Jump to content
BrainDen.com - Brain Teasers
  • 0

Question

This probability problem was born of this forum.

Upon meeting at a gambling establishment, CR and I got into a dispute over the probability of a certain event. CR believed that a particular slot machine paid off 1 time out of 3, when it flickered a blue light, and 3 times out of 7 when it was illuminated pink. I argued that it paid on average 3 times out of 8, no matter what light was on. We both agreed, however, that the blue light is on 9/16 of the time, while pink glows the remaining 7/16 of the time.

The dispute raged on, and neither of us could be swayed by the other's argument. Exasperated, CR offered to beat me with the whiteboard bearing his calculations. But I suggested a wager instead:

We'd set the odds as CR calculated; and I would choose which side to bet on. Thus we'd set our betting odds on "machine payoff" 1 ruble against 2 for a blue light and 3 rubles against 4 for a pink light. And I would, naturally, bet on a "payoff" in the case of "blue", and against a "payoff" in the case of "pink."

This being a popular slot machine and many casino guests waiting in line to get to it, we’d have no shortage of statistical samples.

"This way, each of us can stick to their opinion, but at the end of the day one of us will walk away richer," says I.

Is that bet a fair resolution for the dispute (statistically)? If not, how is it uneven?

Assume for the purposes of this problem that one of us is right and another is wrong (not both wrong). Make no assumptions regarding our respective faculties to estimate probabilities.

Edited by Prime
Link to post
Share on other sites

12 answers to this question

Recommended Posts

  • 0

As far as I can tell, from my old math days, you're both right. The chance of a win with the blue light is 33% (1 out of 3) and a win with the pink light is 43% (3 out of 7) probable. The chance of a win from the machine, regardless of color (your average), is 38% (3 out of 8). Really, the safe bet would be to bet against any pay off, but all this is based on your numbers, assuming the machine actually pays off with these numbers that is.

I may have misunderstood exactly what you were asking for though, if that's the case I'm sure you'll let me know.

Edited by FunkyM0nk
Link to post
Share on other sites
  • 0

If you are betting exactly on machine odds in both cases, how can you possibly expect a net gain or loss? I understand the argument, I understand both points, and I see that they are equivalent, but I don't understand the problem.

Link to post
Share on other sites
  • 0

like funkymonk said, CR's solution has an average payout of 3 out of 8, so if CR is right, then u r right, too.

but then again, CR might be wrong and you might still be right.

Therefore if one of you had to be right and one of you had to be wrong, you would be right and CR wrong.

As for the bet, it's really kind of stupid because you'd lose money if u bet for "payoff" because the odds are against it either way.

CR would walk away richer in the end because you would end up betting for "payoff" more than CR would because the blue light would appear more often than the pink light and you would be betting for "payoff" every time the blue light turns on.

in conclusion, according to the rules of the problem, you would be right and CR would be wrong, yet CR would walk away with more money, therefore your resolution was not very fair.

Link to post
Share on other sites
  • 0
If you are betting exactly on machine odds in both cases, how can you possibly expect a net gain or loss? I understand the argument, I understand both points, and I see that they are equivalent, but I don't understand the problem.

I did say, one of us is right and the other is wrong.

Link to post
Share on other sites
  • 0
CR believed that a particular slot machine paid off 1 time out of 3, when it flickered a blue light, and 3 times out of 7 when it was illuminated pink. I argued that it paid on average 3 times out of 8, no matter what light was on. We both agreed, however, that the blue light is on 9/16 of the time, while pink glows the remaining 7/16 of the time.

I did say, one of us is right and the other is wrong.

Given CR's position that a blue light comes on 9/16 of the time with 1 payout for every 3 bets, and a pink light comes on 7/16 of the time with a payout 3 out of every 7 bets, then the machine as a whole will pay out 3 out of every 8 bets. Both statements agree 100% and given one I could solve for the other. Therefore your other statement that one of you is false is in fact a logical impossibility.

