October 13th, 2014 at 5:06:37 AM
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I don't know where to ask this, so I thought that it fits best in this Math section. (I'm interested in a ballpark.)

I know that on average, the losing team scores 0.54 goals and the winning team scores 1.63 goals per game.

So my correct score picks will always be 2-0 or 2-1 for the favourite team. (2-1 if the game is considered a close call)

So the question would be: - How likely is to predict 7 correct scores (betting like described above) in a row?

Edit: Edited to correct the average figures.

I know that on average, the losing team scores 0.54 goals and the winning team scores 1.63 goals per game.

So my correct score picks will always be 2-0 or 2-1 for the favourite team. (2-1 if the game is considered a close call)

So the question would be: - How likely is to predict 7 correct scores (betting like described above) in a row?

Edit: Edited to correct the average figures.

October 13th, 2014 at 7:04:56 AM
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There's no way to calculate this with any degree of certainty. What matches were used to determine the averages?

Looking at the EPL and English Championship this season, there have been 202 matches, of which 47 have had 2-0 or 2-1 results. Even if the bet was "predict 7 matches where the final result is 2-0 or 2-1 (regardless of who wins)" and each match had a 1/4 chance of having one of those two results, that would be 1 chance in 16,384.

Looking at the EPL and English Championship this season, there have been 202 matches, of which 47 have had 2-0 or 2-1 results. Even if the bet was "predict 7 matches where the final result is 2-0 or 2-1 (regardless of who wins)" and each match had a 1/4 chance of having one of those two results, that would be 1 chance in 16,384.

October 13th, 2014 at 1:53:13 PM
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I was quite wrong about the averages. According to Wiki Answers it is 0.54 for the losing team and 1.63 for the winning team.

I've just done a bit more research right now and found some relevant statistics about correct scores at Soccer by the Numbers where they found that the most frequent scores for EPL and Serie A are 1-1 and 1-0 (for the home team) and that for Bundesliga the most common score is 1-1 (12.5%) and home team win 2-1 (9.7%) and home team win 1-0 (8.7%). (Data included 5 seasons.)

So according to the above data, I believe the most common scores are 1-1, 1-0 and 2-1. - Now this makes me think that I should pick my correct scores as 1-1 for games considered a close call (rated between 2.00-3.00 decimal odds), 2-1 for favourites with an average attack, and 1-0 for favourites with a below average attack (although those are usually considered close call games).

I know the odds of predicting 7 correct scores in a row are very low, however I would like to find the best bet to go for it as a general rule considering all the data we have. And would also like to obtain a rough idea of what the odds are for this occurrence.

This is particularly interesting to me because for a 20 penny ticket I can win £500,000 if I am the only winner (if there are more winners, then the pot splits equally among the winners). Also if I correctly predict the outcomes but not the scores, I get ~£1 back.

I've just done a bit more research right now and found some relevant statistics about correct scores at Soccer by the Numbers where they found that the most frequent scores for EPL and Serie A are 1-1 and 1-0 (for the home team) and that for Bundesliga the most common score is 1-1 (12.5%) and home team win 2-1 (9.7%) and home team win 1-0 (8.7%). (Data included 5 seasons.)

So according to the above data, I believe the most common scores are 1-1, 1-0 and 2-1. - Now this makes me think that I should pick my correct scores as 1-1 for games considered a close call (rated between 2.00-3.00 decimal odds), 2-1 for favourites with an average attack, and 1-0 for favourites with a below average attack (although those are usually considered close call games).

I know the odds of predicting 7 correct scores in a row are very low, however I would like to find the best bet to go for it as a general rule considering all the data we have. And would also like to obtain a rough idea of what the odds are for this occurrence.

This is particularly interesting to me because for a 20 penny ticket I can win £500,000 if I am the only winner (if there are more winners, then the pot splits equally among the winners). Also if I correctly predict the outcomes but not the scores, I get ~£1 back.

October 15th, 2014 at 7:48:04 AM
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This sounds similar to the Sky Super 6 comp. Free to enter, 250k if you predict 6 correct scorelines. I've always assumed the odds of hitting it are similar to winning the lottery, I only do it cos it's free. I think my record is 3 correct, and rarely get more than 1.

I usually go for 1-0, 1-1, 2-0, 2-1 with a more rare 3-2 or 4-2 thrown in, as you have to share it if there are multiple winners.

