Being in a bar, you start chatting to a guy who provides you a difficulty. He hands you 5 red and 2 black cards. After shuffling, you lay them on the bar, deal with down. He bets you that you can not turn over three red cards. And to help you, he explains the chances. When you draw the first card, the odds are 5-2 (5 red cards, two black cards) in favour of choosing a red card. The 2nd draw is 4-2 (or 2-1) and the 3rd draw is 3-2. Each time you draw a card the odds seem in your favour, in that you have more chance of drawing a red card than a black card. So, do you accept the bet? If you answered yes, maybe it's time for you to go over your mathematics. It's a silly bet. The odds offered above are only for a best draw. The real odds of you being able to carry out this accomplishment are actually 5-2 against you. That is, for every single 7 times you play, you'll lose five times.

Let's assume that you took the bet and, nearly undoubtedly, lost cash. read more But this is simply for enjoyable, right? So your new "buddy" recommends a way that you can get your refund. He takes 2 more red cards and hands them to you, so you now have 7 red cards and 2 black cards. You shuffle the nine cards and lay them out, deal with down, in a three by 3 grid. He bets you even money that you can't choose a straight line (vertical, horizontal or diagonal) that has only red cards.

Intuitively, this might sound like a much better bet and the odds are really evens if the 2 black cards are beside each other in a corner (see image). In total there are 8 lines to pick from and four consist of only red cards, and four contain a black card. But that is as excellent as it gets. If the black cards are in opposite corners then you can just win by picking the centre horizontal or vertical row so the odds are 6-2 (or 3-1) versus you winning. Every other layout gives you three winning lines and 5 losing lines. This bet only has 12 ways of being successful, versus 22 methods of you losing. Barely an even-chance bet.

Try to examine the chances for this proposal bet. You shuffle a pack of cards and cut it into three stacks. You are used even money that one of the cards on top of the stacks will be a photo card (a jack, queen or king). That is, if a photo card appears, you lose. Do you think this is a great bet? One way of thinking is that there are only 12 losing cards against 40 winning cards, so the odds look much better than evens? However this is the wrong way of looking at it. It is actually what's known as a combinatorics issue. We need to likewise realise that we are just selecting three cards at random. There are 22,100 ways of picking 3 cards from a 52 card deck. Of these, 12,220 will consist of a minimum of one photo card-- so you lose-- meaning that 9,880 will not include a picture card-- when you win. If you equate this to chances, you will lose fives times out of every 9 times you play (5-4 versus you). The even chance wager you have actually been used is not the great value that you believed it was and you will lose cash if you play a few times.

We can all agree that you have a 50/50 possibility of thinking heads or tails in a coin toss. However if you toss the coin 10 times, would you expect to see 5 heads and five tails? If you were provided odds of 2-1 to attempt this, would you take the bet? You 'd be a sucker if you did. Five heads and five tails will happen regularly than any other mix, however there are lots of other ways that ten flips of a coin can land. In truth, the bet is 5-2 against you. Another name for a proposal bet is the "sucker" bet, and there is not a surprise who the sucker is. However don't feel too bad. We are all usually extremely bad at evaluating real odds. A well-known example is the Monty Hall Problem. Even mathematicians might not agree on the ideal answer to this apparently simple problem.