Your question now is how should one bet given this knowledge? Given that you are betting with even odds against a random machine, I don't see how it makes a difference. It's a more involved coin flip: no one has a better chance of winning.

Link to post
Share on other sites
  • 0
Given CR's position that a blue light comes on 9/16 of the time with 1 payout for every 3 bets, and a pink light comes on 7/16 of the time with a payout 3 out of every 7 bets, then the machine as a whole will pay out 3 out of every 8 bets. Both statements agree 100% and given one I could solve for the other. Therefore your other statement that one of you is false is in fact a logical impossibility.

Your question now is how should one bet given this knowledge? Given that you are betting with even odds against a random machine, I don't see how it makes a difference. It's a more involved coin flip: no one has a better chance of winning.

(9/16)*(1/3) + (7/16)*(3/7) = 3/8 averages the same as my estimated probability of 3/8. But it does not follow that "Both statements agree 100%"...

In fact...

CR's odds 1:2 and 3:4 disagree with mine 3:5 on each and every individual bet.

A side note. In your earlier post, you zeroed in on the solution and placed it in the open. I was a bit disappointed how easy my problem was. Yet, your latter post implies that you did not find the solution, afterall?! After you placed it in the open, not even bothering to hide it inside a "spoiler". I presumed, because my problem wasn't worthy.

Link to post
Share on other sites
  • 0
As far as I can tell, from my old math days, you're both right. The chance of a win with the blue light is 33% (1 out of 3) and a win with the pink light is 43% (3 out of 7) probable. The chance of a win from the machine, regardless of color (your average), is 38% (3 out of 8). Really, the safe bet would be to bet against any pay off, but all this is based on your numbers, assuming the machine actually pays off with these numbers that is.

I may have misunderstood exactly what you were asking for though, if that's the case I'm sure you'll let me know.

Are you overlooking betting odds we set: 1:2 and 3:4?

Link to post
Share on other sites
  • 0
(9/16)*(1/3) + (7/16)*(3/7) = 3/8 averages the same as my estimated probability of 3/8. But it does not follow that "Both statements agree 100%"...

In fact...

CR's odds 1:2 and 3:4 disagree with mine 3:5 on each and every individual bet.

A side note. In your earlier post, you zeroed in on the solution and placed it in the open. I was a bit disappointed how easy my problem was. Yet, your latter post implies that you did not find the solution, afterall?! After you placed it in the open, not even bothering to hide it inside a "spoiler". I presumed, because my problem wasn't worthy.

I apologize for not placing things in spoilers. I was under the impression I was clarifying the problem, not solving it.

Prime's situation: In your statement that the odds of winning are 3/8, you are assuming that one cannot and does not look at the color of the lights until after you place your bet. The probability of winning once you place your bet is as described by CR. Odds are evenly matched this way, so you will come out even.

CR's situation: CR knows the color of the lights. Therefore he is placing one of two different bets. In both cases the odds are even and he comes out even in the long run.

I think a helpful thing for understanding the 3/8 thing is that there are two possible ways for you to end up betting, both with different odds. CR's explanation would be like me going to a roulette wheel and betting conditionally, saying if the number is even I will bet on 16 and if it is odd I will bet on 17 (with the odds adjusted to match). Prime's explanation is like me placing a bet on the adjoining line between 16 and 17.

Link to post
Share on other sites
  • 0
I apologize for not placing things in spoilers. I was under the impression I was clarifying the problem, not solving it.

Prime's situation: In your statement that the odds of winning are 3/8, you are assuming that one cannot and does not look at the color of the lights until after you place your bet. The probability of winning once you place your bet is as described by CR. Odds are evenly matched this way, so you will come out even.

CR's situation: CR knows the color of the lights. Therefore he is placing one of two different bets. In both cases the odds are even and he comes out even in the long run.

