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1 in 3,130.978 (I think)

Quote:beachbumbabs3.19389e-007

I get 6.25469946307465E-007, or 1 in 1,598,797.84

Assume your 24 numbers are painted red, and the other 51 are white. The question becomes, what is the probability that, if you draw 47 balls, all 24 red balls will be included?

There are (75)C(47) ways to draw 47 balls out of 75.

Of these, the sets that have all 24 red balls will each have 23 white balls as well; there are (51)C(23) different sets of 23 white balls.

The probability is (51)C(23) / (75)C(47)

= (51! / (23! x 28!)) / (75! / (47! x 28!))

= (51! / 23!) / (75! / 47!)

= (51 x 50 x 49 x 48 x ... x 24 x 23! / 23!) / (75 x 74 x ... x 48 x 47! / 47!)

= (51 x 50 x 49 x 48 x ... x 24) / (75 x 74 x ... x 48)

Here's something to note; if, instead of 47 balls, 74 are drawn, there is still a 24/75, or 32%, chance that the card will not be filled.

Quote:ThatDonGuyHere's something to note; if, instead of 47 balls, 74 are drawn, there is still a 24/75, or 32%, chance that the card will not be filled.

Ya know what, as I was writing my rebuttal, I realized you're right. I was thinking you were invoking some weird formula that states you'll never get to that point - at least not in a room filled with players.

But in the context of the statement that I quoted, after 74 balls have been called, you DO still have 24 out of 75 chances that one of your numbers hasn't been selected yet.

Quote:ThatDonGuyI get 6.25469946307465E-007, or 1 in 1,598,797.84

Assume your 24 numbers are painted red, and the other 51 are white. The question becomes, what is the probability that, if you draw 47 balls, all 24 red balls will be included?

There are (75)C(47) ways to draw 47 balls out of 75.

Of these, the sets that have all 24 red balls will each have 23 white balls as well; there are (51)C(23) different sets of 23 white balls.

The probability is (51)C(23) / (75)C(47)

= (51! / (23! x 28!)) / (75! / (47! x 28!))

= (51! / 23!) / (75! / 47!)

= (51 x 50 x 49 x 48 x ... x 24 x 23! / 23!) / (75 x 74 x ... x 48 x 47! / 47!)

= (51 x 50 x 49 x 48 x ... x 24) / (75 x 74 x ... x 48)

Here's something to note; if, instead of 47 balls, 74 are drawn, there is still a 24/75, or 32%, chance that the card will not be filled.

Don,

Thanks for showing your work! I just grabbed my number off a third-party bingo calculator; I like how you set it up logically. Not sure why the solutions differ, since I didn't set up the calculation.

Quote:DJTeddyBearBingo math is way to complex for my feeble brain, but I gotta challenge this statement:

Ya know what, as I was writing my rebuttal, I realized you're right. I was thinking you were invoking some weird formula that states you'll never get to that point - at least no in a room filled with players.

But in the context of the statement that I quoted, after 74 balls have been called, you DO still have 24 out of 75 chances that one of your numbers hasn't been selected yet.

Yeah, the fun part comes when dozens of people are waiting on one last number; gets real tense in the ole' parlor on that last big coverall game.

I was gonna say "I didn't even know Revel had Bingo..."Quote:aceofspadesOn my last trip at Revel I shared top prize for a full card bingo!!! ( I got it along with two others )

While I stand by that comment, I wanna say, "Really? That is so unlike the high-end Cosmopolitan-east image Revel was projecting." (Does Cosmopolitan have Bingo?)

Quote:ThatDonGuyI get 6.25469946307465E-007, or 1 in 1,598,797.84

Assume your 24 numbers are painted red, and the other 51 are white. The question becomes, what is the probability that, if you draw 47 balls, all 24 red balls will be included?

There are (75)C(47) ways to draw 47 balls out of 75.

Of these, the sets that have all 24 red balls will each have 23 white balls as well; there are (51)C(23) different sets of 23 white balls.

The probability is (51)C(23) / (75)C(47)

= (51! / (23! x 28!)) / (75! / (47! x 28!))

= (51! / 23!) / (75! / 47!)

