# Winning a Game of Bingo in Nigeria

Filed in Online Casino, PUSSY888 Casino, Recovery on May 4, 2021

Winning a Game of Bingo : Playing bingo online, players can make use of optional features which make playing the game easier, such as auto-daub.

Auto-daub automatically marks off the numbers on cards as they are called, so players don’t have to.

Most software providers support other gaming features as “Best Card Sorting” and “Best Card Highlighting” where players cards are sorted and highlighted by closest to bingo.

#### Read also: How to Play PUSSY888 Casino in Nigeria Winning a Game of Bingo

## Winning a Game of Bingo

There is variety among the different kinds of bingo games that can be played.

For example, some inexpensive game rooms appeal to the player.

Who may want to play for as little as 3 cents or 3 pence per card.

Some bingo games only allow players to purchase the same number of cards.

So they are not competing against the “high rollers” out there who buy many cards for the same game.

## Winning a Game of Bingo

The probability of winning a game of Bingo (ignoring simultaneous winners, making wins mutually exclusive) may be calculated as:

{\displaystyle P(Win)=1-P(Loss)} since winning and losing are mutually exclusive. The probability of losing is the same as the probability of another player winning (for now assuming each player has only one Bingo card). With {\displaystyle n} players taking part: {\displaystyle P(Loss)=P(P_{2}{\mbox{ or }}P_{3}{\mbox{ or }}…P_{n-1}{\mbox{ or }}P_{n})} with {\displaystyle n} players and our player being designated {\displaystyle P_{1}} . This is also stated (for mutually exclusive events) as {\displaystyle P(Loss)=P(P_{2})+P(P_{3})+…+P(P_{n})} .

## Winning a Game of Bingo

If the probability of winning for each player is equal (as would be expected in a fair game of chance), then {\displaystyle P(P_{1})=P(P_{2})=…=P(P_{n})} and thus {\displaystyle P(Loss)=(n-1)P(P_{1})} and therefore {\displaystyle P(Win)=P(P_{1})=1-(n-1)P(P_{1})} . Simplifying yields

{\displaystyle P(P_{1})=1/n} ## Winning a Game of Bingo

For the case where more than one card is bought, each card can be seen as being equivalent to the above players, having an equal chance of winning. {\displaystyle P(C_{1})=1/n_{C}} where {\displaystyle n_{C}} is the number of cards in the game and {\displaystyle C_{1}} is the card we are interested in.

A player ({\displaystyle P_{1}} ) holding {\displaystyle m} cards therefore will be the winner if any of this cards win (still ignoring simultaneous wins):

{\displaystyle P(P_{1})=P(C_{1})+P(C_{2})+…+P(C_{m})=m/n_{C}} A simple way for a player to increase his odds of winning is therefore to buy more cards in a game (increase {\displaystyle m} ).

## Winning a Game of Bingo

Simultaneous wins may occur in certain game types (such as online bingo, where the winner is determined automatically, rather than by shouting “Bingo” for example), with the winnings being split between all simultaneous winners. The probability of our card, {\displaystyle C_{1}} , winning when there is either one or more simultaneous winners is expressed by:

{\displaystyle P(C_{1})=P(w){\frac {w}{n_{C}}}} where {\displaystyle P(w)} is the probability of there being {\displaystyle w} simultaneous winner (a function of the game type and number of players) and {\displaystyle {\frac {w}{n_{C}}}} being the (fair) probability that {\displaystyle C_{1}} is one of the winning cards. The overall expected value for the payout (1 representing the full winning pot) is therefore:

{\displaystyle E={\frac {1}{1}}{\frac {1}{n_{C}}}P(1)+{\frac {1}{2}}{\frac {2}{n_{C}}}P(2)+…+{\frac {1}{n_{C}}}{\frac {n_{C}}{n_{C}}}P(n_{C})} {\displaystyle E={\frac {1}{n_{C}}}(P(1)+P(2)+…+P(n_{C}))} ## Winning a Game of Bingo

Since, for a normal bingo game, which is played until there is a winner, the probability of there being a winning card, either {\displaystyle P(1)} or {\displaystyle P(2)} or … or {\displaystyle P(n_{C})} , and these being mutually exclusive, it can be stated that

{\displaystyle P(1)+P(2)+…+P(n_{C})=1} and therefore that

{\displaystyle E={\frac {1}{n_{C}}}} ## Winning a Game of Bingo

The expected outcome of the game is therefor not changed by simultaneous winners, as long as the pot is split evenly between all simultaneous winners. This has been confirmed numerically.

To investigate whether it is better to play multiple cards in a single game or to play multiple games, the probability of winning is calculated for each scenario, where {\displaystyle m} cards are bought.

{\displaystyle P(win)_{multiplecards}={\frac {m}{m+n-1}}} where n is the number of players (assuming each opposing player only plays one card). The probability of losing any single game, where only a single card is played, is expressed as:

{\displaystyle P(loss)_{game}=1-P(win)=1-{\frac {1}{n}}} The probability of losing {\displaystyle m} games is expressed as:

{\displaystyle P(loss)_{multiplegames}=(1-{\frac {1}{n}})^{m}} ## Winning a Game of Bingo

The probability of winning at least one game out of {\displaystyle m} games is the same as the probability of not losing all {\displaystyle m} games:

{\displaystyle P(win)_{multiplegames}=1-(1-{\frac {1}{n}})^{m}} When {\displaystyle m=1} , these values are equal:

{\displaystyle P(win)_{multiplecards}=P(win)_{multiplegames}}

but is has been shown that that {\displaystyle P(win)_{multiplegames}>P(win)_{multiplecards}} for {\displaystyle m>1} . The advantage of {\displaystyle P(win)_{multiplegames}} grows both as {\displaystyle m} grows and {\displaystyle n} decreases. It is therefore always better to play multiple games rather than multiple cards in a single game, although the advantage diminishes when there are more players in the game.