I think a helpful thing for understanding the 3/8 thing is that there are two possible ways for you to end up betting, both with different odds. CR's explanation would be like me going to a roulette wheel and betting conditionally, saying if the number is even I will bet on 16 and if it is odd I will bet on 17 (with the odds adjusted to match). Prime's explanation is like me placing a bet on the adjoining line between 16 and 17.

I see now that my statement of the problem is too confusing. The essence of the problem is that one man sets the odds, another man chooses which side he bets on. For example, CR says: "The light is blue, the odds for "payoff" event are 1:2". I say: "OK, then I bet my 1 rulbe against your 2 rubles that the machine will spew change next time someone pulls the lever." Or the light is pink and the odds according to CR are 3:4. Then I bet my 4 rubles against CR's 3 that the machine will NOT payoff next time someone pulls the lever.

Link to post
Share on other sites
  • 0
I see now that my statement of the problem is too confusing. The essence of the problem is that one man sets the odds, another man chooses which side he bets on. For example, CR says: "The light is blue, the odds for "payoff" event are 1:2". I say: "OK, then I bet my 1 rulbe against your 2 rubles that the machine will spew change next time someone pulls the lever." Or the light is pink and the odds according to CR are 3:4. Then I bet my 4 rubles against CR's 3 that the machine will NOT payoff next time someone pulls the lever.

Do you agree that CR is correct in his prediction of odds in both cases? If so, the odds he has chosen are fair, and you could bet either way on the outcome and not expect a difference.

In fact, if he is correct in his 1/3 and 3/7 predictions, you could make a statement that the machine will pay off an average of 9 times out of 10 and it would be irrelevant. If CR is correct, then you can't expect to win or lose with those odds no matter which side you bet with which light. The fact that you are also correct in saying the machine pays you off 3/8 times overall is just incidental to the problem.

Link to post
Share on other sites
  • 0

If I'm understanding this right:

It's unfair towards me, because I expect to break even, even if I'm right:

If the blue light is on, I believe the probability of a win is 1/3. At 2:1 odds, I lose $2 1/3 of the time, and win $1 2/3 of the time.

-2*1/3+1*2/3=0

If I'm wrong, and the probability of a win is 3/8, I lose $2 3/8 of the time, and win $1 5/8 of the time.

-2*3/8+1*5/8=-1/

I guess I would only take that bet if I thought the utility I would get from proving my calculations correct would offset the money I could expect to lose if I were wrong

Is that what you were going for?

Link to post
Share on other sites
  • 0
If I'm understanding this right:

It's unfair towards me, because I expect to break even, even if I'm right:

If the blue light is on, I believe the probability of a win is 1/3. At 2:1 odds, I lose $2 1/3 of the time, and win $1 2/3 of the time.

-2*1/3+1*2/3=0

If I'm wrong, and the probability of a win is 3/8, I lose $2 3/8 of the time, and win $1 5/8 of the time.

-2*3/8+1*5/8=-1/

I guess I would only take that bet if I thought the utility I would get from proving my calculations correct would offset the money I could expect to lose if I were wrong

Is that what you were going for?

Your answer is essentially correct. Here is how I'd formulate it:

NO CALCULATIONS ARE NEEDED TO ANSWER THE QUESTION!

If CR's odds are correct, no matter which side I bet on, we'll break even in the long run. (For then I bet fair odds.) If my calculations are correct, each individual bet is in my favor, and I win in the long run. So the outcome of such wager: Either I win, or we break even. No risk for me, some risk for CR.

The numbers are there to create a distruction and lead you astray. But if you insist:

If I were right, I'd win on average 1/8 of a ruble on each blue-light bet, and 3/8 of a ruble on each pink-light bet, with an average of (1/8)*(9/16)+(3/8)*7/16=15/64 of a ruble.

Itsclueless already noted the "break even" possibility in earlier post, but did not describe what happens if CR were wrong. I guess, what threw people off in this problem, no one could even begin to consider the possibility of CR miscalculating the odds. I should have used some fictional character instead. ;)

Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
  • Recently Browsing   0 members

    No registered users viewing this page.

×
×
  • Create New...