= (51 x 50 x 49 x 48 x ... x 24 x 23! / 23!) / (75 x 74 x ... x 48 x 47! / 47!)

= (51 x 50 x 49 x 48 x ... x 24) / (75 x 74 x ... x 48)

Here's something to note; if, instead of 47 balls, 74 are drawn, there is still a 24/75, or 32%, chance that the card will not be filled.

This math looks clear to me. But what happens if you are playing the game with 50 people, should the chance to win the game be the same? I think you are against other 49 people so if everyone get the chance of 3.19e-7 to win the game, but any one other than you calling BINGO will stop the game so how do you modify the math to consider the competition among 50 players?

Quote:konglifyany one other than you calling BINGO will stop the game so how do you modify the math to consider the competition among 50 players?

If the cards and numbers drawn are fair, every card has the same chances of winning. If you want another chance to win, you play more than one card.

If someone calls BINGO, the game is not stopped, it is paused while the card is confirmed If the card has an error, the game is resumed.

Quote:DieterIf the cards and numbers drawn are fair, every card has the same chances of winning. If you want another chance to win, you play more than one card.

If someone calls BINGO, the game is not stopped, it is paused while the card is confirmed If the card has an error, the game is resumed.

Sorry for the misleading word, what I mean is if someone call BINGO and it is valid, the game concluded. Anyway, qualitatively, I know that we need to purchase more card if we want to increase the chance to win a BINGO. But I am looking for the quantitative analysis. Let's said the probability to win the BINGO on 47 numbers are called is P, and including you, there are 8 players playing the game. Does the probability stay the same P? If not, how to figure the new probability?

If there are n cards in play, and you're playing x of them, your chances of winning are x/n.

My mathin' brain isn't ready to do more than that right now, I'm afraid.

Quote:DieterIf there are 8 cards in play, and you're playing 1 of them, and the game must continue until a winner is selected, your chances of winning (or splitting the prize) are 1 in 8.

If there are n cards in play, and you're playing x of them, your chances of winning are x/n.

My mathin' brain isn't ready to do more than that right now, I'm afraid.

Thanks for the reply. But I just have a feeling that the chance doesn't split uniformly so if I playing 1 card and total 8 cards are in play, the chance should be more than 1/8. I don't know how to prove that in math yet but I am trying the simulation. Let's assume there are 2 cards playing and I own 1 card so I against another player. Standard 75 numbers bingo, will draw all 75 numbers until a winner found. As we know the probability of winning any pattern or full coverage will be 1 for a single card before or on the 75th number call. The simulation returns the probability to be 0.516 instead of 0.5 for any pattern winning before or on 75th number call. I run the simulation for 100000000 times.

Quote:konglifyThe simulation returns the probability to be 0.516 instead of 0.5 for any pattern winning before or on 75th number call. I run the simulation for 100000000 times.

Wouldn't that mean that some cards are more likely to win than others? Don't we know that not to be true in a fair game?

I'm guessing that the 100,000,000 runs isn't a big enough sample of the 2.5*10^59 possible cards to be valid to 3 significant places.

Quote:konglifyAs we know the probability of winning any pattern or full coverage will be 1 for a single card before or on the 75th number call. The simulation returns the probability to be 0.516 instead of 0.5 for any pattern winning before or on 75th number call. I run the simulation for 100000000 times.

To get to the 75th draw, in most 1 winner games, you'd need to be going for a coverall (blackout) or some very specific specialty pattern, and all remaining cards would need to be waiting on the same space.

I don't see how we get past draw 70 if we're looking for a straight line bingo (any of 5 rows, 5 columns, 2 major diagonals).

Quote:DieterTo get to the 75th draw, in most 1 winner games, you'd need to be going for a coverall (blackout) or some very specific specialty pattern, and all remaining cards would need to be waiting on the same space.

I don't see how we get past draw 70 if we're looking for a straight line bingo (any of 5 rows, 5 columns, 2 major diagonals).

I am totally lost in the analysis after reading your comment. The math is so confusing to me though I think your reasoning make sense that each of the card should share the same probability of winning the game even I play 1 card while other (n-1) cards are in play against me. So you mean my chance to BINGO is actually 1/n. Consider two cases

1) only straight line bingo allowed, so does it mean if 8 players in play and I am one of them, I will win at the chance 1/8

2) only coverall (all numbers in the card filled), again, my chance to win that is still 1/8

Am I understand that correctly? If that's the case, and if I want to figure out the chance that N winners for one game, should it be (1/8)^N?

