Extra Draw Keno
January 12th, 2012 at 2:07:53 AM
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Hi,
Can anyone explain the Maths for Extra Draw Keno on this page:
I can work out the basic return for 20 balls out of 80, using the equation (x:y)*(80-x:20-y)/(80:20)
But I can't work out the return from the extra 3 balls. I tried using the above equation with 60 and 3 instead of 80 and 20, working out the probability of getting 1, 2 or 3 extra balls and multiplying by the right payout, but it doesn't match the Wizards figures. I also divided the extra return by 2 due to the extra stake.
Thanks for any help!
Can anyone explain the Maths for Extra Draw Keno on this page:
I can work out the basic return for 20 balls out of 80, using the equation (x:y)*(80-x:20-y)/(80:20)
But I can't work out the return from the extra 3 balls. I tried using the above equation with 60 and 3 instead of 80 and 20, working out the probability of getting 1, 2 or 3 extra balls and multiplying by the right payout, but it doesn't match the Wizards figures. I also divided the extra return by 2 due to the extra stake.
Thanks for any help!
January 12th, 2012 at 8:52:52 AM
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Let's just look at the simplest case: 3 spot
1. Using the formula stated above, these are the probabilities of hitting n spots:
2. After the first 20 draws, you have 60 open spots on the board and you have 3 draws (should you choose to do so). The wizard's table only shows that continuing is a good deal if you already hit 2 spots. So, this is the only one I will look at. There are only possible outcomes: you don't get the third spot or you do hit the third spot. The probability of hitting the additional spot is (x:y)*(60-x:3-y)/(60:3), where x and y are both 1 (p=0.05).
3. Calculate the return: r = p(hitting 2 spots in original 20 draws) * [p(hitting 1 additional spot in the next 3 draws) * increase in win - cost of additional wager]
r = 0.138753651 * [0.05 * (30-3) - 1] = 0.048563778
1. Using the formula stated above, these are the probabilities of hitting n spots:
| spots hit | probability |
|---|---|
| 0 | 0.416504382 |
| 1 | 0.430866602 |
| 2 | 0.138753651 |
| 3 | 0.013875365 |
2. After the first 20 draws, you have 60 open spots on the board and you have 3 draws (should you choose to do so). The wizard's table only shows that continuing is a good deal if you already hit 2 spots. So, this is the only one I will look at. There are only possible outcomes: you don't get the third spot or you do hit the third spot. The probability of hitting the additional spot is (x:y)*(60-x:3-y)/(60:3), where x and y are both 1 (p=0.05).
3. Calculate the return: r = p(hitting 2 spots in original 20 draws) * [p(hitting 1 additional spot in the next 3 draws) * increase in win - cost of additional wager]
r = 0.138753651 * [0.05 * (30-3) - 1] = 0.048563778
I heart Crystal Math.
January 16th, 2012 at 12:50:12 AM
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Sorry I've only just had chance to come back to this - thanks for the explanation!
November 1st, 2012 at 5:48:18 AM
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Hi, hopefully this post is still live.
Will it make any difference if there are 2 extra balls instead of 3 for the probability table you made?
Thanks!
Will it make any difference if there are 2 extra balls instead of 3 for the probability table you made?
Thanks!
November 1st, 2012 at 6:02:05 AM
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You just need to change the formula to have a 2 instead of a 3, so the probabilities will change. But, the table above only shows the probabilities before the extra draws, so that part will not change.
I heart Crystal Math.
November 1st, 2012 at 7:03:31 AM
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Thanks for your quick reply. Just seems to be a little more math than I can take ....
I'm trying to find out which strategy would give the best odds. 80 ball keno, 20 ball draw, pick from 4-10 numbers.
From the greatI can play around with the numbers, only leaves me puzzled how these are effected when 2 xtra numbers are drawn.
I'm trying to find out which strategy would give the best odds. 80 ball keno, 20 ball draw, pick from 4-10 numbers.
From the greatI can play around with the numbers, only leaves me puzzled how these are effected when 2 xtra numbers are drawn.
