Consider a game where you have to choose between 1 of 2 payoffs:

Payoff 1. You toss 100 coins.

You get $1 for each Head that appears.

Payoff 2. You toss 10 coins.

You get $10 for each Head that appears.

Which payoff structure would you choose? And why?

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## Comments

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TopNewestI assume each coin toss is independent from one another (so the covariance between each coin toss is zero), and used only fair coins.

Yes, the expected Payoff is the same for ONE and TWO. But their variance differ.

\[ V_1 = \text { Probability } \times \text { Payoff ONE }^2 \times \text { Number of trials } = \frac {1}{2} \times 1^2 \times 100 = 50 \]

\[ V_2 = \text { Probability } \times \text { Payoff TWO }^2 \times\text { Number of trials } = \frac {1}{2} \times 10^2 \times 10 = 500 \]

Because Payoff TWO has a higher variance which is not desired, I will go with Payoff ONE.

\[ \text{ I learned from the best tutor } \]

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What term do we use to describe a person who choses to "minimize variance when expected value is equal"?

Your answer greatly depends on what it is that we wish to maximize / minimize. There could be people who may want to maximize variance, because of the higher likelihood of a huge payoff that they care about.

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What term do we use to describe a person who choses to "minimize variance when expected value is equal"?

Risk-averse investorLog in to reply

It's less of a hassle to toss 10 coins.

If we compare this to stocks, it's easier to keep a portfolio of10 stocks than 100 stocks. And lower commissions. The risk of loss is higher, but the chance of higher returns is better too.

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That's a good point! We might need to factor in convenience / the cost of making a flip!

Can you clarify what you mean by "chance of higher returns is better too"? How do you quantify that?

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Earlier posters have quantified the variance for the smaller vs. the larger set of coins. For the set of 10 coins, the chance of getting more than 9 heads is 1 in 1024. For the set of 100 coins, the chance of getting more than 90 coins is much lower than that. I'd have to get my combinatorics (or statistics) hat on to quantify it, but it's much less. On the order of more than 5-sigma from the mean.

For stocks, a stock club (or newsletter writer) would be silly to buy 1 share each of the 500 largest companies. A number of funds exists to do that with much less expense. If they want to try to beat the performance of the S&P 500, they pick a small number of stocks (using some rational criteria) and often do better than the "market".

Of course, the risk of loss (or underperformance) is higher too.

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Let me try.

Payoff 1. Chance to get head for all coins = 1 out of 10^{200} = Chance to get $100 in terms of $1.

Payoff 2. Chance to get head for all coins = 1 out of 10^{10} = Chance to get $100 in terms of $10.

Thus, I think Payoff 2 is better.

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That's a valid argument if your goal is to "maximize the probability of the highest payoff".

Since getting $99, $98, ... $91 is somewhat similar to getting $100, would this affect your answer? Why, or why not?

Note that the payoff 1 should be "10 ^ {100}"

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The probability of getting payoff between 91$ to 100$ increases in the first case. Hence, if I consider myself satiated by $99, $98, ... $91, I'd prefer case 1.

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This is optimistic way I guess

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what good

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They average out to the same payoff (which is $50.00), so I would flip a coin to decide which to use.

Python:Log in to reply

Expected outcome is same in both cases ,but in case 1 chances are that outcome is more closer to 50$ than in case2, So for more standard outcome I will choose case1.

For suppose if my need demands 60$ I would choose case 2.

CASE1

Chances to get 50$ is 100C50 *.5^{100}=.0786

49$ is 100C49*.5^{100 }=.0780 (same for 51$)

For 60$ it is .0108

By similar calculations(and adding), chances to get outcome from 40$ to 60$ is .946.

where as in case 2 chances reduce to .656 this proves that there are more chances to get outcome closer to 50$ in case 1.

probabilities to get 60$ are 0.108 for case 1

.205 for case 2.

So, if 60$ is my need I will choose case 2

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You bring up a valid point, which is "What is the utility function that I care about". If the utility function is just "I only care if I have more than $60 or not", then your approach shows that we should go for payoff 2.

What kind of utility function makes sense? Do people value all amounts of money equally?

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What are the statistics given that heads side of a coin is slightly heavier than tails? Giving tails more of a chance of facing up? Is that true? I heard that somewhere back maybe in 4th grade haha. Or has that been proven incorrect? if it is true, I would rather chose b since I would want less of a chance of getting tails side up? Can you explain to me how I am wrong (I probably am) like you would explain it to a 3rd grader?

This is why I'm more of an artist than math person.

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That is a very good point. How would your answer change if we didn't have a fair coin? For example, if the coin was totally biased and only gave tails, then both payoffs are the same. Similarly, if the coin was totally biased and only gave heads, then both payoffs are the same. But in between, the payoffs are different. So, does that mean that the answer could change depending on the probability of head/tail? If so, how and why?

How is payoff 2 "less chance of getting tails side up"? Just because of the absolute number? But, the first mostly likely has many more heads appearing, just a smaller payoff each time.

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Less probability of a heads since the gravitational pull of the heavier side. Sorry I missed stated. On the other hand, my fiance makes a good point. The more you flip a coin your results are more likely to reflect the chance of a 50/50 so it is better and more reliable to flip more times than less so option 1 of 100 flips is more reliable then just 10 in case you get a lucky streak of a ten or an unlucky streak with only 2 out of 10.

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Expected Value of both payoffs would be the same, regardless of the probability of landing heads / tails. For example, if the coin is fair, and it will land heads 50% of the time, then the expected payoff in 1 is $50, and the expected payoff in 2 is $50. More generally, if it land heads \(p%\) of the time, then the expected payoff will be \($p \).

As it turns out, theSo, the question then becomes, pick your favorite probability \( p \) of landing heads, which payoff would you want? Is there anything else to consider?

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Payoff 2 as it takes less time to toss 10 coins and the value is same for both

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I will chose payoff 2 because while tossing coins 10 time we may get HEAD more than 1 or 2 times and the pay is more. But in payoff 1 tossing coins 100 times may give a HEAD 20 times and u get 20$. But in payoff 2 you can get more than 20$.

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Payoff 2 because I'll make decent money and i Don't have time to flip 100 coins. Stop over complicating things with math people.

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Pay off 2, because it will likely to get heads more often and earn $10 for each head than 100 coin tosses. Who knows that in a toss of 100 you will only get less than 50% heads.

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In my opinion, it's a matter of risk: payoff 1 has more consistency but less probability of scoring extremely large amounts of money, whereas payoff 2 is a sort of high risk high reward model where you either get a lot or get a bit. Personally, I would choose payoff 1.

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I think payoff 1 is better because chances increses with more no of toss..

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playoff 1

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If i understand the problem correctly, would be indifferent to the payoffs structure, cause both options give the same expected value which would be 50.

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Comment deleted Mar 01, 2015

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Notice the number of coins that are tossed. In the first payoff, if all 100 coins were H, you will get $100.

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I'd go for #2. I'm a risktaker.

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You will still need to define what you mean by "risktaker". There are possible definitions of "risktaker" which would make option 1 more valuable to them.

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Payoff 1 so that I can get more number of chances

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