Let's make it simple, if I want to figure out the chance that two out of 8 players bingo at the same time, can I say it is (1/8)*(1/8)? Here is what confusing me: if (1/8)^2 is the chance for any two out of 8 players winning the bingo, but there are COMBIN(8,2) ways to pick any two players out of 8 and among them, only 7 pairs including my card, so does it mean if I want to figure out the chance that 2 players out of 8 winning the game and I am one of the winner, the chance will be (1/8)^2*COMBIN(7,1)/COMBIN(8,2) ?

Thanks again

Quote:konglify

1) only straight line bingo allowed, so does it mean if 8 players in play and I am one of them, I will win at the chance 1/8

2) only coverall (all numbers in the card filled), again, my chance to win that is still 1/8

Assuming a fair game, yes.

The number of rounds until the bingo is only vaguely interesting for two reasons:

1) A variable prize amount for wins in different rounds

2) You're hosting the game, and trying to plan how many games you can get in per (time unit).

Quote:konglifyLet's make it simple, if I want to figure out the chance that two out of 8 players bingo at the same time, can I say it is (1/8)*(1/8)?

That's a very different question, and it's not simple.

The chances of an n-way tie among x cards should be significantly smaller than (1/x)^n. That's where you get into big numbers.

For a tie to happen, both cards need to have the same number as the last needed spot for a pattern, and then that number needs to be called.

I'd love it if someone wants to improve my understanding of the math, but I think I'm somewhere near close.

Quote:DieterAssuming a fair game, yes.

The number of rounds until the bingo is only vaguely interesting for two reasons:

1) A variable prize amount for wins in different rounds

2) You're hosting the game, and trying to plan how many games you can get in per (time unit).

That's a very different question, and it's not simple.

The chances of an n-way tie among x cards should be significantly smaller than (1/x)^n. That's where you get into big numbers.

For a tie to happen, both cards need to have the same number as the last needed spot for a pattern, and then that number needs to be called.

I'd love it if someone wants to improve my understanding of the math, but I think I'm somewhere near close.

What about this. If my game only draw up to 50th numbers, if no one win before or on the 50th draw, the game concluded. As I known from wizard of odd, for 1 card, the probability to win before or on 50th draw in 0.814392, so if there are 4 players, I am one of them, should my probability of winning the game be 0.814392/4 = 0.203598?

Quote:konglifyWhat about this. If my game only draw up to 50th numbers, if no one win before or on the 50th draw, the game concluded. As I known from wizard of odd, for 1 card, the probability to win before or on 50th draw in 0.814392, so if there are 4 players, I am one of them, should my probability of winning the game be 0.814392/4 = 0.203598?

I could be wrong, but I think your individual chance of winning the game remains the same. You would only divide by 4 if the prize were going to be awarded to someone, no matter how many draws it takes, and that would mean at 25% chance of each of you winning. Each of you has that same (miniscule) chance of getting the coverall in 50 numbers, but the overall chance of the prize being awarded would be additive; with 4 of you playing, there would be 0.814392x4 chance of it (using your number). FWIW, a lot of games keep the big jackpot to, say, the 45 or 50th number, and roll that amount over if it doesn't hit in that amount, awarding a small consolation prize to the eventual cover-all winner.

Quote:beachbumbabsoverall chance of the prize being awarded would be additive; with 4 of you playing, there would be 0.814392x4 chance

How can we have a probability of >1? (3.257568?)

Quote:DieterHow can we have a probability of >1? (3.257568?)

I just read the Bingo FAQ in Wizard of Odds and other published papers, I learn that more players playing the game will increase the chance to win the BINGO, but it doesn't mean that the probability for your card to win get increase, it only means that more cards (players) come into the game, the bingo will appear faster than usual. Following the Bingo FAQ, I obtain the probably that at least one of the 4 cards to win is 0.998813, which is the probability to win before or on the 50th draw. However, my question what is the probability for my card to win if I am playing with other 3 players and if the game draw at max to 50th numbers.

