Tips for Tikay

[tikay]

Calling Tikay

I'm ahead of them both here horsey, worry not, I've got it totally sussed.

[tikay]

Leaving aside all this bio-whatsit & reverting to Level One, presumably betting the draw on local derbies such as Man U - Man C, Liverpool - Everton etc (all other things being equal) must generally be value, as these matches are very heavily traded, & the fan bases are especially partisan?

Presumably that was Lesson 7, on Page 2 of Fred.

Bear with me, I'll get there eventually.

[Eso Kral]

yes chis v ando sorry to not confirm the actual match

Would love eso to coach me on how to run like him.  I would be retired by now!

Welcome back fish, stick around pls as this time next year...

[redarmi]

I have used binomial distribution for these before and they come out as value for sure but can someone with better maths knowledge verify that the approach is valid here.  Like Mark says with the throw an individual player is actually more like 65/35 so by calculating on the basis of the average price it isn't strictly speaking correct but now, thinking about it, of more importance is the fact that you calculate binomial distribution on the basis of a known number of trials (in this case 12) but that isn't the case here because if a player reaches 7 games they stop and also we aren't allowing for things like momentum especially when you get to 5-0, 6-1, 6-0 type scores or the fact it isn't actually a flip in this spot (I would guess the legs are more like 70/30 and 60/40).  I think they probably are value but I don't think the correct price is 7/2.  At best it is a shade under 4/1 I think which is basically the price at the moment.

agree with this and would be open to more advanced maths feedback to confirm the above although the lack of exactly 12 trials surely doesnt matter because we are betting on 6 of each so once 7 of one happens our bet is lost.   I agree though the bigger factor is the momentum factor at 5-0 6-1 etc as to whether the legs are still priced as pre game.

Yeah I realised this as soon as I posted it but then I decided to have a simplistic play around with some formats.  If we assume it was just a two legged match then using binomial distribution as a true flip then the chance of 1-1 would be 50%.  If we assume for ease that it is 65/35 to both players with the throw.  Then the chance of either player winning 2-0 is (0.65*0.35)*2=45.5% of a player winning 2-0 so actually in that spot the draw would be more likely than the binomial distribution but if you change the advantage for the throw it changes the probabilities quite drastically so if the advantage for the throw is 55/45 then it becomes (0.55*0.45)*2 which is closer to even.  Irrespective I am not sure that binomial distribution works if it won't work for such a simple two legs approach.

[Doobs]

and in order to pay for the offer they have now made half the field in the race bottom price on oddschecker.

http://www.oddschecker.com/cheltenham-festival/supreme-novices-hurdle/

Suppose there has to be a bit of give and take for them to do the offer, here's the t and c's

Cheltenham Money-Back Special
We will refund stakes on the 2nd, 3rd & 4th placed horses in the Supreme Novices Hurdle!
   Applies to 13:30 at Cheltenham on March 11th.
   Applies to 'Win or Each Way' market only. Applies to Win part of Each Way bets only
   Applies to singles only.
   Maximum refund £/€50 per customer.
   Does not include tote bets or in-running bets.
   Applies to bets placed from 8pm on February 26th.
   Paddy Power horse racing rules apply.
   Paddy Power reserves the right in its discretion to exclude and/or limit the availability of the offer to certain customers.

They are also NRNB, although they have gone 9/4 surely Irving is the bet for those that can? Thoughts?

Yes, even at 9/4 Irving is by far the best bet for this offer.

I wish to retract that. This is a little more complex than the standard refund if offers. I'll take a look at it later.

It is a no brainer, it is better than the standard free bet on 2nd.  I have pretty much said as much already.

You get slightly lower odds but money back on 2nd, 3rd and 4th.  The maths is just going to be so compelling I really don't need to do it.  Obviously don't go over the £50 limit.

The only problem Tikay is going to have is if he is so restricted and/or doesn't get the offer.  They restricted me a bit, so guess they'll give Tikay £15 or so.  Paddy Power seem to really hammer me on some of their offers, but weren't so bad here.

  

Yes the offer is obviously great. But is there more EV in Irving 3.25/3.8 or Wicklow Brave 7/8 ?
Your thought would be appreciated please Doobs

Yes I think so, I think Irving is more likely to finish 2nd to 4th than Wicklow Brave.  Probably 5/4 vs 7/4 at a guess.   I don't think there is that much in it in this case, as you lose more on the win side with Irving, but gain more on the place side.   I think the place bit is more than likely big enough to outweigh the win side.  Even without this, I'd always go the lower variance route if there wasn't much between them, which would be Irving.  

