean that the pocket pair loses. There are lots of other things that could happen, which come together to make the underpair about 55% favourite in the case of 55 v A9.

Also...

You are correct, the probability of 5 heads in a row is not 2% (it's 3.125%), but it would be if the coin were biased around 54% in favour of tails.

Goethe

In Response to Re: Anyone tell me the odds of.....:

... Yes, he has 6 outs and 5 chances to hit, but the probability is not simply 6/50 x 5. Firstly, there are 48 cards left, not 50.  .  . .
Posted by Pysterman

Yeah . . . but if you raise or call All-In pre-flop there are fifty unseens to take account of at the point of commitment (you only know what your own hole cards are at that stage).

You're right . . . the rationale isn't mathematically spot on, but is simplified to make it easier to do some sums at the table (don't know about you but I struggle to multiple two figures to four decimal places together on the fly), and they don't take account of every possibility (in the case above I've focussed on the four common outcomes - player 1 makes a set, player 2 makes an overpair, both players draw to a set/overpair, nothing hits the board). So there's going to be a margin of error in there, and your ten outs example highlights this. But materially, it won't be a huge margin of error, and I'd be surprised if it affected anyone's decision making (even where they calculate it). You've used a computer to calculate the answer, and I've just answered the query applying a workable rationale, typed straight in and allowing for some rounding at the end. Perhaps I ought to put a note at the end stating there some +/- in future (although I've never worked out what the margin of error actually is - I'd need a computer for that!)

On the last bit, I've had a bollo##ing off of my wife for being pedantic over the decimal places, and in hindsight she's right (as she always is) - so apologies for that, and to everyone who may have been misled by my post.

Good cards.