Should it be 0.998813/4? If so, how do you estimate the payback for bingo in that case?

Quote:konglifyShould it be 0.998813/4? If so, how do you estimate the payback for bingo in that case?

That sounds close.

How is the prize structure arranged? Is there a quorum requirement?

A lot of the games around here are fixed prizes, which means the payback percentage varies based on the number of participants. Some bingo games award variable prizes based on the number of participants.

(The last bingo I played was a charity thing at a summer fair... 25c/card, $5/round prize. There were a "lot" of players (guessing around 55-60, it seemed about 1/4 were playing more than one card at a time), the house was raking it in (~$13+ buy per game, $5 pay per game, about 15 games/hour, + beer sales). As is usually my luck, I was always 1 spot short of a bingo.)

So, not very good odds for you at all. Not sure where you're getting the .8 number, but it's a long way from correct.

Estimating the jackpot payback point where it becomes playable is beyond my abilities. (Which mostly are looking up numbers on tables.) But the Wizard has calculated that info on the WoO website. You might look there some more.

Quote:DieterThat sounds close.

How is the prize structure arranged? Is there a quorum requirement?

A lot of the games around here are fixed prizes, which means the payback percentage varies based on the number of participants. Some bingo games award variable prizes based on the number of participants.

(The last bingo I played was a charity thing at a summer fair... 25c/card, $5/round prize. There were a "lot" of players (guessing around 55-60, it seemed about 1/4 were playing more than one card at a time), the house was raking it in (~$13+ buy per game, $5 pay per game, about 15 games/hour, + beer sales). As is usually my luck, I was always 1 spot short of a bingo.)

The pay structure is simple. Let's assume each card cost 1 credit, if 1 player (1 card) win, he/she will get 2 credits back. If more than 1 players win at the same round, 1 credit will be split. Let's also assume only 4 players at max are playing the game.

Quote:beachbumbabsIn 50 number coverall (cutoff at 50 numbers), the probability of any one card (your card) winning is 0.00000471508, or 1 in 212085.4789. If there are 1000 cards in play, and only 50 numbers are called, the chance that someone in that 1000 will win is 0.004704, or 1 in 212.5850.

So, not very good odds for you at all. Not sure where you're getting the .8 number, but it's a long way from correct.

Estimating the jackpot payback point where it becomes playable is beyond my abilities. (Which mostly are looking up numbers on tables.) But the Wizard has calculated that info on the WoO website. You might look there some more.

Hi beachbumbabs, let's make sure that we are in the same page on the terms and game rules here. So I assume we use 75 numbers in the game, 12 configuration (not including the 4-corner configuration), the game will draw one number at a time until the 50th number drawn then no matter if any one win or not, the game will conclude. So is this the same thing as what you mean by "50 number coverall"? It looks like to me that you are talking all numbers on a card are filled by "coverall" but my question is asking any line win not full coverage.

Next is to consider the probability that winning a bingo for one card (my card) before or on the last number called (i.e. before the 50th number called), I've got 0.81 something, which was from calculation, but looking up the table given here

https://wizardofodds.com/games/bingo/probabilities/1/

the 50th row gives the same number.

In addition, in the same page above, you could compare the probability when 1 player and 2 players are playing in the game. The probability at/before 50 draws is proportional to the number of players.

Quote:konglifyHi beachbumbabs, let's make sure that we are in the same page on the terms and game rules here. So I assume we use 75 numbers in the game, 12 configuration (not including the 4-corner configuration), the game will draw one number at a time until the 50th number drawn then no matter if any one win or not, the game will conclude. So is this the same thing as what you mean by "50 number coverall"? It looks like to me that you are talking all numbers on a card are filled by "coverall" but my question is asking any line win not full coverage.

Next is to consider the probability that winning a bingo for one card (my card) before or on the last number called (i.e. before the 50th number called), I've got 0.81 something, which was from calculation, but looking up the table given here

https://wizardofodds.com/games/bingo/probabilities/1/

the 50th row gives the same number.

In addition, in the same page above, you could compare the probability when 1 player and 2 players are playing in the game. The probability at/before 50 draws is proportional to the number of players.