I wouldn't worry about getting exactly the right bet, as either is going to be very good value.  You could probably do £25 on each if you wanted very low variance and weren't restricted.

Thank you Doobs

My thought were that for the refund if 2nd offers we pick low odds close to Betfair.

For the refund on lose offers we chose high odds close to Betfair.

This offer appears to sit midway between the two

Been in and out all day.  

A 5/2 horse is significantly more likely to come 2nd than the 7/1 horse.  By about 4th place they are probably equally likely to finish there.  I'd include a couple if pictures of distributions here, but some of the members get a little overexcited if they see too much stats porn in one day.  I don't think it would be fair on Jeeves to have to tidy up after that.

[doubleup]

I have used binomial distribution for these before and they come out as value for sure but can someone with better maths knowledge verify that the approach is valid here.  Like Mark says with the throw an individual player is actually more like 65/35 so by calculating on the basis of the average price it isn't strictly speaking correct but now, thinking about it, of more importance is the fact that you calculate binomial distribution on the basis of a known number of trials (in this case 12) but that isn't the case here because if a player reaches 7 games they stop and also we aren't allowing for things like momentum especially when you get to 5-0, 6-1, 6-0 type scores or the fact it isn't actually a flip in this spot (I would guess the legs are more like 70/30 and 60/40).  I think they probably are value but I don't think the correct price is 7/2.  At best it is a shade under 4/1 I think which is basically the price at the moment.

agree with this and would be open to more advanced maths feedback to confirm the above although the lack of exactly 12 trials surely doesnt matter because we are betting on 6 of each so once 7 of one happens our bet is lost.   I agree though the bigger factor is the momentum factor at 5-0 6-1 etc as to whether the legs are still priced as pre game.

Yeah I realised this as soon as I posted it but then I decided to have a simplistic play around with some formats.  If we assume it was just a two legged match then using binomial distribution as a true flip then the chance of 1-1 would be 50%.  If we assume for ease that it is 65/35 to both players with the throw.  Then the chance of either player winning 2-0 is (0.65*0.35)*2=45.5% of a player winning 2-0 so actually in that spot the draw would be more likely than the binomial distribution but if you change the advantage for the throw it changes the probabilities quite drastically so if the advantage for the throw is 55/45 then it becomes (0.55*0.45)*2 which is closer to even.  Irrespective I am not sure that binomial distribution works if it won't work for such a simple two legs approach.

wouldn't a tennis set probability of going to a tie break be exactly the calc?  That must be somewhere on the net

[redarmi]

I have used binomial distribution for these before and they come out as value for sure but can someone with better maths knowledge verify that the approach is valid here.  Like Mark says with the throw an individual player is actually more like 65/35 so by calculating on the basis of the average price it isn't strictly speaking correct but now, thinking about it, of more importance is the fact that you calculate binomial distribution on the basis of a known number of trials (in this case 12) but that isn't the case here because if a player reaches 7 games they stop and also we aren't allowing for things like momentum especially when you get to 5-0, 6-1, 6-0 type scores or the fact it isn't actually a flip in this spot (I would guess the legs are more like 70/30 and 60/40).  I think they probably are value but I don't think the correct price is 7/2.  At best it is a shade under 4/1 I think which is basically the price at the moment.

agree with this and would be open to more advanced maths feedback to confirm the above although the lack of exactly 12 trials surely doesnt matter because we are betting on 6 of each so once 7 of one happens our bet is lost.   I agree though the bigger factor is the momentum factor at 5-0 6-1 etc as to whether the legs are still priced as pre game.

Yeah I realised this as soon as I posted it but then I decided to have a simplistic play around with some formats.  If we assume it was just a two legged match then using binomial distribution as a true flip then the chance of 1-1 would be 50%.  If we assume for ease that it is 65/35 to both players with the throw.  Then the chance of either player winning 2-0 is (0.65*0.35)*2=45.5% of a player winning 2-0 so actually in that spot the draw would be more likely than the binomial distribution but if you change the advantage for the throw it changes the probabilities quite drastically so if the advantage for the throw is 55/45 then it becomes (0.55*0.45)*2 which is closer to even.  Irrespective I am not sure that binomial distribution works if it won't work for such a simple two legs approach.

wouldn't a tennis set probability of going to a tie break be exactly the calc?  That must be somewhere on the net

Possibly but you would have to have the stats on two players that are equally matched which would be harder to find.  Also think serve strength is much more likely to vary than who throws first which should confer relatively equal advantage.

