A gambler who makes 100 bets of $1, each at payoff odds of 8 to 1. He wins 10 of these bets and loses 90. How many dollars has the gambler gained overall?
I don't seem to understand what "odds of 8 to 1" means. Can someone please explain this to me?
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Odds of 8 to 1 means that for every \$1 you wager, you could either
- win \$8 in addition to receive your \$1 wager back
- lose your \$1 wager
So from his 10 wins, he gets \$80, and from his 90 losses, he is down \$90. Net, he is down \$10.
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Odds of 8 to 1 mean you make a bet of \$1, then you win \$8. You also receive your original \$1 back, so if you win an 8 to 1 bet, you pay \$1 and receive back \$9.
In regards to your problem, the gambler pays in $\$100$, and gets back $\$90$, so all in all, he has lost $\$10$.
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"8 to 1" is a not very exact, but it means in general that you win 1 times while you lose 8 times. Thus, you have around 1 times from 9 tries. It is like throwing a 9-sided dice:
In the context of this task, it means that you win 9 times of your bet, if you win. It is a different thing. The win is always smaller as your chance.
The problem doesn't say, what is the chance, but it is not needed here, because it says, how many times he won.
Because he won 10 times, he gets \$90 from the wins. He also lost 100 in the bets. His total result is a \$10 loss.
P.s. Don't gamble. Never ever. You are a very beginner in probability theory - if you wouldn't, you would understand why you shouldn't ever gamble. No, there is no some "tricky formula" with what you could win. And it is not a complex proof, it is a very-very basic thing in the probability theory. You could ask a new question from the details.
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$\begingroup$ @JohnDoe Thanks - maybe you could read also the text below the picture. $\endgroup$ – peterh May 10 '17 at 3:03
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$\begingroup$ Yes, the edit improves the answer. However it is still slightly wrong - 8/1 actually means (in a theoretical sense) that you win $\frac19$ of the time. This is because at 8/1 odds, you get back \$9 if you win a bet (\$8 winnings and your \$1 returned). So if you make 9 bets at 8/1, with probability of winning being $\frac19$, you would expect to win 1 and so get back what you paid in, which makes sense (in practice bookies hold a slight advantage however). In the answer you offered, you forgot to factor in that in the 10 wins he has, he also gets back his original stake of \$1. $\endgroup$ – John Doe May 10 '17 at 3:09
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1$\begingroup$ @alexis I am not sure, what are you talking about. I didn't use the "payoff odds" terminology in the answer. I think the probability theory part is now okay, this is how the answer was focused on. I also googled for "payoff odds", I didn't find any reasonable, although I think I am enough good in probability theory to answer this question. Could you please detail, what is the problem with the answer now? I am ready and willing to fix/improve any mathematical problem in it, if there is still any. $\endgroup$ – peterh Sep 24 '17 at 12:48
