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Death Mage
Raving Lunatic

 Posted: Sun Feb 22, 2009 11:18 pm    Post subject: 1 A video bingo simulation: A card with the numbers 1-24 randomly placed, once each, in a 5x5 grid, with the center square a free space. 10 random numbers, 1-24, are each drawn once. Prizes are given out based on how quickly someone scores a bingo, paid out directly based on how much was put in, \$1-\$500. Draws to Bingo - Payout 4 - x1000 5 - x250 6 - x100 7 - x50 8 - x10 9 - x5 10 - x2 Is the advantage house, or player? What is the advantage? How can the player maximize their payout? I do NOT have the answer to these questions. It's been way too long since I did probability work. I'm curious to see the answers though, and I hope someone here finds this challenge entertaining._________________* These senseless ramblings brought to you by Insanity™. If you just can't figure the dang thing out, it must be Insanity™. [YOUR AD HERE!]
Zag
Tired of his old title

 Posted: Sun Feb 22, 2009 11:56 pm    Post subject: 2 Here's a histogram of 1,000,000 trials: Wins in 4 balls: 383 Wins in 5 balls: 1745 Wins in 6 balls: 4677 Wins in 7 balls: 10343 Wins in 8 balls: 19508 Wins in 9 balls: 33675 Wins in 10 balls: 53886 Wins in 11 balls: 79160 Wins in 12 balls: 107548 Wins in 13 balls: 135197 Wins in 14 balls: 153042 Wins in 15 balls: 151846 Wins in 16 balls: 125272 Wins in 17 balls: 80310 Wins in 18 balls: 34629 Wins in 19 balls: 8194 Wins in 20 balls: 585
Zag
Tired of his old title

 Posted: Sun Feb 22, 2009 11:58 pm    Post subject: 3 And 10,000,000 Wins in 4 balls: 3800 Wins in 5 balls: 16945 Wins in 6 balls: 46753 Wins in 7 balls: 103343 Wins in 8 balls: 196849 Wins in 9 balls: 339000 Wins in 10 balls: 537402 Wins in 11 balls: 789202 Wins in 12 balls: 1076467 Wins in 13 balls: 1349240 Wins in 14 balls: 1527809 Wins in 15 balls: 1517855 Wins in 16 balls: 1254991 Wins in 17 balls: 805610 Wins in 18 balls: 347772 Wins in 19 balls: 81322 Wins in 20 balls: 5640
Death Mage
Raving Lunatic

 Posted: Mon Feb 23, 2009 12:15 am    Post subject: 4 I was hoping to see the math, rather than brute forcing it. Does that include the chance for double bingos?_________________* These senseless ramblings brought to you by Insanity™. If you just can't figure the dang thing out, it must be Insanity™. [YOUR AD HERE!]
MatthewV
Daedalian Member :_

 Posted: Mon Feb 23, 2009 12:46 am    Post subject: 5 Zag's numbers stop after a win happens. The win may have completed more than one bingo. The scoring method appears to more than double the players money. I recommend a payout structure of 1000x, 100x, 25x, 10x, 5x, 2x, 1x. With these numbers, a player can maximize their payout by not playing! The theoretical chance of hitting a bingo with just four numbers called is 0.0003764... (16/24 * 3/23 * 2/22 * 1/21). Zag's emulation program was a little lucky. Higher numbers would be too complicated for my slow methods.
Zag
Tired of his old title

 Posted: Mon Feb 23, 2009 2:19 am    Post subject: 6 I think I misunderstood the problem. I was assuming that there was only one player with one card, and he got paid based on how many draws it took to make a bingo with that card. I did not consider multiple bingos on a draw, since I figured that either you got bingo or you didn't. I can see several different ways to change this -- You might payout double or triple if the one card makes more than one bingo on a single draw. (This would seem odd, to me.) Also, you might allow for more than one player at a time (or for the player to have more than one card), increasing the chance that somebody is going to get a bingo sooner. However, this doesn't seem to fit with the idea of a video bingo game.
Death Mage
Raving Lunatic

 Posted: Mon Feb 23, 2009 2:40 am    Post subject: 7 Getting two bingos with one number (8+) would net in double payout._________________* These senseless ramblings brought to you by Insanity™. If you just can't figure the dang thing out, it must be Insanity™. [YOUR AD HERE!]
lostdummy
Daedalian Member

