Re: [vpFREE] Re: A Keno Math Problem

 

This is what happens when you hit a solid 4 in FME/FE/FM modes when the meters are showing:

Free Games....8
Multiplier..........8
Extra Draws....8

The key is knowing what the extra draws are worth. At 20 draws the payscale is worth 50% or half a bet. But with 28 draws the payscale is worth 1.6 bets. Then there is the adjustment for the number of games you are in each mode. Per 78.6 cycles you are in FME mode 31.8% of the time; in FE mode 31.8% of the time; and FM mode 36.4% of the time. So getting the average values looks like this:

fFree Games-Multiplier-Extra Draws
1.6X8X8x31.8 = 3256.32

If you hit in Free Games-Extra Draws mode you get a 2X multiplier.

1.6X2X8X31.8 = 814.08

If you hit in Free Games-Multiplier mode you get 2 extra draws which puts the base payscale at 68.7%.

.687X8X8X36.4 = 1600.44

3256.32 plus 814.08 plus 1600.44 totals to 5670.84.

5670.84/100 means the average return is 56.7 bets. Since the cost to play the game off is 65 bets this would be an 8.29 bet deficit.

8.29/369 = 2.25% which would put the game at 97.75%.But can I play an 8-8-8 anyway? I'll get to that in the next post.

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Posted by: mickeycrimm@yahoo.com
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