Yes, we are talking 2 different games. I thought you've been asking about a coverall the whole time, and you're asking about any bingo. Sorry for the confusion.

ANY bingo in 50 numbers, with 4 players? A virtual certainty by then, though not impossible to have an unresolved game. I'm not qualified to give you numbers, so I'll just shut up about specifics.

Quote:beachbumbabsYes, we are talking 2 different games. I thought you've been asking about a coverall the whole time, and you're asking about any bingo. Sorry for the confusion.

ANY bingo in 50 numbers, with 4 players? A virtual certainty by then, though not impossible to have an unresolved game. I'm not qualified to give you numbers, so I'll just shut up about specifics.

Thanks for the reply. Well, actually, it doesn't have to be 4 but I just want to make it simple for me to understand the math. Since it looks quite complicate to analyze the multiplayer game, I try the simulation on 50 players (so my card is against other 49 cards)., at max draw 50 numbers out of 75. I got the probability for MY CARD to win is about 0.021 something.

From the table when 50 cards consdiered, I find that the probability for at least 1 card out of 50 to win before or on the 50 draw is so close to P(50)=1. This "P(50)=1" is not for my card but for all 50 cards. Dieter mentioned before the fair game should have identical probability for all players so I am thinking my simulation result 0.021 is actually coming from P(50)/50=1.0/50 = 0.02. Hence, if I pay 2 for any bingo winning, can I say the payback is 2*0.02*100 = 4%?

Quote:konglifyThe pay structure is simple. Let's assume each card cost 1 credit, if 1 player (1 card) win, he/she will get 2 credits back. If more than 1 players win at the same round, 1 credit will be split. Let's also assume only 4 players at max are playing the game.

This is not a good game to play with 4 players. It also sounds like the prize per player gets cut in 4 with a 2 player tie - more not good. (1 credit will be split? I'm guessing the entire prize would be split n ways between n winners in a tie... that's not so bad.)

If we assume that 1/2 of the buy in goes to the prize pool for the game, I'd guess it becomes reasonable for most people to play at around 200 players.

No matter how you slice it, it's -EV.

If you can find a +EV bingo game - say a 1 unit buy in, 100 (or n) unit prize, 100 (or n) or fewer cards in play, game goes until a winner is selected... then well done you. Typically a game like this will have a quorum of 100+ (or n+) cards, just to keep it +EV for the house. (I wouldn't be surprised at a quorum of n*1.5.)

Quorum isn't so necessary with a limited number of calls, but I'd expect a 30-35 ball limit for an "any SLB" and a quorum of at least 10 players.

Quote:konglifyI am thinking my simulation result 0.021 is actually coming from P(50)/50=1.0/50 = 0.02

Yep.

1/n + a small chance of getting a tie and splitting a prize. I normally don't think about the tie possibilities, since they're... small.

Quote:DieterThis is not a good game to play with 4 players. It also sounds like the prize per player gets cut in 4 with a 2 player tie - more not good. (1 credit will be split? I'm guessing the entire prize would be split n ways between n winners in a tie... that's not so bad.)

If we assume that 1/2 of the buy in goes to the prize pool for the game, I'd guess it becomes reasonable for most people to play at around 200 players.

No matter how you slice it, it's -EV.

If you can find a +EV bingo game - say a 1 unit buy in, 100 (or n) unit prize, 100 (or n) or fewer cards in play, game goes until a winner is selected... then well done you. Typically a game like this will have a quorum of 100+ (or n+) cards, just to keep it +EV for the house. (I wouldn't be surprised at a quorum of n*1.5.)

Quorum isn't so necessary with a limited number of calls, but I'd expect a 30-35 ball limit for an "any SLB" and a quorum of at least 10 players.

I know it doesn't sound good game. But I am not designing a game, I want to learn the math only. To make it simple, I still want to use the pay table like

if winning any bingo, pay 2

if winning coverall, pay 5

Like what we discuss above, if we draw all 75 numbers, the probability of at least 1 card will be win is 1.0; also the probability of at least 1 card will win the coverall still 1. The game is fair, let assume the probability of winning any bingo for my card is 1/8 (if there is 8 cards), the probability of winning any coverall for my card is also 1/8.