 Posted: Thu Mar 12, 2009 9:31 am    Post subject: 8 when someone score such bingo? I never played it, so I'm not sure what are conditions to score , for example, bingo in 4 numbers. Does it means that any of two rows with 4 numbers on my card (horizontal and vertical, intersecting at empty center) needs to be filled? Or diagonal lines are also accepted? What in that case about short diagonals? Just going by intuition, if only 2 rows are accepted for 4-bingo, I get chance for bingo in 4 numbers at 2/C(24,4)=0.000188 , half of that shown by Zag or Matthew. Therefore I suspect I'm missing some additional rows acceptable for bingo, and seeing how Matthew started with 16/24, i can guess that somehow 4 rows with 16 numbers are acceptable for 4-bingo. In that case, assuming 4 rows can match 4-bingo, and 8 rows can match 5-bingo, mathematical chances are: if p4= 1/C(24,4) and p5=1/C(24,5) : P(in 4)=4*p4 P(in 5)=8*p5 + 4*P(in 4) P(in 6)= 5*8*p5+C(5,3)*4*p4 or in general: P(in N)= C(N-1,4)*8*p5+C(N-1,3)*4*p4, or: P(in N)= C(N-1,4)*8/C(24,5) + C(N-1,3)*4/C(24,4) That formula result in probabilities (x10mil to compare to Zags numbers) of: 4: 3764 5: 16940 6: 47054 7: 103520 8: 197628 9: 342556 10: 553360 With given payouts, expected return for 1\$ invested would be 2.27\$ - which is not good from gameowner point, since usually it tend to be slightly under 1\$. One payout option that could do that is: 300,90,30,15,9,3,2 -> results in 0.95\$ expected gain. But above formula also do not consider if someone score two bingos in one number.
lostdummy
Daedalian Member

 Posted: Thu Mar 12, 2009 10:35 am    Post subject: 9 one approximate math formula including double bingos could be: p5(N)=C(N-1,4)*8/C(24,5) p4(N)=C(N-1,3)*4/C(24,4) P_double(in N)= 16/24*(p5(N)^2 + p5(N)*p4(N)) P_single(in N)= (1-P_double(N))*( p5(N)+p4(N) ) That "16/24" is there because there is only 16 cells where 5rows intersect other 5 rows, while also there is 16 cells where 5 rows intersect 4 rows (if there are 4 rows as i guessed) Above is not exact formula ( I would need more of "(1-px)*" for that, even on original non-double-bingo formula), but should be close enough approximation. And, as expected, end result is not greatly changed by double bingo chances - now expected payoff for 1\$ invested is 2.278357664\$, compared to previous 2.267645398\$ - difference of about 1cent ;p
marcusI
Daedalian Member

 Posted: Fri Mar 13, 2009 10:01 pm    Post subject: 10 lostdummy, does your new result include that double bingo is only possible with N of 9 and 10? It seems high. Also, since only one number can overlap between the two bingos, they must be not be parallel to each other and cannot both be crossing the free square. This severely limits the possible combinations.
lostdummy
Daedalian Member

 Posted: Fri Mar 13, 2009 10:21 pm    Post subject: 11 It covers requirement that they need to be crossing , but it is good point about needing at least 9 numbers - that was not explicitly covered. Its easy to do it though, just : P_double(for N<9) =0 P_double(N>=9)= 16/24*(p5(N)^2 + p5(N)*p4(N)) P_single(in N)= (1-P_double(N))*( p5(N)+p4(N) ) As for "seems high", expected value with double bingo included barely changed value if double bingo is not included. As I said, only with singles it was 2.267\$, and with double bingo it was 2.278\$ - which is under half percent difference. Now with excluded doubles under 9 numbers , that difference is even less: new expected payoff per \$ with double bingo included is 2.274\$ , which is only 0.3% more than if double bingo is ignored. Of course, expected payoff in itself looks high (2.3\$ >> 1\$), but that is due to high payoff values.
marcusI
Daedalian Member

 Posted: Fri Mar 13, 2009 11:06 pm    Post subject: 12 Thanks lostdummy. I don't really understand the math, but that sounds better since 9 and 10 pay very low amounts.
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