Nerves of Steel (a dice game)

A carnival offers a game, called Nerves of Steel. It is played with a standard die, except that the "six" has been replaced with a skull.

You start by paying $7, then rolling the die. Each time you roll a number, that value is added to your running total. After each roll, you can either stop, receiving your running total in dollars, or you can roll again. However, if you ever roll a skull, the game immediately ends while you walk away seven dollars lighter. So, if you rolled a 5, then a 3, then stopped, you would win \$1 (5+3, minus the \$7 fee). If you rolled a 4, then a 4, then a 3, then a skull, you would lose \$7.

What strategy maximizes your expected winnings?

If you want, you can also try to find what your expected winnings will actually be when you play optimally, but I wouldn't recommend doing this without a computer.

• What if if you roll once and didn't get skull (say, got 5). Can you walk away by paying $7 -$5 ?
– PCM
Dec 18 '15 at 20:13
• @PCM Yes you can. Dec 18 '15 at 20:19
• The game has a positive payoff if the fee is <= $6.17... (for an optimal strategy) – smci May 13 '16 at 18:36 4 Answers Roll until you hit 15 dollars. If you consider the game without an initial fee, your goal is to maximize the amount taken. At every dollar point below 15 dollars, you stand to gain more on average from another roll than what you'd lose from rolling a 6. OLD The average amount of a roll that doesn't skull you is 3. If you roll once, 5/6 of the time you make 3 (-4 overall), and 1/6 of the time you make 0 (-7 overall). If you roll twice, 25/36 of the time you make 6 (-1 overall) and 11/36 you make 0 (-7). If you roll thrice, 125/216 you make 9 (+2 overall) and 91/216 you make 0 (-7). Four rolls, 625/1296 you make 12 (+5 overall) and 671/1296 you make 0 (-7). In every scenario, you lose money (on average) per roll, so the best way to maximize your earnings is to not play. That's averages, so let's look at discreet numbers. If you roll a 1, a subsequent roll will 5/6 of the time average to 4 (-3), and 1/6 to 0 (-7), and you can't get positive. If you roll a 5, a subsequent roll will 5/6 of the time average to 8 (+1), and 1/6 to 0 (-7). So, for every roll after the first, your expected earning is 15/6 and your expected loss is (total)/6. So, if you could somehow always get to 15, you should stop, since your expected roll after that is negative. This is the same as other answers but this is (to me) an easier way to see it. You are at a point where you will get \$$S if you quit right now. For instance, before your first roll, S=7. Your expected profit by rolling again is:$$1*\frac 1 6+2*\frac 1 6+3*\frac 1 6+4*\frac 1 6+5*\frac 1 6-S*\frac 1 6 =(1+2+3+4+5-S)/6 = (15-S)/6$$So: If$S<15$you earn a profit on average by rolling again so should. If$S>15$you lose money if you roll again so shouldn't. If$S=15$you are cost neutral so you are wasting time by rolling again. To maximize rate of profit generation: 1. Pay your$7 to play
2. Roll once
3. Quit
4. Repeat from step 1 indefinitely
• Maybe I misunderstood something, but with your strategy of just one roll you always lose money, even if it rolls a 5 every time, you will lose $2 each game Dec 21 '15 at 12:11 • @BianBBB I read the question's 3rd paragraph and understood "win" ot mean your profits. That means you keep the \$7 if you don't roll a skull. If I am wrong, then the accepted answer is slightly off. If \$7 fee is already lost, it does not count towards the \$15. Dec 21 '15 at 14:19

A quick Monte Python simulation (hey, I never saw that before!) indicates that @JonTheMon's idea works, at least for those kinds of strategies.

This graph shows the average payoff after 100,000 games played, where the strategy is to stop if your score is more that or equal to the $x$-axis number (labelled strategy). This shows a clear peak for a strategy of 15.

Here's the data if you want:

4 -3.24325
5 -2.74864
6 -2.23449
7 -1.94655
8 -1.68452
9 -1.43732
10 -1.27297
11 -1.09576
12 -1.01448
13 -0.92683
14 -0.86798
15 -0.82838
16 -0.85695
17 -0.87132
18 -0.94816
19 -0.95924
20 -1.00816

Using Markov Chains, you can figure out the probabilities after $n$ throws exactly. Here's a graph showing the probability of different payouts after $n$ throws for $1\leq n\leq10$.

Disclaimer: I've never been particularly good with probability, so please correct me if I'm wrong at any point.

I wrote a short Python script to generate the expected value after a given number of rolls. The expected value is calculated as the sum of the product of each outcome and its corresponding probability. For example, the expected value after one roll is given by $E=-7+1*1/6+2*1/6+3*1/6+4*1/6+5*1/6+0*1/6=-4.5$. That is, playing this game a large number of times with a single roll would—on average—net you -$4.50 on each game. Here's my code: import itertools sides = range(1, 7) for i in xrange(1, 10): rolls = itertools.product(sides, repeat=i) results = [] count = 0 for item in rolls: count += 1 if 6 in item: results.append(-7) else: results.append(-7 + sum(item)) print '%i rolls:$%.2f' % (i,  float(sum(results))/len(results))


The output of this script is as follows:

1 rolls: $-4.50 2 rolls:$-2.83
3 rolls: $-1.79 4 rolls:$-1.21
5 rolls: $-0.97 6 rolls:$-0.97
7 rolls: $-1.14 8 rolls:$-1.42
9 rolls: $-1.77  It looks like the maximum possible payout—if forced to play the game—is 5 or 6 rolls. However, these are both negative, so your best course of action is to just not play. • It looks like if the initial payment were \$6, you could eventually make money (average \\$0.03/game) by rolling 5 or 6 times every game. Dec 18 '15 at 22:01