JOKER DRAW - A weekly game played at a pub in Cambodia

Mr Wizard, really need your expertise regarding this question.

It involves a single deck of 53 playing cards (including 1 Joker). A new deck of 53 cards are thoroughly mixed face down on a table. After thoroughly mixed, the cards are brought together as a complete deck of cards. Then 1 by 1, from the top of the deck, each card is placed face down and affixed to the Joker drawing board. Then the Joker Draw begins. Each week a player has the opportunity to randomly draw from the Joker board. Obviously, the probability is 53-1 the first draw will unveil the Joker. In the event the Joker is not drawn, that card is placed face up and re-affixed to the board. Each subsequent week the odds of drawing the Joker go from 52-1, 51-1, 50-1, accordingly. So here is the question. What are the odds or probability that the Joker draw will go the entire 53 weeks until the Joker is finally unveiled?

Most of my friends say the probability or odds remain the same at 53-1 after 53 weeks. To me, this logic does not seem right. I think, the initial probability of 53-1 will substantially change at the end of the 53 week cycle.

Mr Wizard, can you or any BM provide a clear answer to this question? Also, provide the formula of probability increase if being the case. Thank you very much.

SS

If I understand the rules correctly I believe there is a 98.1132% chance the joker will not be the last card remaining.

Humorously, I made an Excel table with some formulas to derive this. Looking at the answer though it does indeed appear that 98.1132% equals....52/53 ;)

SS, what you're really asking here is what are the odds a card from the NotJoker pool will be drawn instead of a card from the Joker pool. Initially there are 52 NoJokers and 1 Joker. For the Joker to not be drawn week 1 you need to pick 1 of 52 out of 53 cards. Then in week 2 you have to pick 1 of 51 out of 52 cards, etc. As you can guess your initial odds of not picking the joker are quite good, so that for example you have a 96% chance of not picking it after two weeks. The 50/50 mark is, as expected, after 26 weeks. Anyways, the odds of picking 52 out of 53, 51 out of 52....2 out of 3, 1 out of 2, leaving the joker is, if my math is right, 1.8868%, or 1/53.

Incidentally if this game was instead sometime like a search for black suited cards your intuition would be correct. As the class of red cards would be rapidly falling below the number of black cards remaining, you have 99.96 chance of finding a red card in the first ten draws (versus less than 20% for a joker in the actual game), and the chances of winding up with all 26 black cards left is something like .000000000000002%.

Quote: SS

Most of my friends say the probability or odds remain the same at 53-1 after 53 weeks. To me, this logic does not seem right. I think, the initial probability of 53-1 will substantially change at the end of the 53 week cycle.

Start making bets with your friends!!

Interesting. I too had to make an Excel chart. Here's what I got:

Week / Draw	Cards Remaining	No Joker	Combined
1	53	0.9811	0.9811
2	52	0.9808	0.9623
3	51	0.9804	0.9434
4	50	0.9800	0.9245
5	49	0.9796	0.9057
6	48	0.9792	0.8868
7	47	0.9787	0.8679
8	46	0.9783	0.8491
9	45	0.9778	0.8302
10	44	0.9773	0.8113
11	43	0.9767	0.7925
12	42	0.9762	0.7736
13	41	0.9756	0.7547
14	40	0.9750	0.7358
15	39	0.9744	0.7170
16	38	0.9737	0.6981
17	37	0.9730	0.6792
18	36	0.9722	0.6604
19	35	0.9714	0.6415
20	34	0.9706	0.6226
21	33	0.9697	0.6038
22	32	0.9688	0.5849
23	31	0.9677	0.5660
24	30	0.9667	0.5472
25	29	0.9655	0.5283
26	28	0.9643	0.5094
27	27	0.9630	0.4906
28	26	0.9615	0.4717
29	25	0.9600	0.4528
30	24	0.9583	0.4340
31	23	0.9565	0.4151
32	22	0.9545	0.3962
33	21	0.9524	0.3774
34	20	0.9500	0.3585
35	19	0.9474	0.3396
36	18	0.9444	0.3208
37	17	0.9412	0.3019
38	16	0.9375	0.2830
39	15	0.9333	0.2642
40	14	0.9286	0.2453
41	13	0.9231	0.2264
42	12	0.9167	0.2075
43	11	0.9091	0.1887
44	10	0.9000	0.1698
45	9	0.8889	0.1509
46	8	0.8750	0.1321
47	7	0.8571	0.1132
48	6	0.8333	0.0943
49	5	0.8000	0.0755
50	4	0.7500	0.0566
51	3	0.6667	0.0377
52	2	0.5000	0.0189
53	1	-	-

The third column indicates the odds of picking anything other than the Joker on the particular draw. The fourth column is the combined odds of having not picked it so far. The fourth column is the prior fourth column times the current third column.

You'll note that 1 / 53 = 0.0189.

Note that this could be simplified by stating: Take a 53 card deck (including joker). Shuffle it.
If you draw cards individually, what are the odds that the joker is the last card?
Or even more simply:
What's the odds that the joker is shuffled into the last spot?

No matter how you slice it, it's still 1/53.

Quote: SS

What are the odds or probability that the Joker draw will go the entire 53 weeks until the Joker is finally unveiled?

Most of my friends say the probability or odds remain the same at 53-1 after 53 weeks. To me, this logic does not seem right. I think, the initial probability of 53-1 will substantially change at the end of the 53 week cycle.

These are two different questions.
The probability that a joker will be the last of a fairly shuffled deck of cards is 1/53.
However, based the conditional probability does not remain the same after 53 weeks (presumably you mean after 52 weeks, just before the 53rd draw) because over time your knowledge of the drawn cards increases. After 52 weeks one of two things is true:
1) You've already drawn the joker, in which case the probability of the joker being the last card is 0.
2) You haven't yet drawn the joker, in which case the probability of the joker being the last card is 1.

In general, if you haven't drawn the joker after N cards, then at that point in time the conditional probability that the joker is the last card has increased to 1/(53-N). However, this is a practically useless fact if you cannot bet on the game after the drawings begin.

Based upon M.E.'s response, the third column in my chart above (No Joker) is the "Conditional Probability."

The best way is to look at the series 52/53 * 51/52 ... 2/3 * 1/2 note that the Numerator (top) is 52*51*...*2*1 and the Denominator (bottom) is 53*52*...*3*2 so all but the 1/53 get crossed out.

There was a similar stunt in a bingo hall a few years ago where 15 boxes had a key for the car. One lucky punter opened one box each Sunday. Amazingly(?) it took all 15 weeks before the car was given away and the bingo hall was packed towards the end.

Quote: charliepatrick

There was a similar stunt in a bingo hall a few years ago where 15 boxes had a key for the car.
Amazingly(?) it took all 15 weeks before the car was given away.

Wow. That brings back memories.

In 1986, I was the DJ at an oldies club. We were giving away a 1957 Chevy. Nice looking car.

The night of the give-away, all 50 finalists showed up to pick a key from a bucket, and try it in a lock we had set up in the club.

When it got down to about 7 people left, I looked into the bucket, and could see that one key was indeed cut differently.

When it got to the last person, the manager declared that person the winner, but said to go ahead and try the key.

The problem was, somehow, there were still two keys in the bucket, and the last person also picked a losing key.

Because the manager had opened his big dumb mouth, that last person demanded the car. But the other finalists rightfully complained.

It got ugly quick.

We ended up letting that person keep the car, and gave away a second car.