So can I say the payback is 2*(1/8)+5*(1/8) = 87.5%? I know your reasoning and I agree with you but it is still shocking that the chance to win any bingo is the same as the chance to win coverall at 1/8.

What not convincing myself is the following. If every one got 1/8 chance to win the coverall, and we multiply 1/8 with the pay to get the payback (or EV), what happen if someone else make coverall before I did, do I have to include that odd into the calculation. For example, we have 8 players, I got 1/8 chance to win coverall, but that mean I got 7/8 not winning the game or just be tie. So I don't have to consider that 7/8 into my calculating the payback? (Sorry for keep asking this, the more I think about that, the more confusion I have)

Quote:konglifyThis math looks clear to me. But what happens if you are playing the game with 50 people, should the chance to win the game be the same? I think you are against other 49 people so if everyone get the chance of 3.19e-7 to win the game, but any one other than you calling BINGO will stop the game so how do you modify the math to consider the competition among 50 players?

The way I see it, the probability of you being the only winner = (the probability of you filling your card in exactly 24 numbers x the probability no one else does it in 24 or fewer) + (the probability of you filling your card in exactly 25 numbers x the probability no one else does it in 25 or fewer) + ... + (the probability of you filling your card in exactly 47 numbers x the probability no one else does it in 47 or fewer). If I punched the numbers into Excel correctly, I get 1 / 3,267,141.

Quote:DJTeddyBearI was gonna say "I didn't even know Revel had Bingo..."

While I stand by that comment, I wanna say, "Really? That is so unlike the high-end Cosmopolitan-east image Revel was projecting." (Does Cosmopolitan have Bingo?)

It was a special invite only event…not a normal occurrence :-)

Quote:ThatDonGuyThe way I see it, the probability of you being the only winner = (the probability of you filling your card in exactly 24 numbers x the probability no one else does it in 24 or fewer) + (the probability of you filling your card in exactly 25 numbers x the probability no one else does it in 25 or fewer) + ... + (the probability of you filling your card in exactly 47 numbers x the probability no one else does it in 47 or fewer). If I punched the numbers into Excel correctly, I get 1 / 3,267,141.

This makes sense to me. But I didn't get the same number as yours. Here is what I did. I first copy the probability for a single card coverall table in https://wizardofodds.com/games/bingo/probabilities/1/