I usually go for 1-0, 1-1, 2-0, 2-1 with a more rare 3-2 or 4-2 thrown in, as you have to share it if there are multiple winners.

October 22nd, 2014 at 1:10:49 AM
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Until now I was playing only 2-0 and 2-1 as I said in my first post, and hit 3 correct scores out of 7 many times and even 4 correct scores a couple of times. But now I have also introduced 1-0 scores on my list. Let's see if this improves anything. If I am to win, I don't think I will mind splitting the 500K with 2-3 other guys, as that money will still be about a small fortune for me at any time.

Yes, I am comparing the odds with winning the (6/49) lottery too (1/13M). But I think predicting 7 scores are much more likely?

Yes, I am comparing the odds with winning the (6/49) lottery too (1/13M). But I think predicting 7 scores are much more likely?

October 22nd, 2014 at 2:47:29 PM
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Say the 12,5% you quote for 1:1 is accurate . OR you have 12,5% probability for each selection.

Then the EV of the Bet is = 12,5%^7 *$500.000 = $0.23 compared to a Cost of $0,20 for the Bet.

Ie there is Positive EV of $0,03 per Bet.

Th extra EV on the $1 win is miniscule.

But this is only if you win $500.000.

If it is often that a few players win (say on average 2) then actually you lose half a Bet on average.

Then the EV of the Bet is = 12,5%^7 *$500.000 = $0.23 compared to a Cost of $0,20 for the Bet.

Ie there is Positive EV of $0,03 per Bet.

Th extra EV on the $1 win is miniscule.

But this is only if you win $500.000.

If it is often that a few players win (say on average 2) then actually you lose half a Bet on average.

October 22nd, 2014 at 5:15:04 PM
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Thanks a lot AceTwo ! That would mean a win every 2,100,840 (if I did the equation correctly) and ONLY if the chances would be 12.5% for each score to hit. But as actually probably nobody will pick 1:1 on all 7 scores (because some games will have large favourites where a 1-1 may be very unlikely), I think the chances are roughly between 10% and 12.5% to hit each correct score in part. - So with that in mind, I think the real chances of hitting 7 correct scores are 1 in about 3,000,000 (more or less) probably.

That's anyway about 4 times more likely than winning at the 6/49 lottery (1 in 13M+). On overall, must be still a lousy bet I agree.

By the way ShineyShine the Super 6 Comp sounds great as it seems a 6 correct score prediction should occur 1 in about 200,000!

That's anyway about 4 times more likely than winning at the 6/49 lottery (1 in 13M+). On overall, must be still a lousy bet I agree.

By the way ShineyShine the Super 6 Comp sounds great as it seems a 6 correct score prediction should occur 1 in about 200,000!

October 22nd, 2014 at 7:29:54 PM
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Even better this week, double jackpot 500k... £ Sterling too, none of that weak $ rubbish! Google 'Sky Sports Super 6' to play it.

Still a long shot, but less than i'd guessed, if you're maths is correct.

If you're regularly getting 3/7, you're doing really well. I struggle to get 2/6.

Still a long shot, but less than i'd guessed, if you're maths is correct.

If you're regularly getting 3/7, you're doing really well. I struggle to get 2/6.

October 29th, 2014 at 2:16:13 PM
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The best approach to maxime the chances of winning is NOT to bet on 1:1 which as you say has the highest Average probability of 12,5% on matches in general.

The best approach is to use market information to bet according to the highest probability outcome.

ie, to check waht Odds Bookmakers give on each score (saying using betfair as very good market information) and bet each match according to the score which gives the worst Odds (= Highest Probability).

And using that information you can calculate the probability of winning.

Say Betfair the Worst Odds for a particular match is for 1:0 with Odds 5 for 1 (decimal odds as in Betfair). The probability is the inverse of the decimal odds.

ie 1/5 = 20%.

The best approach is to use market information to bet according to the highest probability outcome.

ie, to check waht Odds Bookmakers give on each score (saying using betfair as very good market information) and bet each match according to the score which gives the worst Odds (= Highest Probability).

And using that information you can calculate the probability of winning.

Say Betfair the Worst Odds for a particular match is for 1:0 with Odds 5 for 1 (decimal odds as in Betfair). The probability is the inverse of the decimal odds.

ie 1/5 = 20%.

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