What is the expected value of the following game?

Question

You are offered a game in which you select one bill at random from 4 bills: $1, $5, $10, and $20. You are given the choice to either keep the bill you drew or draw again with replacement, but if you draw the second time, you must keep it. What is the expected value of this game? How does the expected value change if you can choose to redraw a second time?

I got $12 for the first game, but was not sure how to tackle the variation with a second redraw

Answer 1

First of all the expected value for the first pick is $(1+5+10+20)/4 = 9$

Then it depends on whether the bill is returned to be selected from.

If it's returned it's quite straight forward. The strategy must be that if you don't pick the 10 or 20 bill you try again. So in 25% of the cases you settle for your 20, 25% for the 10 and otherwise try again with an expected value of 9. So the result would be $20/4 + 10/4 + 9/2 = 12$.

If it's not returned the expected result of the second draw is dependent on the first bill:

• 1 means that the expected value of the next is $(5+10+20)/3 = 35/3$, it's correct to draw the second here
• 5 means $(1+10+20)/3 = 31/3$ that is draw the second here too
• 10 means $(1+5+20)/3 = 26/3$ that is we do not draw the second
• 20 means $(1+5+10)/3 = 16/3$ that is we do not draw the second.

So the result is that in 25% of the case we settle for 20, 25% we settle for 10, 25% we draw a second bill with the expected value of 31/3, and 25% we draw a second with the expected value of 35/3. So the result is

$${20 + 10 + 31/3 + 35/3\over 4} = {60+30+31+35\over12} = 13$$

Answer 2

Expected value for first draw = $(1+5+10+20)/4 = 9$

The question clearly specifies "draw again with replacement" for the 2nd draw, and it has also been clarified that we see what was first drawn.

Since the expected value of the redraw is also necessarily $9$, we ahould be satisfied if we got $10$ or $20$ first time, else redraw.
Expected value with this strategy = $\dfrac14(10+20) + \dfrac12\cdot9 = 12$

Answer 3

I also get $12.

1. If you have just 1 attempt the expected value is \$9. I guess it's clear.

2. Now you need to define a strategy for deciding if you make the second attempt or keep the bill you got on the first attempt. Strategy is simple: if the expected value of the single attempt is higher than the bill you already have you should make the second attempt. So, if you got \$1 or \$5 you do make the second attempt.

3. Now let's analyze possible outcomes.

   • case1: first attempt gives you \$20. Probability 1/4, game is over, outcome \$20.
   • case2: first attempt gives you \$10. Probability 1/4, game is over, outcome \$10.
   • case3: first attempt gives you either \$1 or \$5. Probability is 1/2. You decide to make a second attempt and this case gets splited into 4 subcases (you get \$1, \$5, \$10 or \$20).

If you get into case 3 the probability of all the 4 subcases is equal and is 1/4. So, the probability that you get in some particular subcase during the game is (probability you get into case3) * 1/4 = 1/8.

Now you are ready to calculate the expected outcome of entire game: you have 6 possible game scenarios (case1, case2, 4 subcases of case3), you know probability of each scenario and outcome. I'll use a small shortcut however:

1/4 \$20 + 1/4\$10 + 1/2 * \$9 = \$12.

(\$9 here is an expected outcome of a single-draw game).

BTW, what if you have 3 attempts?

My answer would be:

1/4 * \$20 + 3/4 * \$12 = \$14