24 0.00000000000000000004 0.00000000000000000004

25 0.00000000000000000093 0.00000000000000000097

26 0.00000000000000001164 0.00000000000000001261

27 0.00000000000000010086 0.00000000000000011347

28 0.00000000000000068079 0.00000000000000079426

29 0.00000000000000381245 0.00000000000000460671

30 0.00000000000001842684 0.00000000000002303355

31 0.00000000000007897218 0.00000000000010200573

32 0.00000000000030601718 0.00000000000040802291

33 0.00000000000108806109 0.00000000000149608400

34 0.00000000000359060160 0.00000000000508668560

35 0.00000000001109822313 0.00000000001618490874

36 0.00000000003236981747 0.00000000004855472621

37 0.00000000008963949453 0.00000000013819422074

38 0.00000000023690437841 0.00000000037509859915

39 0.00000000060015775864 0.00000000097525635779

40 0.00000000146288453669 0.00000000243814089448

41 0.00000000344208126279 0.00000000588022215727

42 0.00000000784029620969 0.00000001372051836696

43 0.00000001733118109511 0.00000003105169946207

44 0.00000003726203935449 0.00000006831373881656

45 0.00000007807284436178 0.00000014638658317834

46 0.00000015969445437637 0.00000030608103755472

47 0.00000031938890875275 0.00000062546994630747

48 0.00000062546994630747 0.00000125093989261493

49 0.00000120090229691033 0.00000245184218952526

50 0.00000226323894417717 0.00000471508113370243

51 0.00000419118322995771 0.00000890626436366014

52 0.00000763394088313726 0.00001654020524679740

53 0.00001368844572148750 0.00003022865096828490

54 0.00002418292077462790 0.00005441157174291290

55 0.00004212508780096470 0.00009653665954387760

56 0.00007240249465790830 0.00016893915420178600

57 0.00012286483941948100 0.00029180399362126600

58 0.00020597928961501200 0.00049778328323627800

59 0.00034133710850487700 0.00083912039174115500

60 0.00055941359449410300 0.00139853398623526000

61 0.00090715718026070800 0.00230569116649597000

62 0.00145622599989219000 0.00376191716638816000

63 0.00231502594854655000 0.00607694311493471000

64 0.00364616586896083000 0.00972310898389554000

65 0.00569157599057300000 0.01541468497446850000

66 0.00880839141398202000 0.02422307638845060000

67 0.01351985658890260000 0.03774293297735320000

68 0.02058705435128360000 0.05832998732863680000

69 0.03110932657527300000 0.08943931390390970000

70 0.04666398986290940000 0.13610330376681900000

71 0.06949955937029060000 0.20560286313711000000

72 0.10280143156855500000 0.30840429470566500000

73 0.15105516475379500000 0.45945945945945900000

74 0.22054054054054100000 0.68000000000000000000

75 0.32000000000000000000 1.00000000000000000000

The second column is the probability to win a coverall at that draw, the 3rd column is the probability to win a coverall on or before that draw. So based on your express, I try the following

p means my winning probability at that draw, q means the probability that any other won't win on or before that draw

nthdraw p q p*q^49

--------------------------------------------------------------------------------------------------------

24 0.00000000000000000004 1-0.00000000000000000004 0.0000000000000000000400000000

25 0.00000000000000000093 1-0.00000000000000000097 0.0000000000000000009300000000

26 0.00000000000000001164 1-0.00000000000000001261 0.0000000000000000116400000000

27 0.00000000000000010086 1-0.00000000000000011347 0.0000000000000001008600000000

...

47 0.00000031938890875275 1-0.00000062546994630747 0.0000003193791202596720000000

By adding the last column, I got 0.0000006254570451581000000000 which is not same as 1/3267141. So what's wrong with my math. Thanks again for giving the hint :)

Quote:konglifyBy adding the last column, I got 0.0000006254570451581000000000 which is not same as 1/3267141. So what's wrong with my math. Thanks again for giving the hint :)

There is nothing wrong with your math. I was calculating p * (1 - q

^{49}) on my spreadsheet instead of p * ((1-q)

^{49}).

Quote:ThatDonGuyThere is nothing wrong with your math. I was calculating p * (1 - q

^{49}) on my spreadsheet instead of p * ((1-q)^{49}).

Thanks for get back with the correction. I think I got the idea now. I just have one more question. Following your idea, I am trying to calculate the probability that I am the only one winning coverall against other 49 players but with all numbers drawing (i.e. up to 75th numbers). Again, I add all numbers and get 0.0161323041644552000000000000. As my discussion with others in the forum, if we consider the game is fair, each player in the game has the same chance to win, so if there is 50 players, each one winning coverall before or on 75th calling should be 1/50=0.02, correct?

The number we calculate above is for the case when I am the only winner. If I want to prove 0.02 is the correct one, do I have to count, all the case, when

1) I am the only winner

2) both myself as well as 1 of the 49 players are winners

3) both myself as well as 2 of the 49 players are winners

4) both myself as well as 3 of the 49 players are winners

5) both myself as well as 4 of the 49 players are winners

...

50) all 50 players are winner

Do you think this logic is correct to prove that 0.02?

Quote:konglifywinning coverall before or on 75th calling should be 1/50=0.02, correct?

If it gets to the 75th call, all cards in play will tie. That would require all the cards in play to have 1 number in common, and that to be the last number called.

Quote:DieterIf it gets to the 75th call, all cards in play will tie. That would require all the cards in play to have 1 number in common, and that to be the last number called.

Yes if it gets to the 75th call. But someone might win before 75th call so in ThatDonGuy's method, the winning probability at 1st, 2nd, 3rd ... will be added. I am not asking the win probability on 75th call, but that on or before 75th call. So I did the math as follow

let p(k) be the winning probability (for coverall) on (exactly on not on or before) the kth call, q(k) is the probability not winning on or before kth call, 50 players in total, m means m people wins at the same time

P(m) = sum over { p(k)^m*q^(50-m) } with k = 24 to 75

Total winning probability = sum over {P(m)} with m=1 to 50

then I got 0.016254 but not 0.02. So what's wrong with my math?