[arbboy]

I have used binomial distribution for these before and they come out as value for sure but can someone with better maths knowledge verify that the approach is valid here.  Like Mark says with the throw an individual player is actually more like 65/35 so by calculating on the basis of the average price it isn't strictly speaking correct but now, thinking about it, of more importance is the fact that you calculate binomial distribution on the basis of a known number of trials (in this case 12) but that isn't the case here because if a player reaches 7 games they stop and also we aren't allowing for things like momentum especially when you get to 5-0, 6-1, 6-0 type scores or the fact it isn't actually a flip in this spot (I would guess the legs are more like 70/30 and 60/40).  I think they probably are value but I don't think the correct price is 7/2.  At best it is a shade under 4/1 I think which is basically the price at the moment.

agree with this and would be open to more advanced maths feedback to confirm the above although the lack of exactly 12 trials surely doesnt matter because we are betting on 6 of each so once 7 of one happens our bet is lost.   I agree though the bigger factor is the momentum factor at 5-0 6-1 etc as to whether the legs are still priced as pre game.

Yeah I realised this as soon as I posted it but then I decided to have a simplistic play around with some formats.  If we assume it was just a two legged match then using binomial distribution as a true flip then the chance of 1-1 would be 50%.  If we assume for ease that it is 65/35 to both players with the throw.  Then the chance of either player winning 2-0 is (0.65*0.35)*2=45.5% of a player winning 2-0 so actually in that spot the draw would be more likely than the binomial distribution but if you change the advantage for the throw it changes the probabilities quite drastically so if the advantage for the throw is 55/45 then it becomes (0.55*0.45)*2 which is closer to even.  Irrespective I am not sure that binomial distribution works if it won't work for such a simple two legs approach.

wouldn't a tennis set probability of going to a tie break be exactly the calc?  That must be somewhere on the net

Possibly but you would have to have the stats on two players that are equally matched which would be harder to find.  Also think serve strength is much more likely to vary than who throws first which should confer relatively equal advantage.

very much this in tennis.  The model is much more simple in darts. 

[doubleup]

I was thinking more of a formula where you input the prob of each player holding serve.  Sure it must be out there.

[Tal]

Forget Rubbish's scantily clad ball girls and tickly gentleman vegetable jokes. Feast your eyes on this top shelf literature:

http://heavytopspin.com/category/serve-statistics/

http://plus.maths.org/content/anyone-tennis

[Tal]

If that was a suggestive look across a crowded room, this is the full scale stuff that gets your hard drive confiscated:

http://www.tennisabstract.com/

[RobS]

I have used binomial distribution for these before and they come out as value for sure but can someone with better maths knowledge verify that the approach is valid here.  Like Mark says with the throw an individual player is actually more like 65/35 so by calculating on the basis of the average price it isn't strictly speaking correct but now, thinking about it, of more importance is the fact that you calculate binomial distribution on the basis of a known number of trials (in this case 12) but that isn't the case here because if a player reaches 7 games they stop and also we aren't allowing for things like momentum especially when you get to 5-0, 6-1, 6-0 type scores or the fact it isn't actually a flip in this spot (I would guess the legs are more like 70/30 and 60/40).  I think they probably are value but I don't think the correct price is 7/2.  At best it is a shade under 4/1 I think which is basically the price at the moment.

agree with this and would be open to more advanced maths feedback to confirm the above although the lack of exactly 12 trials surely doesnt matter because we are betting on 6 of each so once 7 of one happens our bet is lost.   I agree though the bigger factor is the momentum factor at 5-0 6-1 etc as to whether the legs are still priced as pre game.

Yeah I realised this as soon as I posted it but then I decided to have a simplistic play around with some formats.  If we assume it was just a two legged match then using binomial distribution as a true flip then the chance of 1-1 would be 50%.  If we assume for ease that it is 65/35 to both players with the throw.  Then the chance of either player winning 2-0 is (0.65*0.35)*2=45.5% of a player winning 2-0 so actually in that spot the draw would be more likely than the binomial distribution but if you change the advantage for the throw it changes the probabilities quite drastically so if the advantage for the throw is 55/45 then it becomes (0.55*0.45)*2 which is closer to even.  Irrespective I am not sure that binomial distribution works if it won't work for such a simple two legs approach.

