Blindfold Bingo

Question

Congratulations, you have been selected from among the audience to compete in the fabulous Blindfold Bingo challenge! (cue annoying, yet catchy theme music.)

Here you see our perfectly ordinary Bingo machine:

Well, almost ordinary! In addition to having just one output tube, it also has two input tubes. (click the image to see the full schematic)

Let's recap the rules: (cue near-silent suspense-building music)

There are 75 White bingo balls. One of them has been randomly preselected to be the ODDBALL. The balls are numbered, so they can be distinguished by looking at them, but not otherwise. All the white balls, including the oddball, are shuffled, and placed into the white input tube in a random order. 75 balls, incidentally, is the maximum number of balls an input tube can contain.

Then there's the yellow input tube filled with 75 yellow bingo balls, indistinguishable from the white balls except by looking.

The bingo cage is initially empty. (music fades out)

You are allowed to use these two actions, as many times you want, in any order, and there's no time limit:

1: choose an input tube, and drop one ball from its bottom end into the bingo cage. The balls will come out of the input tube in the same order that they were put in. (Actions A and B in the schematic)

2: shuffle the balls that are in the cage, randomly pick one, and place it into the input tube of your choice at the top end of that tube. You can only place a ball into a tube that has 74 or fewer balls in it. (Actions C and D in the schematic)

After you are done, all the balls must be back in the input tubes, and if the ODDBALL is now in the YELLOW INPUT TUBE, you win the car! (cue over-the-top sound effects, ridiculous camerawork and scantily clad gentlemen making the cheapo car seem outright luxurious)

Now, let's check your blindfold.. yes, good .. aaaaand GO! (cue lights dimming, dramatically suspenseful music..)

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Daily Hint 0:

@TwoBitOperation has posted a strategy that gives better than 50% chance of winning. However, a strategy that wins more than 60% of the time does exist.

Daily hint 1:

@noedne has posted a strategy that gives better than 60% chance of winning. However, a strategy that wins more than 70% of the time does exist.

Daily hint 2:

As of yet, no-one has posted a strategy that gives better than 70% chance of winning. However, a strategy that wins more than 80% of the time does exist.

Daily hint 3:

As of yet, no-one has posted a strategy that gives better than 80% chance of winning. However, a strategy that wins more than 90% of the time does exist.

Daily hint 4:

The intended solution wins about 93% of the time. In a 10000 run simulation, it got, on average, about 70 white balls (70.12) into the yellow tube, and vice versa.

Answer 1

I have a solution with a success rate of 93.5%, according to my simulations.

Start by dropping a single yellow ball into the cage.
Next, drop a single ball from the white tube into the cage, then pull out one ball and put it back in the white tube. Do the above step 75 times, once for every ball in the white tube. The ball in the cage is very likely to be white at this point. Put it in the yellow tube.
Repeat from the beginning 75 times, once for each yellow ball.

The reason this solution works so well is

The yellow balls tend to end up towards the "beginning" of the white tube, the balls that get dropped into the cage just after the yellow ball is added. As a result, we're still likely to pull out a white ball even after many yellow balls are in the white tube.

Here's my code that I used to verify my solution:

Try it online!

Answer 2

One strategy:

Drop all white balls. For 75 times, drop a yellow ball and add a random ball to the yellow tube. Add all balls to the white tube.

Probability of success:

The desired white ball does not end up in the yellow tube if and only if it failed on each draw, with probability $75/76$ each time. Then the probability of failure over 75 trials is $(75/76)^{75}$, so the probability of success is $1-(75/76)^{75}\approx1-1/e$.

Answer 3

I don't know how to calculate probability - it is not over 90%, but:

1 Drop all from white tube and one from yellow tube. Take one random back to yellow tube and rest back to white tube.
2 Now drop 74 from yellow tube (since last one is most likely white and I want to keep it in yellow tube for rest of the game), and one from white tube. Now place one random back to white tube and rest to yellow tube. Rotate yellow tube by 1, so first is now last. (and last should now be my only white in yellow tube) (not sure if rotating decrease chances, comment below)
3 Now drop 74 from white tube and one from yellow tube, take one random back to yellow tube and rest back to white tube. Rotate white by 1.
4 Now drop 73 from yellow tube, and one from white tube. Take one back to white tube, rest to yellow tube and rotate yellow tube by 2 (since two were most likely white).
And so on.

However:

If I don't rotate and just drop always 75 "likely" white against 1 yellow (when i have 74y and 1w in yellow tube and i want to increase chances of picking two white to yellow tube) and vice versa - I don't believe it increases chances, but maybe someone could figure it out how to calculate this probability. I think this could be that (1-1/e).

But mainly I'm writing, because I wanna see correct solution.

Answer 4

This is probably not a perfect answer, but it's a start:

I would perform the following steps:

1. Starting state

2. Drop all yellow balls from yellow tube into the cage

3. Alternate dropping one white ball from white tube into cage and pulling one ball from the cage into the white tube. Do this 37 times.
4. Drop 38 remaining white balls from white tube into the cage

5. Move 75 balls from the cage into the yellow tube.

Visual:

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This makes my odds of winning >50% for sure, but I need to work on the exact math. I'm also thinking the process may be iterable in some way.

ODDS:

a) probability oddball is in cage by step 4:
$(37/75)*((75/76)^{37}) + (38/75) ≈{0.81}$
b) probability oddball put into yellow tube assuming ball in cage on step 4:
$(1- (112!/113!)/(37!/38!)) ≈ {.665}$
probability of success: ${0.81}* {.665} ≈ 0.55 $

Answer 5

I've come up with this which seems to be an improvement upon the other answers a duplicate of Jan's answer:

1. Drop all 75 balls from white tube.
2. Replace 1 ball from the yellow tube.
3. The ball in position 74(0 indexed) of the yellow tube now has odds 75/76 of being white.
4. Place all balls from basket back into the white tube

From N = 1 to 74 loop

1. Drop balls 0-74-N from yellow tube.
2. Replace 1 white tube ball.
3. Put all remaining balls in the basket back in the yellow tube.
4. Drop 1 yellow tube ball and add it back N times. (Moves the highest chance white ball(s) to the back)
5. Drop balls 0-74-N from the white tube.
6. Replace 1 yellow ball.
7. Put all remaining balls in the basket back in the white tube.
8. Drop 1 white tube ball and add it back N times. (Moves the highest chance yellow ball(s) to the back)

Once my code is finished executing the runs I'll post the accuracy.