# Disproving the Gambler's Paradox (part 2)

A certain slot machine is rigged to pay out, on average, once every 10 games. It costs 1 dollar to play, and the machine pays out 11 dollars if you win. On average, then, you will be up one dollar every ten games.

However, there is a catch. The slot machine is an up-and-coming machine and takes a credit card. Due to losing money on the machine, the casino will only accept up to 10 games per card, so you can only play 10 times.

Now, a certain person has only one credit card. Suppose they've played the slot machine three times already. They are three dollars down, with 7 games remaining. Is it profitable to keep playing until they win? Was it profitable when they first started playing?

• What have you figured out so far? Where are you stuck? – ConMan Oct 11 '18 at 22:25
• According to my math (which may be wrong), they are more likely to lose money now that they've played three games and lost. But that wasn't true at the beginning. Therefore, I am uncertain as to whether they should keep playing or not. I've tried and I cannot seem to express the math on paper. – Jonathan Graef Oct 11 '18 at 22:29
• Do you mean "they're more likely to have an overall loss, including the \$3 already spent" or "they're more likely to make a loss on the remaining \$7 compared to if they just cash out now"? Because the first might be true (I haven't run the sums), but the second shouldn't be. – ConMan Oct 11 '18 at 22:32
• I was under the impression that, according to current theory, they would run a loss both ways (both including and excluding the -\$3). However I do not know how to prove it. But at the beginning, they stood to run a profit. – Jonathan Graef Oct 11 '18 at 22:38
• Linearity of expectation. – Doug M Oct 11 '18 at 23:02

This is a tricky problem but here's my best shot.

Profitable implies a wise business decision. The dilemma here is being limited to only another $$7$$ plays. One can look at the expected winnings as a summation of the product of probabilities of winning times the amount won for all $$7$$ plays and compare it to the expected loss which is a much simpler $$0.9^7\cdot 10 = \4.78$$.

Expected winnings (already $$\3$$ down) are:

$$E(x) = (0.1\cdot 8) + (0.9\cdot 0.1\cdot 7) + (0.9^2\cdot 0.1\cdot 6) + (0.9^3\cdot 0.1\cdot 5) + (0.9^4\cdot 0.1\cdot 4) + (0.9^5\cdot 0.1\cdot 3) + (0.9^6\cdot 0.1\cdot 2) = \2.83$$.

Here is the confusing consideration which may lead to making an incorrect decision. As $$\2.83 < \4.78$$ it appears continuing isn't a profitable proposition. But because one is going from $$-\3$$ to +$$\2.83$$ one is actually gaining $$\5.83 > \4.78$$ so it makes sense to continue.

Also, there is a psychological phenomenon called loss aversion, whereby people take a slightly higher risk in wiping out a loss (which isn't the case here). So if one just looks at the bare probability of winning versus losing there is a slightly higher probability of winning $$1 - 0.9^7 = 0.5217$$ versus losing of $$0.4783$$ so one may opt to continue simply based on this slightly higher probability.

From the start, with $$10$$ plays available the expected loss is $$0.9^{10}\cdot 10 = \3.49$$ versus..............$$\4.79$$ in winnings. So this makes sense from a profit perspective. But this is where the paradox comes in to confuse the issue. That is, it never makes sense to bail on an initially profitable series of $$10$$ plays because one hasn't won yet.