I think the probability distribution required is the Poisson Binomial Distribution.

However, if with two evenly matched players 65% on their throw, if the first leg thrower goes 3-0 up in legs (which will happen 15% of the time with the assumptions), then there is no way the opponent is close to 65% to win the fourth leg is there?

[arbboy]

I have used binomial distribution for these before and they come out as value for sure but can someone with better maths knowledge verify that the approach is valid here.  Like Mark says with the throw an individual player is actually more like 65/35 so by calculating on the basis of the average price it isn't strictly speaking correct but now, thinking about it, of more importance is the fact that you calculate binomial distribution on the basis of a known number of trials (in this case 12) but that isn't the case here because if a player reaches 7 games they stop and also we aren't allowing for things like momentum especially when you get to 5-0, 6-1, 6-0 type scores or the fact it isn't actually a flip in this spot (I would guess the legs are more like 70/30 and 60/40).  I think they probably are value but I don't think the correct price is 7/2.  At best it is a shade under 4/1 I think which is basically the price at the moment.

agree with this and would be open to more advanced maths feedback to confirm the above although the lack of exactly 12 trials surely doesnt matter because we are betting on 6 of each so once 7 of one happens our bet is lost.   I agree though the bigger factor is the momentum factor at 5-0 6-1 etc as to whether the legs are still priced as pre game.

Yeah I realised this as soon as I posted it but then I decided to have a simplistic play around with some formats.  If we assume it was just a two legged match then using binomial distribution as a true flip then the chance of 1-1 would be 50%.  If we assume for ease that it is 65/35 to both players with the throw.  Then the chance of either player winning 2-0 is (0.65*0.35)*2=45.5% of a player winning 2-0 so actually in that spot the draw would be more likely than the binomial distribution but if you change the advantage for the throw it changes the probabilities quite drastically so if the advantage for the throw is 55/45 then it becomes (0.55*0.45)*2 which is closer to even.  Irrespective I am not sure that binomial distribution works if it won't work for such a simple two legs approach.

I think the probability distribution required is the Poisson Binomial Distribution.

However, if with two evenly matched players 65% on their throw, if the first leg thrower goes 3-0 up in legs (which will happen 15% of the time with the assumptions), then there is no way the opponent is close to 65% to win the fourth leg is there?

why not?  the other guy has held 'serve' in 2 out of the 3 legs and broke the other guy once.  Why does that suddenly change the likelihood of him holding in the 4th leg 3-0 down?

[RobS]

I think the probability distribution required is the Poisson Binomial Distribution.

However, if with two evenly matched players 65% on their throw, if the first leg thrower goes 3-0 up in legs (which will happen 15% of the time with the assumptions), then there is no way the opponent is close to 65% to win the fourth leg is there?
[/quote]

why not?  the other guy has held 'serve' in 2 out of the 3 legs and broke the other guy once.  Why does that suddenly change the likelihood of him holding in the 4th leg 3-0 down?
[/quote]

Tilt, Lack of confidence, the fact the opponent is showing better form on this particular evening, some combination of these factors?

Anderson for example often looks like and throws like he would rather be back in his pub once he goes significantly behind in a match.

It has to be some number lower than 65%, how much lower I don't know, others will have a better estimate of it than me.

[arbboy]

I think the probability distribution required is the Poisson Binomial Distribution.

However, if with two evenly matched players 65% on their throw, if the first leg thrower goes 3-0 up in legs (which will happen 15% of the time with the assumptions), then there is no way the opponent is close to 65% to win the fourth leg is there?

why not?  the other guy has held 'serve' in 2 out of the 3 legs and broke the other guy once.  Why does that suddenly change the likelihood of him holding in the 4th leg 3-0 down?
[/quote]

Tilt, Lack of confidence, the fact the opponent is showing better form on this particular evening, some combination of these factors?

Anderson for example often looks like and throws like he would rather be back in his pub once he goes significantly behind in a match.

It has to be some number lower than 65%, how much lower I don't know, others will have a better estimate of it than me.
[/quote]

Not looking for a ruck here btw just putting it out there i would argue he is as likely to be more focused as he knows if he doesn't win this leg he is 65% likely to be 5-0 down.  I just think these legs are pretty much independent at the early stages of the match.